Lòng khoan dung

\(\left(x-3\right)\left(y+5\right)=9\)

Do \(y\in N\Rightarrow\) \(\left(y+5\right)\)\(\ge5\)

\(x\in N\Rightarrow x-3\ge-3\)

\(\Rightarrow\left\{x-3;y+5\right\}=\left\{1;9\right\}\\ \Rightarrow\left\{x;y\right\}=\left\{4;4\right\}\) 

\(4\left(x+5\right)^3-7=101\\ \Rightarrow4\left(x+5\right)^3=108\\ \Rightarrow\left(x+5\right)^3=27\Rightarrow x+5=3\\ \Rightarrow x=-2\)

\(a.\left(b-1\right)=3\), \(Ta\) \(lập\) \(bảng\):

\(Vậy\) \(\left\{a;b\right\}\in\left\{1;4\right\};\left\{-1;-2\right\};\left\{3;2\right\};\left\{-3;0\right\}\)

\(Ư\left(91\right)=\left\{1,7,13,91\right\}\)

\(Đáp\) \(án\):\(B\)

\(Ta\) \(có\) \(S_{ABCD}=6.6=36\left(cm^2\right)\)

\(S_{EFGH}=\dfrac{1}{2}S_{ABCD}=\dfrac{1}{2}.36=18\left(cm^2\right)\)

-20

Bài 34:

\(a,9.\left(x+28\right)=0\Rightarrow x+28=0\Rightarrow x=-28\\ b,\left(27-x\right)\left(x+9\right)=0\Rightarrow\left\{{}\begin{matrix}27-x=0\\x+9=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=27\\x=-9\end{matrix}\right.\\ c,\left(-x\right)\left(x-43\right)=0\Rightarrow\left\{{}\begin{matrix}-x=0\\x-43=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=43\end{matrix}\right.\)

D

\(Xét\)  \(\Delta ABD\) \(và\) \(\Delta ACD\) \(có:\)

\(AD\) \(chung\)

\(AB=AC\)

\(\Rightarrow\Delta ABD=\Delta ACD\left(c.h;c.g.v\right)\)

\(\left(đpcm\right)\)