Bài tập theo tuần Toán 8 - Tuần 4

$\begin{array}{l}{\left(\mathrm{x}-\mathrm{y}\right)}^3={\mathrm{x}}^3-3{\mathrm{x}}^2\mathrm{y}+3{\mathrm{xy}}^2-{\mathrm{y}}^3\end{array}$

$8{\mathrm{x}}^3-{\mathrm{y}}^3=\left(2\mathrm{x}-\mathrm{y}\right)\left(4{\mathrm{x}}^2+2\mathrm{xy}+{\mathrm{y}}^2\right)$

$64+{\mathrm{x}}^3=4^3+{\mathrm{x}}^3=\left(\mathrm{x}+4\right)\left({\mathrm{x}}^2-4\mathrm{x}+16\right)$

$\left(\mathrm{y}+3\right)\left({\mathrm{y}}^2-3\mathrm{y}+9\right)={\mathrm{y}}^3+27$

Tìm x:

$\begin{array}{l}{\left(2\mathrm{x}-7\right)}^3-27=0\\ {\left(2\mathrm{x}-7\right)}^3-3^3=0\iff \left(2\mathrm{x}-10\right)\left(4{\mathrm{x}}^2-28\mathrm{x}+49+6\mathrm{x}+21+9\right)=0\\ \left(2\mathrm{x}-10\right)\left(4{\mathrm{x}}^2-22\mathrm{x}+79\right)=0\\ \mathrm{Do}4{\mathrm{x}}^2-22\mathrm{x}+79={\left(2\mathrm{x}-\frac{11}{2}\right)}^2+\frac{195}{4}>0\\ \Rightarrow 2\mathrm{x}-10=0\Rightarrow \mathrm{x}=5\end{array}$

Ta có: ${\mathrm{x}}^3+{\mathrm{y}}^3+{\mathrm{z}}^3$

$\begin{array}{l}={\left(\mathrm{x}+\mathrm{y}\right)}^3-3\mathrm{xy}\left(\mathrm{x}+\mathrm{y}\right)+{\mathrm{z}}^3\\ ={\left(\mathrm{x}+\mathrm{y}\right)}^3+{\mathrm{z}}^3-3\mathrm{xy}.\left(-\mathrm{z}\right)(\mathrm{do}\mathrm{x}+\mathrm{y}+\mathrm{z}=0\Rightarrow \mathrm{x}+\mathrm{y}=-\mathrm{z})\\ =\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left[{\left(\mathrm{x}+\mathrm{y}\right)}^2-\left(\mathrm{x}+\mathrm{y}\right)\mathrm{z}+{\mathrm{z}}^2\right]+3\mathrm{xyz}\\ =3\mathrm{xyz}\left(\mathrm{do}\mathrm{x}+\mathrm{y}+\mathrm{z}=0\right)\end{array}$

$\begin{array}{l}\left(\mathrm{a}+\mathrm{b}\right)\left({\mathrm{a}}^2-\mathrm{ab}+{\mathrm{b}}^2\right)+\left(\mathrm{a}-\mathrm{b}\right)\left({\mathrm{a}}^2+\mathrm{ab}+{\mathrm{b}}^2\right)\\ ={\mathrm{a}}^3+{\mathrm{b}}^3+{\mathrm{a}}^3-{\mathrm{b}}^3=2{\mathrm{a}}^3=\mathrm{VP}\left(\mathrm{dfcm}\right)\end{array}$

$\begin{array}{l}\left(\mathrm{a}+\mathrm{b}\right)\left[{\left(\mathrm{a}-\mathrm{b}\right)}^2+\mathrm{ab}\right]=\left(\mathrm{a}+\mathrm{b}\right)\left({\mathrm{a}}^2-2\mathrm{ab}+{\mathrm{b}}^2+\mathrm{ab}\right)\\ =\left(\mathrm{a}+\mathrm{b}\right)\left({\mathrm{a}}^2-\mathrm{ab}+{\mathrm{b}}^2\right)={\mathrm{a}}^3+{\mathrm{b}}^3=\mathrm{VP}\left(\mathrm{dfcm}\right)\end{array}$

$\begin{array}{l}{\left(\mathrm{a}+\mathrm{b}\right)}^3-3\mathrm{ab}\left(\mathrm{a}+\mathrm{b}\right)=\left(\mathrm{a}+\mathrm{b}\right)\left[{\left(\mathrm{a}+\mathrm{b}\right)}^2-3\mathrm{ab}\right]\\ =\left(\mathrm{a}+\mathrm{b}\right)\left({\mathrm{a}}^2+2\mathrm{ab}+{\mathrm{b}}^2-3\mathrm{ab}\right)=\left(\mathrm{a}+\mathrm{b}\right)\left({\mathrm{a}}^2-\mathrm{ab}+{\mathrm{b}}^2\right)={\mathrm{a}}^3+{\mathrm{b}}^3=\mathrm{VP}\left(\mathrm{dfcm}\right)\end{array}$

Ta có: $a^3+b^3+c=3abc$

$\begin{array}{l}{\left(a+b\right)}^3-3ab\left(a+b\right)+c^3-3abc=0\\ {\left(a+b\right)}^3+c^3-3ab\left(a+b+c\right)=0\\ \left(a+b+c\right)\left[{\left(a+b\right)}^2+\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\\ \left(a+b+c\right)\left[{\left(a+b\right)}^2+\left(a+b\right)c+c^2-3abc\right]=0\\ Do{\left(a+b\right)}^2+\left(a+b\right)c+c^2-3abc>0\\ \Rightarrow a+b+c=0\Rightarrow \left\{\begin{array}{l}a+b=-c\\ a+c=-b\\ b+c=-a\end{array}\right.\end{array}$

Ta có: $B=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}.\frac{b+c}{a}.\frac{a+c}{b}=\frac{\left(-c\right)\left(-a\right)\left(-b\right)}{abc}=-1$

Vậy B = -1

Ta có:
$\begin{array}{l}{a}^2-2a+b^2+4b+4c^2-4c+6=0\\ \left(a^2-2a+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)=0\\ {\left(a-1\right)}^2+{\left(b+2\right)}^2+{\left(2c-1\right)}^2=0\\ Do{\left(a-1\right)}^2\ge 0,{\left(b+2\right)}^2\ge 0;{\left(2c-1\right)}^2\ge 0\\ \Rightarrow \left\{\begin{array}{l}a-1=0\\ b+2=0\\ 2c-1=0\end{array}\right.\iff \left\{\begin{array}{l}a=1\\ b=-2\\ c=\frac{1}{2}\end{array}\right.\end{array}$

Vậy $a=1;b=-2;c=\frac{1}{2}$

a) Xét $\Delta AHB$ có $ME\perp AC,BH\perp AC\Rightarrow ME//BH$

b) $\Delta ABH$ có M là trung điểm $AB,ME//BH\left(cmt\right)\Rightarrow E$ là trung đểim AH

=> ME là đường trung bình $\Delta ABH\Rightarrow ME=\frac{1}{2}BH$

Chứng minh tương tự => NP là đường trung bình $\Delta BHC\Rightarrow NP=\frac{1}{2}BH$

$\Rightarrow ME=NP\left(=\frac{1}{2}BH\right)$ và (cùng // BH)