\(a,ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm1\end{cases}}\)

Sao phân số thứ 2 là \(\frac{1-2}{1+x}\) .Bạn chép đề thật chuẩn mới trả lời đúng nhé

a) ĐKXĐ: \(x\ne-2;x\ne2\), rút gọn:

\(A=\left[\frac{3\left(x-2\right)-2x\left(x+2\right)+2\left(2x^2+3\right)}{2\left(x-2\right)\left(x+2\right)}\right]\div\frac{2x-1}{4\left(x-2\right)}\)

\(A=\frac{3x-6-2x^2-4x+4x^2+6}{2\left(x-2\right)\left(x+2\right)}\cdot\frac{4\left(x-2\right)}{2x-1}=\frac{4\left(2x^2-x\right)}{x\left(x+2\right)\left(2x-1\right)}=\frac{4x\left(2x-1\right)}{x\left(x+2\right)\left(2x-1\right)}=\frac{4}{x+2}\)

b) Ta có: \(\left|x-1\right|=3\Leftrightarrow\hept{\begin{cases}x-1=3\\x-1=-3\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\left(n\right)\\x=-2\left(l\right)\end{cases}}}\)

=> Khi \(x=4\)thì \(A=\frac{4}{4+2}=\frac{4}{6}=\frac{2}{3}\)

c) \(A< 2\Leftrightarrow\frac{4}{x+2}< 2\Leftrightarrow4< 2x+4\Leftrightarrow0< 2x\Leftrightarrow x>0\)Vậy \(A< 2,\forall x>0\)

d) \(\left|A\right|=1\Leftrightarrow\left|\frac{4}{x+2}\right|=1\Leftrightarrow\hept{\begin{cases}\frac{4}{x+2}=1\\\frac{4}{x+2}=-1\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\left(l\right)\\x=-6\left(n\right)\end{cases}}}\)Vậy \(\left|A\right|=1\)khi và chỉ khi x = -6

bài 2 : ĐKXĐ : \(x\ge0\) và \(x\ne1\) 

Rút gọn :\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{5\sqrt{x}-1}{x-1}\)

               \(B=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{5\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

                \(B=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-5\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

               \(B=\frac{-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

                \(B=\frac{-1}{\sqrt{x}+1}\)

giúp vs

a)  ĐKXĐ:  \(x\ne1\)

b)  \(A=\frac{2}{x-1}+\frac{2\left(x+1\right)}{x^2+x+1}+\frac{x^2-10x+3}{x^3-1}\)

\(=\frac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x^2-10x+3}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\frac{2x^2+2x+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x^2-10x+3}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\frac{5x^2-8x+3}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{\left(x-1\right)\left(5x-3\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{5x-3}{x^2+x+1}\)

\(A=\frac{2x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+4}{3-x}\)

a) ĐKXĐ : \(\hept{\begin{cases}x\ne2\\x\ne3\end{cases}}\)

\(A=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{x+3}{x-2}+\frac{2x+4}{x-3}\)

\(=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\frac{\left(2x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{x^2-9}{\left(x-2\right)\left(x-3\right)}+\frac{2x^2-8}{\left(x-2\right)\left(x-3\right)}\)

\(=\frac{2x-9-x^2+9+2x^2-8}{\left(x-2\right)\left(x-3\right)}\)

\(=\frac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}=\frac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}=\frac{x+4}{x-3}\)

b) Ta có : \(A=\frac{x+4}{x-3}=\frac{x-3+7}{x-3}=1+\frac{7}{x-3}\)

Để A đạt giá trị nguyên thì \(\frac{7}{x-3}\)đạt giá trị nguyên

=> 7 ⋮ x - 3

=> x - 3 ∈ Ư(7) = { ±1 ; ±7 }

So với ĐKXĐ ta thấy x = 4 , x = 10 , x = -4 thỏa mãn 

Vậy với x ∈ { ±4 ; 10 } thì A đạt giá trị nguyên

(....) dùng để nhìn được chữ số ở phân số cuối cùng thôi, ko dùng để làm gì.

( ác ) là từ ( các ) 

(gia strij) là từ ( giá trị )