\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)

a, \(Đkxđ:x\ne-3;x\ne2\)

b,\(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)

\(=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{1}{x-2}\)

\(=\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{x+3}{\left(x+3\right)\left(x-2\right)}\)

\(=\dfrac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)

\(=\dfrac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}\)\(=\dfrac{x-4}{x-2}\)

c,\(A=-\dfrac{3}{4}\) khi \(\dfrac{x-4}{x-2}=-\dfrac{3}{4}\)

\(\Leftrightarrow\left(x-4\right).4=-3\left(x-2\right)\)

\(\Leftrightarrow4x-16=-3x+6\)

\(\Leftrightarrow7x=22\)

\(\Leftrightarrow x=\dfrac{22}{7}\)

Vậy khi \(x=\dfrac{22}{7}\) thì \(A=-\dfrac{3}{4}\)

a) ĐKXĐ : \(\left\{{}\begin{matrix}x+3\ne0\\2-x\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne-3\\x\ne2\end{matrix}\right.\)

b) \(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}-\dfrac{1}{x-2}\)

\(A=\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{x+3}{\left(x+3\right)\left(x-2\right)}\)

\(A=\dfrac{-x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\dfrac{x-4}{x-2}\)

c) Để \(A=\dfrac{-3}{4}\) thì :

\(A=\dfrac{x-4}{x-2}=\dfrac{-3}{4}\)

\(\Rightarrow\dfrac{x-4}{x-2}+\dfrac{3}{4}=0\)

\(\Rightarrow\dfrac{4\left(x-4\right)}{4\left(x-2\right)}+\dfrac{3\left(x-2\right)}{4\left(x-2\right)}=0\)

\(\Rightarrow4x-16+3x-6=0\)

\(\Rightarrow7x+22=0\)

\(\Rightarrow x=\dfrac{-22}{7}\)

d) Ta có : \(A=\dfrac{x-4}{x-2}=\dfrac{x-2-2}{x-2}=1-\dfrac{2}{x-2}\)

Vì \(1\in Z\) để \(A\in Z\) thì \(\dfrac{2}{x-2}\in Z\)

\(\Rightarrow x-2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Có : \(\left\{{}\begin{matrix}x-2=1=>x=3\\x-2=-1=>x=1\\x-2=2=>x=4\\x-2=-2=>0\end{matrix}\right.\)

Vậy để A nhận gt nguyên thì x \(\in\left\{3;1;4;0\right\}\)

e) \(x^2-9=0\)

\(\Rightarrow\left(x+3\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(loại\right)\\x=3\end{matrix}\right.\)

Thay vào A ta có :

\(A=\dfrac{x-4}{x-2}=\dfrac{3-4}{3-2}=-1\)

d, \(A\) nguyên thì \(\dfrac{x-4}{x-2}\) nguyên.

Ta có: \(\dfrac{x-4}{x-2}=\dfrac{x-2-2}{x-2}=1-\dfrac{2}{x-2}\)

Mà \(1\) nguyên, để \(A\) nguyên thì \(\dfrac{2}{x-2}\in Z\)

\(\Rightarrow x-2\inƯ\left(2\right)=\left\{2;-2;1;-1\right\}\)

Ta có bảng sau:

Vậy A nguyên khi \(x\in\left\{4;0;3;1\right\}\)

e, \(x^2-9=0\)

\(\Leftrightarrow x^2-3^2=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-3\left(ktm\right)\end{matrix}\right.\)

Khi \(x=3\) thì \(A=\dfrac{3-4}{3-2}=-1\)

bn chưa ngủ ak 

h giái giúp mk đy

A=\(x^5-100x^4+100x^3-100x^2+100x-9\)

  =\(x^5-\left(99+1\right)x^4+\left(99+1\right)x^3-\left(99+1\right)x^2+\left(99+1\right)x-9\)

Thay 99=x vào ta được:

A=\(x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-9\)

  =\(x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-9\)

  =x-9

thay x=99 vào ta được:

A=99-9=90

Vậy giá trị của biểu thức A tại x=99 là 90

C=-2x^2+2x-2= -(2x^2-2x+2)= -(x-1)² =>C luôn âm

A= -x² +24x-4= -(x² -4x+4)= -(x-1)²  =>ko có gía trị x nào để biểu thức nhận giá trị dương

Chắc vậy :((

Mk nghĩ cái này giống 7 hàng đẳng thức nhưng mk ms học lp 7 nên ko bít làm có đúng ko nữa,nếu sai cho mk xl bn nha :)

M² = b(a - c) - c(a - b)

= ab - bc - ac + bc

= ab - ac

= a(b - c)

Thay a = -20; b - c = -5 vào biểu thức M² ta có :

-20 x (-5)

= 20 x 5

= 100

\(\Rightarrow M^2=100\)

\(\Rightarrow M=\pm10\)