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4 tháng 12 2018

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

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

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

\(P=\frac{-6}{\left(x+2\right)\left(x-2\right)}:\frac{2}{x+2}\)

\(P=\frac{-6}{\left(x+2\right)\left(x-2\right)}\cdot\frac{x+2}{2}\)

\(P=\frac{-3}{x-2}.\)

Câu 1:

Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)

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

Câu 3: 

Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)

\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)

\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)

\(=\sqrt{a}\left(\sqrt{a}-2\right)\)

\(=a-2\sqrt{a}\)

15 tháng 7 2016

(x+1)^2(x-1)^2(x-1)(x+1)

(x+1)^3(x-1)^3

28 tháng 6 2015

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

\(=\left(x^2-1\right)\left(x^2+1\right)\left(x^2+4\right)\left(x^2-4\right)=\left(x^4-1\right)\left(x^4-16\right)\)

29 tháng 9 2019

(x + 1)^4 - 6(x + 1)^2 - (x^2 - 2)(x^2 + 2)

= (x^2 + 2x + 1)(x^2 + 2x + 1) - 6(x^2 + 2x + 1) - (x^2 - 2)(x^2 + 2)

= x^2.(x^2 + 2x + 1) + 2x.(x^2 + 2x + 1) + x^2 + 2x + 1 - (x^2 - 2)(x^2 + 2)

= x^4 + 2x^3 + x^2 + 2x^3 + 4x^2 + 2x + x^2 + 2x + 1 - 6x^2 - 12x - 6 - x^2 + 2^2

= 4x^3 - 8x - 1

29 tháng 9 2019

\(\left(x+1\right)^4-6\left(x+1\right)^2-\left(x^2-2\right)\left(x^2+2\right)\)

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

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

\(=4x^3-8x-1\)

27 tháng 10 2021

Bài 1: 

a: \(Q=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\left(x+\sqrt{x}\right)\)

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

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

Bạn vào biểu tượng \(\Sigma\) để nhập biểu thức cho chính xác nhé

9 tháng 7 2017

a) (x+1)(x^2-x+1)-x^3 

= x^3+1 - x^3 =1

b) (x-2)^2 -x(x+2)

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

=-6x+4

=-2(3x-2)