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Bài 2: 

a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x-2}\right):\left(\dfrac{x^2-4+16-x^2}{x+2}\right)\)

\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)

\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)

\(=\dfrac{x-x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{12}=\dfrac{-1}{6\left(x-2\right)}\)

b: Thay x=1/2 vào B, ta được:

\(B=\dfrac{-1}{6\cdot\left(\dfrac{1}{2}-2\right)}=\dfrac{-1}{6\cdot\dfrac{-3}{2}}=\dfrac{1}{9}\)

Thay x=-1/2 vào B, ta được:

\(B=\dfrac{-1}{6\cdot\left(-\dfrac{1}{2}-2\right)}=-\dfrac{1}{15}\)

c: Để B=2 thì \(\dfrac{-1}{6\left(x-2\right)}=2\)

=>6(x-2)=-1/2

=>x-2=-1/12

hay x=23/12

17 tháng 1 2021

Đk: x \(\ne\)0; x \(\ne\)\(\pm\)3

Ta có: A = \(\left(\frac{1}{3}+\frac{3}{x^2-3x}\right):\left(\frac{x^2}{27-3x^2}+\frac{1}{x+3}\right)\)

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

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

A = \(\frac{-\left(x+3\right)}{x}\)

Để A < -1 <=> \(-\frac{\left(x+3\right)}{x}< -1\) <=> \(\frac{-x-3}{x}+1< 0\)

<=> \(\frac{-x-3+x}{x}< 0\) <=> \(-\frac{3}{x}< 0\) 

Do -3 <0 => x> 0

Vậy Để A < -1 <=> x > 0 và x khác 3

12 tháng 5 2017

a/ \(P=\left(\frac{3}{\sqrt{x}-1}+\frac{\sqrt{x}-3}{x-1}\right):\left(\frac{x+2}{x+\sqrt{x}-2}-\frac{\sqrt{x}}{\sqrt{x}+2}\right).\)

\(P=\left(\frac{3}{\sqrt{x}-1}+\frac{\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2}{\sqrt{x^2}-1+\sqrt{x}-1}-\frac{\sqrt{x}}{\sqrt{x}+2}\right).\)

\(P=\left(\frac{3\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}}{\sqrt{x}+2}\right).\)

\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\)

\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\)

\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\)

\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{1}{\left(\sqrt{x}-1\right)}\right)\)

\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\frac{\sqrt{x}-1}{1}\right)\)

=> \(P=\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)}\)

b/ \(P=\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)}=\sqrt{x}-1\)

<=> \(4\sqrt{x}=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)

<=> \(4\sqrt{x}=x-1\). Bình phương 2 vế, ta được:

<=> 16x=(x-1)2

<=> 16x=x2-2x+1

<=> x2-18x+1=0

\(\Delta'=81-1=80=>\sqrt{\Delta'}=4\sqrt{5}\)

=> \(x_1=9-4\sqrt{5}\)

\(x_2=9+4\sqrt{5}\)

19 tháng 6 2019

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

\(=\frac{1}{2}x^2.6x+\frac{1}{2}x^2.\left(-3\right)+\left(-x\right).x^2+\left(-x\right).\frac{1}{2}+\frac{1}{2}.x+\frac{1}{2}.4\)

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

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

\(=2x^3-\frac{3}{2}x^2+2\)

19 tháng 6 2019

\(a,\)\(\frac{1}{2}x^2\left(6x-3\right)-x\left(x^2+\frac{1}{2}\right)+\frac{1}{2}\left(x+4\right)\)

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

\(=2x^3-\frac{3}{2}x^2+2\)

\(b,\)\(2x\left(3x^3-x\right)-4x^2\left(x-x^2+1\right)+\left(x-3x^2\right)x\)

\(=6x^4-2x^2-4x^3+4x^4-4x^2+x^2-3x^3\)

\(=10x^4-7x^3-5x^2\)