a: |x-1|=3

=>x-1=3 hoặc x-1=-3

=>x=-2(nhận) hoặc x=4(loại)

Khi x=-2 thì \(A=\dfrac{4+4}{-2-4}=\dfrac{8}{-6}=\dfrac{-4}{3}\)

b: ĐKXĐ: x<>4; x<>-4

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

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

=-4x/x-4

c: A+B

=-4x/x-4+x^2+4/x-4

=(x-2)^2/(x-4)
A+B>0

=>x-4>0

=>x>4

MN ƠI GIÚP EM VS 15PHÚT NX EM PK NỘP R =(((

Quy đồng vế phải:

\(VP=\dfrac{a\left(x+2\right)+b\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(a+b\right)x+2a-2b}{x^2-4}\)

Đồng nhất tử số vế phải và vế trái ta được:

\(\left\{{}\begin{matrix}a+b=0\\2a-2b=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{1}{4}\\b=-\dfrac{1}{4}\end{matrix}\right.\)

\(a,ĐK:x\ne0;x\ne1;x\ne\pm2\\ b,A=\left[\dfrac{2+x}{2-x}-\dfrac{2-x}{2+x}+\dfrac{4x^2}{\left(2-x\right)\left(x+2\right)}\right]\cdot\dfrac{x\left(2-x\right)}{x-1}\\ A=\dfrac{x^2+4x+4-x^2+4x-4+4x^2}{\left(2-x\right)\left(x+2\right)}\cdot\dfrac{x\left(2-x\right)}{x-1}\\ A=\dfrac{4x\left(x+1\right)\cdot x}{\left(x+2\right)\left(x-1\right)}=\dfrac{4x^2}{x+2}\)

1:Sửa đề: \(A=\left(\frac{2x+1}{x\cdot\sqrt{x}-1}-\frac{\sqrt{x}}{x-\sqrt{x}}\right)\left(x-\frac{x-4}{\sqrt{x}-2}\right)\)

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

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

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

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

2: Thay x=2 và y=1 vào y=ax+b, ta được:

\(a\cdot2+b=1\)

=>b=1-2a

Thay x=1 và y=2 vào y=ax+b, ta được:

\(a\cdot1+b=2\)

=>a+b=2

=>a+1-2a=2

=>1-a=2

=>a=1-2=-1

b=1-2a=1-2*(-1)=1+2=3

a: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

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

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

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

a) ĐKXĐ: \(x\notin\left\{0;-5\right\}\)

Ta có: \(B=\dfrac{x^2+2x}{2x+10}+\dfrac{x-5}{x}-\dfrac{5x-50}{2x^2+10x}\)

\(=\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{x-5}{x}-\dfrac{5x-50}{2x\left(x+5\right)}\)

\(=\dfrac{x^3+2x^2}{2x\left(x+5\right)}+\dfrac{2\left(x+5\right)\left(x-5\right)}{2x\left(x+5\right)}-\dfrac{5x-50}{2x\left(x+5\right)}\)

\(=\dfrac{x^3+2x^2+2x^2-50-5x+50}{2x\left(x+5\right)}\)

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

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

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

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

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

\(=\dfrac{x-1}{2}\)

b) Để B=0 thì \(\dfrac{x-1}{2}=0\)

\(\Leftrightarrow x-1=0\)

hay x=1(nhận)

Vậy: Để B=0 thì x=1

Để \(B=\dfrac{1}{4}\) thì \(\dfrac{x-1}{2}=\dfrac{1}{4}\)

\(\Leftrightarrow4\left(x-1\right)=2\)

\(\Leftrightarrow4x-4=2\)

\(\Leftrightarrow4x=6\)

hay \(x=\dfrac{3}{2}\)(nhận)

Vậy: Để \(B=\dfrac{1}{4}\) thì \(x=\dfrac{3}{2}\)

c) Thay x=3 vào biểu thức \(B=\dfrac{x-1}{2}\), ta được:

\(B=\dfrac{3-1}{2}=\dfrac{2}{2}=1\)

Vậy: Khi x=3 thì B=1

d) Để B<0 thì \(\dfrac{x-1}{2}< 0\)

\(\Leftrightarrow x-1< 0\)

\(\Leftrightarrow x< 1\)

Kết hợp ĐKXĐ, ta được: 

\(\left\{{}\begin{matrix}x< 1\\x\notin\left\{0;-5\right\}\end{matrix}\right.\)

Vậy: Để B<0 thì \(\left\{{}\begin{matrix}x< 1\\x\notin\left\{0;-5\right\}\end{matrix}\right.\)

Để B>0 thì \(\dfrac{x-1}{2}>0\)

\(\Leftrightarrow x-1>0\)

hay x>1

Kết hợp ĐKXĐ, ta được: x>1

Vậy: Để B>0 thì x>1

a: ĐKXĐ: x∉{2;-2;0;3}

b: \(A = \left(\dfrac{4x}{x+2}+\dfrac{8x^2}{4-x^2}\right):\left(\dfrac{x-1}{x^2-2x}-\dfrac{2}{x}\right)\)

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

\(=\frac{4x^2-8x-8x^2}{\left(x+2\right)\left(x-2\right)}\cdot\frac{x\left(x-2\right)}{x-1-2x+4}\)

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

c: \(x^2+2x=15\)

=>\(x^2+2x-15=0\)

=>(x+5)(x-3)=0

=>x=-5(nhận) hoặc x=3(loại)

Khi x=-5 thì \(A=\frac{4\cdot\left(-5\right)^2}{-5-3}=\frac{4\cdot25}{-8}=\frac{-100}{8}=-\frac{25}{2}\)

d: |A|>A

=>A<0

=>\(\frac{4x^2}{x-3}<0\)

=>x-3<0

=>x<3

Kết hợp ĐKXĐ, ta được: x<3 và x∉{-2;0;2}

a: ĐKXĐ: x<>2; x<>-2; x<>0; x<>3

b: \(P=\left(\dfrac{-\left(x+2\right)}{x-2}+\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right)\cdot\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)

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

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

c: 2(x-1)=6

=>x-1=3

=>x=4

Thay x=4 vào P, ta đc:

\(P=\dfrac{-4\cdot4^2\cdot\left(4-2\right)}{\left(4+2\right)\left(4-3\right)}=\dfrac{-64\cdot2}{6}=\dfrac{-128}{6}=-\dfrac{64}{3}\)

\(a,\)

\(B=\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}-\dfrac{4}{1-x^2}\) (Điều kiện xác định: \(x\ne\pm1\))

\(=\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)}+\dfrac{4}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{x^2-2x+1-\left(x^2+2x+1\right)+4}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{x^2-2x+1-x^2-2x-1+4}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{-4x+4}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{-4\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)

\(=-\dfrac{4}{x+1}\)

\(b,\)

\(x^2-x=0\)

\(\Leftrightarrow x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Với \(x=0\Rightarrow B=-\dfrac{4}{0+1}=-4\)

Với \(x=1\Rightarrow B=-\dfrac{4}{1+1}=-2\)

\(c,\)

\(B=-3\Rightarrow-\dfrac{4}{x+1}=-3\)

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

\(\Leftrightarrow-3x-3+4=0\)

\(\Leftrightarrow-3x+1=0\)

\(\Leftrightarrow-3x=-1\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

\(d,\)

\(B< 0\Rightarrow-\dfrac{4}{x+1}< 0\)

\(\Leftrightarrow x+1>0\)

\(\Leftrightarrow x>-1\)

Kết hợp điều kiện \(x\ne\pm1\) 

\(\Rightarrow-1< x< 1\)