Bài tập theo tuần Toán 8 - Tuần 2

$\begin{array}{l}{\left(\mathrm{x}-2\right)}^2={\mathrm{x}}^2-2.\mathrm{x}.2+2^2={\mathrm{x}}^2-4\mathrm{x}+4\end{array}$

${\left(4-\mathrm{y}\right)}^2=4^2-\mathrm{2.4.}\mathrm{y}+{\mathrm{y}}^2=16-8\mathrm{y}+{\mathrm{y}}^2$

${\left(\mathrm{z}-\frac{1}{3}\right)}^2={\mathrm{z}}^2-2.\mathrm{z}.\frac{1}{3}+{\left(\frac{1}{3}\right)}^2={\mathrm{z}}^2-\frac{2}{3}\mathrm{z}+\frac{1}{9}$

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

${\left(4+\frac{1}{\mathrm{x}}\right)}^2=4^2+\mathrm{2.4.}\frac{1}{\mathrm{x}}+{\left(\frac{1}{\mathrm{x}}\right)}^2=16+\frac{8}{\mathrm{x}}+\frac{1}{{\mathrm{x}}^2}$

${\mathrm{x}}^2+2\mathrm{x}+1={\left(\mathrm{x}+1\right)}^2$

${\mathrm{x}}^2-8\mathrm{x}+16={\mathrm{x}}^2-2.\mathrm{x}.4+4^2={\left(\mathrm{x}-4\right)}^2$

${\mathrm{x}}^2+\mathrm{x}+\frac{1}{4}={\mathrm{x}}^2+2.\mathrm{x}.\frac{1}{2}+{\left(\frac{1}{2}\right)}^2={\left(\mathrm{x}+\frac{1}{2}\right)}^2$

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

$\begin{array}{l}49{\mathrm{x}}^2-1=0\\ \left(7\mathrm{x}-1\right)\left(7\mathrm{x}+1\right)=0\\ \Rightarrow \left[\begin{array}{l}7\mathrm{x}-1=0\\ 7\mathrm{x}+1=0\end{array}\right.\Rightarrow \left[\begin{array}{l}\mathrm{x}=\frac{1}{7}\\ \mathrm{x}=-\frac{1}{7}\end{array}\right.\\ \end{array}$

$\begin{array}{l}{\left(2\mathrm{x}-1\right)}^2-\left(4\mathrm{x}+1\right)\left(\mathrm{x}-3\right)=-3\\ 4{\mathrm{x}}^2-4\mathrm{x}+1-4{\mathrm{x}}^2+11\mathrm{x}+3=-3\\ -4\mathrm{x}+11\mathrm{x}=-3-3-1\\ 7\mathrm{x}=-7\\ \mathrm{x}=-1\end{array}$

$\begin{array}{l}9\mathrm{x}\left(\mathrm{x}-1\right)-{\left(3\mathrm{x}-1\right)}^2=11\\ 9{\mathrm{x}}^2-9\mathrm{x}-9{\mathrm{x}}^2+6\mathrm{x}-1=11\\ -3\mathrm{x}=12\Rightarrow \mathrm{x}=-4\end{array}$

$\begin{array}{l}{\left(2\mathrm{x}+3\right)}^2-{\left(\mathrm{x}-1\right)}^2=0\\ \left[\left(2\mathrm{x}+3\right)-\left(\mathrm{x}-1\right)\right].\left[\left(2\mathrm{x}+3\right)+\left(\mathrm{x}-1\right)\right]=0\\ \left(\mathrm{x}+4\right)\left(3\mathrm{x}+2\right)=0\Rightarrow \left[\begin{array}{l}\mathrm{x}+4=0\\ 3\mathrm{x}+2=0\end{array}\right.\Rightarrow \left[\begin{array}{l}\mathrm{x}=-4\\ \mathrm{x}=-\frac{2}{3}\end{array}\right.\end{array}$

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

$\begin{array}{l}2\left(\mathrm{x}-\mathrm{y}\right)\left(\mathrm{x}+\mathrm{y}\right)+{\left(\mathrm{x}+\mathrm{y}\right)}^2+{\left(\mathrm{x}-\mathrm{y}\right)}^2\\ ={\left(\mathrm{x}+\mathrm{y}+\mathrm{x}-\mathrm{y}\right)}^2={\left(2\mathrm{x}\right)}^2=4{\mathrm{x}}^2\end{array}$

$\begin{array}{l}\left(\mathrm{x}-3\right)\left(\mathrm{x}+3\right)-{\left(\mathrm{x}-5\right)}^2\\ ={\mathrm{x}}^2-9-{\mathrm{x}}^2+10\mathrm{x}-25=10\mathrm{x}-34\end{array}$

$\begin{array}{l}\mathrm{A}={100}^2-{99}^2+{98}^2-{97}^2+\mathrm{.....}+2^2-1^2\\ =\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+\mathrm{.....}+\left(2-1\right)\left(2+1\right)\\ =199+195+\mathrm{....}+3=5050\end{array}$

$\begin{array}{l}\mathrm{B}=3.\left(2^2+1\right)\left(2^4+1\right)\mathrm{.....}\left(2^{64}+1\right)+1\\ =\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\mathrm{.....}\left(2^{64}+1\right)+1\\ =\left(2^4-1\right)\left(2^4+1\right)\mathrm{.....}\left(2^{64}+1\right)+1\\ =\mathrm{.....}\\ =2^{128}-1+1=2^{128}\end{array}$

a) Ta có: $AB=AC,BM=NC\left(gt\right)\Rightarrow AB-BM=AC-NC\Rightarrow AM=AN\Rightarrow \Delta AMN$

cân tại A $\Rightarrow \widehat{M}=\frac{{180}^0-\widehat{A}}{2}\left(1\right)$mà $\Delta ABC$cân tại $A\Rightarrow B=\frac{{180}^0-\widehat{A}}{2}\left(2\right)$