$33.(-5).8<0$ (Vì chứa lẻ số dấu "-" )

-7.(-24).9 > 0 (Vì chứa chẵn số"-")

-6.(-35).(-27) < 0 ( Vì chưa lẻ số dấu "-")

5.35 > 0 ( Vì tích hai số dương)

Do đó -6.(-35).(-27) < 5.35

a) b là một số nguyên âm (Vì  a âm mà a.b dương)

b) b là một số nguyên dương (Vì a âm mà a.b âm)

a)  b là một số 0 (Vì a âm mà a.b =0)

Thay x = 7 vào biểu thức ta được:

(7 - 5).(7 + 12) = 2.19 = 38

Thay x = 8 vào biểu thức ta được:

$\left(-55\right).\left(-25\right)\left(-8\right)=-11000.$

Thay x = 10 vào biểu thức ta được:

$\left(-1\right).\left(-2\right).\left(-3\right).\left(-4\right)\left(-5\right).10=-120.10=-1200.$

$\mathrm{x}.\left(\mathrm{y}-3\right)=-12$

Ta có bảng sau:

$$\begin{gathered} \left(\mathrm{x}-3\right)\left(\mathrm{y}-3\right)=9\iff \left[\begin{array}{l}\left\{\begin{array}{l}\left(\mathrm{x}-3\right)=\pm 3\\ \left(\mathrm{y}-3\right)=\pm 3\end{array}\right.\\ \left\{\begin{array}{l}\left(\mathrm{x}-3\right)=\pm 1\\ \left(\mathrm{y}-3\right)=\pm 9\end{array}\right.\\ \left\{\begin{array}{l}\left(\mathrm{x}-3\right)=\pm 9\\ \left(\mathrm{y}-3\right)=\pm 1\end{array}\right.\end{array}\right. \\ \mathrm{TH}1: \left\{\begin{array}{l}\left(\mathrm{x}-3\right)=3\\ \left(\mathrm{y}-3\right)=3\end{array}\right.\iff \left\{\begin{array}{l}\mathrm{x}=3+3\\ \mathrm{y}=3+3\end{array}\right.\iff \left\{\begin{array}{l}\mathrm{x}=6\\ \mathrm{y}=6\end{array}\right. \\ \mathrm{TH}2: \left\{\begin{array}{l}\left(\mathrm{x}-3\right)=-3\\ \left(\mathrm{y}-3\right)=-3\end{array}\right.\iff \left\{\begin{array}{l}\mathrm{x}=-3+3\\ \mathrm{y}=-3+3\end{array}\right.\iff \left\{\begin{array}{l}\mathrm{x}=0\\ \mathrm{y}=0\end{array}\right. \\ \mathrm{TH}3: \left\{\begin{array}{l}\left(\mathrm{x}-3\right)=9\\ \left(\mathrm{y}-3\right)=1\end{array}\right.\iff \left\{\begin{array}{l}\mathrm{x}=9+3\\ \mathrm{y}=1+3\end{array}\right.\iff \left\{\begin{array}{l}\mathrm{x}=12\\ \mathrm{y}=4\end{array}\right. \\ \mathrm{TH}4: \left\{\begin{array}{l}\left(\mathrm{x}-3\right)=-9\\ \left(\mathrm{y}-3\right)=-1\end{array}\right.\iff \left\{\begin{array}{l}\mathrm{x}=-9+3\\ \mathrm{y}=-1+3\end{array}\right.\iff \left\{\begin{array}{l}\mathrm{x}=-6\\ \mathrm{y}=2\end{array}\right. \\ \mathrm{TH}5: \left\{\begin{array}{l}\left(\mathrm{x}-3\right)=1\\ \left(\mathrm{y}-3\right)=9\end{array}\right.\iff \left\{\begin{array}{l}\mathrm{x}=1+3\\ \mathrm{y}=9+3\end{array}\right.\iff \left\{\begin{array}{l}\mathrm{x}=4\\ \mathrm{y}=12\end{array}\right. \\ \mathrm{TH}6: \left\{\begin{array}{l}\left(\mathrm{x}-3\right)=-1\\ \left(\mathrm{y}-3\right)=-9\end{array}\right.\iff \left\{\begin{array}{l}\mathrm{x}=-1+3\\ \mathrm{y}=-9+3\end{array}\right.\iff \left\{\begin{array}{l}\mathrm{x}=2\\ \mathrm{y}=-6\end{array}\right. \\ \left(\mathrm{x};\mathrm{y}\right)\in \left\{\left(2;-6\right);\left(6;6\right);\left(0;0\right);\left(12;4\right);\left(-6;2\right);\left(4;12\right)\right\} \end{gathered}$$

Ta thấy: 

$\left(\overline{\mathrm{x}\text{x}}+\overline{\mathrm{yy}}\right).\mathrm{xy}=(11\text{x}+11\mathrm{y})\mathrm{xy}=11(\mathrm{x}+\mathrm{y})\mathrm{xy}=1980\Rightarrow .(\mathrm{x}+\mathrm{y})\mathrm{xy}=180$

Vì $0\le \mathrm{x}\le 9;0\le \mathrm{y}\le 9$ nên suy ra x = 4; y = 5 hoặc x= 5; y = 4

$\left(-8\right).{\left(-3\right)}^3.\left(+125\right)={\left(-2\right)}^3.{\left(-3\right)}^3{.5}^3={\left[-2.\left(-3\right).5\right]}^3={30}^3$

$27.{\left(-2\right)}^3.\left(-7\right).\left(+49\right)=3^3.{\left(-2\right)}^3\left(-7\right){\left(-7\right)}^2=3^3.{\left(-2\right)}^3.{\left(-7\right)}^3={\left[3.\left(-2\right).\left(-7\right)\right]}^3={42}^3.$

$$\begin{gathered} \mathrm{VT}=\mathrm{a}(\mathrm{b}+\mathrm{c})-\mathrm{b}(\mathrm{a}-\mathrm{c})=\mathrm{ab}+\mathrm{ac}-\mathrm{ba}+\mathrm{bc}=(\mathrm{ab}-\mathrm{ab})+(\mathrm{ac}+\mathrm{bc})=0+\left(\mathrm{a}+\mathrm{b}\right).\mathrm{c}=\mathrm{VP} \\ \mathrm{Vậy} \mathrm{a}(\mathrm{b}+\mathrm{c})-\mathrm{b}(\mathrm{a}-\mathrm{c})=(\mathrm{a}+\mathrm{b}).\mathrm{c} \end{gathered}$$

$$\begin{gathered} \mathrm{VT}=(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})={\mathrm{a}}^2-\mathrm{a}.\mathrm{b}+\mathrm{b}.\mathrm{a}-{\mathrm{b}}^2={\mathrm{a}}^2-{\mathrm{b}}^2+\left(\mathrm{ab}-\mathrm{ab}\right) \\ ={\mathrm{a}}^2-{\mathrm{b}}^2+0={\mathrm{a}}^2-{\mathrm{b}}^2=\mathrm{VP}. \\ \mathrm{Vậy} (\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})={\mathrm{a}}^2-{\mathrm{b}}^2 \end{gathered}$$

$(-4)\mathrm{.13.}(-250)=(-4).(-250).13=1000.13=13000$

$(-37).84+37.(-16)=(-37).84+(-37).16=(-37).(84+16)=(-37).100=-3700$

$25.(-5)\mathrm{.4.}(-20)=\mathrm{25.4.}(-5).(-20)=100.100=10000$

$(-48).72+36.(-304)=(-48)\mathrm{.2.36}+36.(-304)=(-96).3+36.(-304)$

$(-125).(-5)\mathrm{.8.}(-2)=(-125)\mathrm{.8.}(-5).(-2)=-1000.10=-10000$

$$\begin{gathered} (-127).(1-582)-582.127=(-127).(-581)-582.127=127.581-582.127 \\ =127(581-582)=127.(-1)=-127 \end{gathered}$$

$$\begin{gathered} (43-13).(-3)+27.(-14-16)=30.(-3)+27.(-30)=(-30).3+27.(-30) \\ =(-30).(3+27)=(-30).30=-90 \end{gathered}$$

$125.(-61).{(-2)}^3.{(-1)}^{2\mathrm{n}}=125.(-62).(-8).1=125.(-8).(-62)=-1000.(-62)=62000$

$(-25).11=(-25).(10+1)=-25.10+(-25).1=-250+(-25)=-275$

$31.(-99)=31.(-100+1)=31.(-100)+31.1=-3100+31=-3069$

$(-8).130=(-8).(125+5)=(-8).125+(-8).5=-1000+(-40)=-1040$

$25.(-14)=25.\left[(-10)+(-4)\right]=25.(-10)+25.(-4)=-250+(-100)=-350$

(-225).(-336) > 0

(-112).2.(-26).(-321) < 0

Vì $\left\{\begin{array}{l}{(-2)}^3.125.32.(-76)>0\\ \mathrm{12.74.}{(-3)}^4.(-395)<0\end{array}\right.\Rightarrow {(-2)}^3\mathrm{.125.32.}(-76)>\mathrm{12.74.}{(-3)}^4.(-395)$

Vì $\left\{\begin{array}{l}(-1).(-2).(-3)\mathrm{...}(-20)>0\\ (-3).(-4).(-5)\mathrm{...}(-23)<0\end{array}\right.\Rightarrow (-1).(-2).(-3)\mathrm{...}(-20)>(-3).(-4).(-5)\mathrm{...}(-23)$

$\mathrm{5.5.5.5.5}=5^5$

$(-3).(-3).(-3)\mathrm{.3.3}=(-3).(-3).(-3).(-3).(-3)={(-3)}^5$

$\mathrm{2.2.2.}(-5).(-5).(-5)=\left[2.(-5)\right].\left[2.(-5)\right].\left[2.(-5)\right]=(-10).(-10).(-10)={(-10)}^3$

$(-8).{(-5)}^3.64={(-2)}^3.{(-5)}^3{.4}^3={\left[(-2).(-5).4\right]}^3={40}^3$

$\mathrm{x}(\mathrm{x}+6)=0\iff \left[\begin{array}{l}\mathrm{x}=0\\ \mathrm{x}+6=0\end{array}\right.\iff \left[\begin{array}{l}\mathrm{x}=0\\ \mathrm{x}=-6\end{array}\right.$

Vậy x = 0 hoặc x = -6

$(\mathrm{x}-3).(\mathrm{y}+7)=0\iff \left[\begin{array}{l}\mathrm{x}-3=0\\ \mathrm{y}+7=0\end{array}\right.\iff \left[\begin{array}{l}\mathrm{x}=3\\ \mathrm{y}=-7\end{array}\right.$

Vậy x = 3; y = -7

$(\mathrm{x}-2)({\mathrm{x}}^2+2)=0\iff \left[\begin{array}{l}\mathrm{x}-2=0\\ {\mathrm{x}}^2+2=0\end{array}\right.\iff \left[\begin{array}{l}\mathrm{x}=2\\ {\mathrm{x}}^2=-2\left(\mathrm{L}\right)\end{array}\right.$

Vậy x = 2

$$\begin{gathered} \begin{array}{l}xy+3x-7y=21\\ \iff xy+3x-7y-21=0\\ \iff x(y+3)-7(y+3)=0\end{array} \\ \begin{array}{l}\iff (x-7).(y+3)=0\\ \Rightarrow \left[\begin{array}{l}x-7=0\\ y+3=0\end{array}\right.\iff \left[\begin{array}{l}x=7\\ y=-3\end{array}\right.\end{array} \end{gathered}$$

Thay x = 4 vào biểu thức ta được 

$(-75).(-25).4=(-75).(-100)=7500$

Thay x = 2, y = -5 vào biểu thức ta đươc:

$\left[2+(-5)\right]\left[2-(-5)\right]=\left(-3\right).7=-21$

Thay a = -2, b = 4 vào biểu thức ta được:

${(-2)}^2+2.(-2).4+4^2-1=4+(-16)+16-1=3$

Thay x =4 vào biểu thức ta được:

$4.(-234)+(-4).16=(-4).234+(-4).16=(-4).(234+16)=(-4).250=-1000$

Ta có: 

${\mathrm{A}}^2=\mathrm{b}(\mathrm{a}-\mathrm{c})-\mathrm{c}(\mathrm{a}-\mathrm{b})=\mathrm{ab}-\mathrm{bc}-\mathrm{ca}+\mathrm{bc}=\mathrm{ab}-\mathrm{ca}=\mathrm{a}(\mathrm{b}-\mathrm{c})$

Thay a = -20, b - c = -5 vào biểu thức ta được:

$\begin{array}{l}{\mathrm{A}}^2=\mathrm{a}(\mathrm{b}-\mathrm{c})=(-20).(-5)=100={10}^2\\ \Rightarrow \mathrm{A}=10\end{array}$

$\begin{array}{l}\mathrm{VT}={\left(\mathrm{a}+\mathrm{b}\right)}^2=\left(\mathrm{a}+\mathrm{b}\right).\left(\mathrm{a}+\mathrm{b}\right)\\ =\mathrm{a}(\mathrm{a}+\mathrm{b})+\mathrm{b}(\mathrm{a}+\mathrm{b})={\mathrm{a}}^2+\mathrm{ab}+\mathrm{ba}+{\mathrm{b}}^2\\ ={\mathrm{a}}^2+2\mathrm{ab}+{\mathrm{b}}^2=\mathrm{VP}\left(\mathrm{dpcm}\right)\end{array}$

$\begin{array}{l}\mathrm{VT}=\left(\mathrm{a}-\mathrm{b}\right).\left(\mathrm{a}+\mathrm{b}\right)\\ =\mathrm{a}(\mathrm{a}+\mathrm{b})-\mathrm{b}(\mathrm{a}+\mathrm{b})={\mathrm{a}}^2+\mathrm{ab}-\mathrm{ba}-{\mathrm{b}}^2\\ ={\mathrm{a}}^2-{\mathrm{b}}^2=\mathrm{VP}\left(\mathrm{dpcm}\right)\end{array}$

$\begin{array}{l}\mathrm{VT}=\mathrm{a}(\mathrm{b}+\mathrm{c})-\mathrm{b}(\mathrm{a}-\mathrm{c})=\mathrm{ab}+\mathrm{ac}-\left(\mathrm{ba}-\mathrm{bc}\right)\\ =\mathrm{ab}+\mathrm{ac}-\mathrm{ba}+\mathrm{bc}=\mathrm{ac}+\mathrm{bc}=\mathrm{c}(\mathrm{a}+\mathrm{b})=\mathrm{VP}\left(\mathrm{dpcm}\right)\end{array}$

Tích của ba số bất kì là một số âm nên trong ba số đó có ít nhất 1 số âm.

Ta tách riêng số âm đó ra, còn lại 15 số.

Ta chia 15 số này thành 5 nhóm, mỗi nhóm 3 số.

Tích 3 số trong mỗi nhóm đó đều là một số âm.

Vậy tích của 5 nhóm với một số âm để tách riêng ra là tích của 6 số âm, do đó tích của chúng là  một số dương.