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a: ĐKXĐ: x<>0; x<>5; x<>5/2; x<>-5

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

\(=\dfrac{x^2-x^2+10x-25}{x\left(x-5\right)\left(x+5\right)}\cdot\dfrac{x\left(x+5\right)}{2x-5}=\dfrac{1}{x-5}\)

a) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

b) Ta có: \(B=\left(\dfrac{2x+1}{x-1}+\dfrac{8}{x^2-1}-\dfrac{x-1}{x+1}\right)\cdot\dfrac{x^2-1}{5}\)

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

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

\(=\dfrac{2x^2+3x+9-x^2+2x-1}{5}\)

\(=\dfrac{x^2+5x+8}{5}\)

Ta có: \(x^2+5x+8\)

\(=x^2+2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}+\dfrac{7}{4}\)

\(=\left(x+\dfrac{5}{2}\right)^2+\dfrac{7}{4}\)

Ta có: \(\left(x+\dfrac{5}{2}\right)^2\ge0\forall x\)

\(\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}>0\forall x\)

\(\Leftrightarrow x^2+5x+8>0\forall x\)

\(\Leftrightarrow\dfrac{x^2+5x+8}{5}>0\forall x\) thỏa mãn ĐKXĐ(đpcm)

28 tháng 6 2023

Xem lại biểu thức P.

28 tháng 6 2023

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Mình phải đi ăn nên chiều mình làm nốt câu d nhé

3 tháng 2 2022

a) ĐKXĐ: \(x\ne-10;x\ne0;x\ne-5\)

b) \(P=\dfrac{x^2+2x}{2x+20}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)

\(=\dfrac{x^2+2x}{2\left(x+10\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)

\(=\dfrac{x\left(x^2+2x\right)\left(x+5\right)}{2x\left(x+10\right)\left(x+5\right)}+\dfrac{2\left(x-5\right)\left(x+10\right)}{2x\left(x+10\right)\left(x+5\right)}+\dfrac{\left(50-5x\right)\left(x+10\right)}{2x\left(x+5\right)\left(x+10\right)}\)

\(=\dfrac{x^4+7x^3+10x^2+2x^2+10x-100+500-5x^2}{2x\left(x+10\right)\left(x+5\right)}\)

\(=\dfrac{x^4+7x^3+7x^2+10x+400}{2x\left(x+10\right)\left(x+5\right)}\)

c) \(P=0\Rightarrow x^4+7x^3+7x^2+10x+400=0\Leftrightarrow...\)

Số xấu thì câu c, d làm cũng như không. Bạn xem lại đề.

11 tháng 1 2021

[2x-2=0=>x=1

x-1=0=>x=1

x+1=0=>x=-1

5=0=>x=5

15 tháng 12 2021

a) ĐKXĐ: \(x\ne-3,x\ne2\)

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

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

30 tháng 12 2021

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

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

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

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

19 tháng 11 2021

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

a: ĐKXĐ:\(x\notin\left\{2;0\right\}\)

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

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

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

c: Thay x=2017 vào C, ta được:

\(C=\dfrac{2017+1}{2\cdot2017}=\dfrac{1009}{2017}\)