cái này nó hơi khó 1 tí nên chú ý chút khác lên lever :>

a, \(A=\left(\frac{4x}{x^2+2x}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right):\frac{x+1}{x-2}\)ĐK : x khác 0 ; 2 ; -2

\(=\left(\frac{4x}{x\left(x+2\right)}+\frac{2}{x-2}-\frac{6-5x}{\left(2-x\right)\left(x+2\right)}\right):\frac{x+1}{x-2}\)

\(=\left(\frac{4x\left(x-2\right)}{MTC}+\frac{2x\left(x+2\right)}{MTC}+\frac{\left(6-5x\right)x}{MTC}\right):\frac{x+1}{x-2}\)

\(=\left(\frac{4x^2-8x+2x^2+4x+6x-5x^2}{MTC}\right):\frac{x+1}{x-2}\)

\(=\frac{x^2+2x}{x\left(x+2\right)\left(x-2\right)}.\frac{x-2}{x+1}=\frac{1}{x+1}\)

b, Ta có : \(x^2-2x=8\Leftrightarrow x^2-2x-8=0\)

\(\left(x-4\right)\left(x+2\right)=0\)<=> \(x=4;-2\)

TH1 : Thay x = 4 ta được : \(\frac{1}{4+1}=\frac{1}{5}\)

TH2 : Thay x = -2 ta được : ( ktmđkxđ ) 

\(A=\left(\frac{4x}{x^2+2x}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right)\div\frac{x+1}{x-2}\)

a)\(=\left(\frac{4x}{x\left(x+2\right)}+\frac{2}{x-2}+\frac{6-5x}{x^2-4}\right)\times\frac{x-2}{x+1}\)

\(=\left(\frac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{6-5x}{\left(x-2\right)\left(x+2\right)}\right)\times\frac{x-2}{x+1}\)

\(=\left(\frac{4x-8+2x+4+6-5x}{\left(x-2\right)\left(x+2\right)}\right)\times\frac{x-2}{x+1}\)

\(=\frac{x+2}{\left(x-2\right)\left(x+2\right)}\times\frac{x-2}{x+1}\)

\(=\frac{1}{x+1}\)

b) x² - 2x = 8

<=> x² - 2x - 8 = 0

<=> x² - 4x + 2x - 8 = 0

<=> x( x - 4 ) + 2( x - 4 ) = 0

<=> ( x - 4 )( x + 2 ) = 0

<=> x = 4 ( tm ) hoặc x = -2 ( ktm )

Với x = 4 ( tm ) => A = 1/5

Với x = -2 ( ktm ) => A không xác định

\(\frac{x^2+6x}{x-7}\div\frac{x^2-36}{x^2-14x+49}\)

\(=\frac{x\left(x+6\right)}{x-7}\times\frac{\left(x-7\right)^2}{\left(x-6\right)\left(x+6\right)}\)

\(=\frac{x\left(x-7\right)}{x-6}\)

\(\frac{x^2+6x}{x-7}\div\frac{x^2-36}{x^2-14x+49}\)

\(=\frac{x^2+6x}{x-7}.\frac{x^2-14x+49}{x^2-36}\)

\(=\frac{x\left(x+6\right)\left(x-7\right)^2}{\left(x-7\right)\left(x-6\right)\left(x+6\right)}\)

\(=\frac{x\left(x-7\right)}{x-6}\)

gọi giao của (d₂) và (d₃) là A(x,y) suy ra x, y thỏa mãn hệ \(\hept{\begin{cases}2x-y=-1\\-3x-2y=5\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\y=-1\end{cases}}\Leftrightarrow A\left(-1.-1\right)\in\left(d\right)}\)

thay vào ta được \(-5m-6m+6=4m+6\Rightarrow m=0\)

vậy m=0 thỏa mãn đề bài

\(\left(x+1\right)\left(x+3\right)\left(x+4\right)\left(x+6\right)=720\)

\(\Leftrightarrow\left(x+1\right)\left(x+6\right)\left(x+3\right)\left(x+4\right)=720\)

\(\Leftrightarrow\left(x^2+7x+6\right)\left(x^2+7x+12\right)=720\)

Đặt \(x^2+7x+6=t\left(t>0\right)\)

\(t\left(t+6\right)=720\Leftrightarrow t^2+6t=720\)

\(\Leftrightarrow t^2+6t-720=0\)giải delta ta được : 

\(\Leftrightarrow t_1=-30\left(ktm\right);t_2=24\left(tm\right)\)

thay lại đi :> 

 Ta có: a= a

a⁵=a.a.a.a.a

=> a và a⁵ có chữ số tận cùng là a

=> đpcm