Rút gọn biểu thức : \(\frac{x^2-1}{x+1}\)
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H9
HT.Phong (9A5)
CTVHS
5 tháng 8 2023
\(A=\left(\dfrac{1}{\sqrt{x-1}}+\dfrac{1}{\sqrt{x-1}}\right)^2\cdot\dfrac{x^2-1}{2}-\sqrt{x^2-1}\) (ĐK: \(x>1\))
\(A=\left(\dfrac{2}{\sqrt{x-1}}\right)^2\cdot\dfrac{x^2-1}{2}-\sqrt{x^2-1}\)
\(A=\dfrac{4}{x-1}\cdot\dfrac{\left(x+1\right)\left(x-1\right)}{2}-\sqrt{x^2-1}\)
\(A=2\left(x+1\right)-\sqrt{\left(x+1\right)\left(x-1\right)}\)
\(A=\sqrt{x+1}\left(2\sqrt{x+1}-\sqrt{x-1}\right)\)
HQ
Hà Quang Minh
Giáo viên
5 tháng 8 2023
\(A=\left(\dfrac{1}{\sqrt{x-1}}+\dfrac{1}{\sqrt{x+1}}\right)^2\cdot\dfrac{x^2-1}{2}-\sqrt{x^2-1}\\ \Rightarrow A=\left(\dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x^2-1}}\right)^2\cdot\dfrac{x^2-1}{2}-\sqrt{x^2-1}\\ \Rightarrow A=\dfrac{\left(\sqrt{x+1}+\sqrt{x-1}\right)^2}{2}-\sqrt{x^2-1}\\ \Rightarrow A=\dfrac{2x+2\sqrt{x^2-1}-2\sqrt{x^2-1}}{2}\\ \Rightarrow A=x\)
29 tháng 11 2016
\(A=\frac{\left[x\left(x^2-x+1\right)\right]-\left[\left(x+1\right)\left(3-3x\right)\right]+\left[x+4\right]}{x^3+1}\)
\(A=\frac{\left(x^3-x^2+x\right)+3\left(x^2-1\right)+\left(x+4\right)}{x^3+1}=\frac{x^3+2x^2+2x+1}{x^3+1}\)
\(A=\frac{\left(x^3+1\right)+2x\left(x+1\right)}{x^3+1}=1+\frac{2x}{x^2-x+1}\)
29 tháng 11 2016
\(A=\frac{x}{x+1}-\frac{3-3x}{x^2-x+1}+\frac{x+4}{x^3+1}\)
\(A=\frac{x}{x+1}-\frac{3-3x}{x^2-x+1}+\frac{x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(A=\frac{x\left(x^2-x+1\right)-\left(3+3x\right)\left(x+1\right)+\left(x+4\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(A=\frac{x^3-x^2+x-9x-3-3x^2+x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(A=\frac{x^3-x^2-3x^2+x-9x+x+3+4}{x^3+1}\)
\(A=\frac{x^3+2x^2-4x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)
Y
7 tháng 8 2023
\(Q=\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}+\dfrac{2\sqrt{x}}{x-4}\left(dk:x\ge0,x\ne4\right)\\ =\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\\ =\dfrac{2\left(2-\sqrt{x}\right)+2+\sqrt{x}-2\sqrt{x}}{4-x}\\ =\dfrac{4-2\sqrt{x}+2+\sqrt{x}-2\sqrt{x}}{4-x}\\ =\dfrac{-3\sqrt{x}+6}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\\ =\dfrac{-3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{3}{\sqrt{x}+2}\)
\(b,Q=\dfrac{6}{5}\Leftrightarrow\dfrac{3}{\sqrt{x}+2}=\dfrac{6}{5}\Rightarrow15-6\left(\sqrt{x}+2\right)=0\Rightarrow15-6\sqrt{x}-12=0\)
\(\Rightarrow-6\sqrt{x}=-3\Rightarrow\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\left(tm\right)\)
Vậy \(x=\dfrac{1}{4}\)thỏa mãn đề bài.
13 tháng 8 2020
\(A=\left(\frac{1}{1-x}+\frac{2}{x+1}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}\left(x\ne\pm1;x\ne\frac{1}{2}\right)\)
\(\Leftrightarrow A=\left(\frac{-1}{x-1}+\frac{2}{x+1}+\frac{5-x}{x^2-1}\right)\cdot\frac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(\Leftrightarrow A=\left[\frac{-x-1}{\left(x-1\right)\left(x+1\right)}+\frac{2x-2}{\left(x-1\right)\left(x+1\right)}+\frac{5-x}{\left(x-1\right)\left(x+1\right)}\right]\cdot\frac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(\Leftrightarrow A=\frac{-x-1+2x-2+5-x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(x-1\right)\left(x+1\right)}{2}\)
\(\Leftrightarrow A=\frac{2\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}=1\)
vậy \(A=1\left(x\ne\pm1;x\ne\frac{1}{2}\right)\)
A
13 tháng 8 2020
\(A=\left(\frac{1}{1-x}+\frac{2}{x+1}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}.\)
\(A=\left(\frac{x+1}{\left(1-x\right)\left(x+1\right)}+\frac{2\left(1-x\right)}{\left(x+1\right)\left(1-x\right)}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}.\)
\(A=\left(\frac{x+1}{\left(1-x\right)\left(x+1\right)}+\frac{2\left(1-x\right)}{\left(x+1\right)\left(1-x\right)}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}.\)
\(A=\left(\frac{x+1+2-2x-5+x}{1-x^2}\right):\frac{1-2x}{x^2-1}.\)
\(A=\left(\frac{-2}{1-x^2}\right):\frac{1-2x}{x^2-1}.\)
\(A=\frac{2}{x^2-1}:\frac{1-2x}{x^2-1}.\)
\(A=\frac{2}{x^2-1}\cdot\frac{^2-1}{1-2x}=\frac{2}{1-2x}\)ĐK: x khác 1/2
10 tháng 8 2021
Bài 1 : Với : \(x>0;x\ne1\)
\(P=\left(1+\frac{1}{\sqrt{x}-1}\right)\frac{1}{x-\sqrt{x}}=\left(\frac{\sqrt{x}}{\sqrt{x}-1}\right).\sqrt{x}\left(\sqrt{x}-1\right)=x\)
Thay vào ta được : \(P=x=25\)
10 tháng 8 2021
Bài 2 :
a, Với \(x\ge0;x\ne1\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}=\frac{x+\sqrt{x}-2\sqrt{x}+2-2}{x-1}\)
\(=\frac{x-\sqrt{x}}{x-1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
Thay x = 9 vào A ta được : \(\frac{3}{3+1}=\frac{3}{4}\)
5 tháng 8 2016
\(A=\frac{1}{x-2}+\frac{1}{x+2}+\frac{x^2}{x^2-4}\)
\(ĐKXĐ:x\ne\pm2\)
\(a,A=\frac{x+2}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}+\frac{x^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x+2+x-2+x^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2x+x^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x\left(2+x\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x}{x-2}\)
\(b,A=\frac{x}{x-2}\)
\(=\frac{x-2+2}{x-2}\)
\(=\frac{x-2}{x-2}+\frac{2}{x-2}\)
\(=1+\frac{2}{x-2}\)
\(\text{Để A có giá trị nguyên thì:2⋮ x-2}\)
\(\text{hay }x-2\inƯ\left(2\right)=\left\{-1;1;-2;2\right\}\)
\(\Rightarrow x\in\left\{1;3;0;4\right\}\left(tm\right)\)
\(\text{Vậy }x\in\left\{1;3;0;4\right\}\) \(\text{thì A có giá trị nguyên.}\)
28 tháng 7 2017
\(P=\frac{x+2}{\sqrt{x}^3-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
\(P=\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
2,
\(A=\frac{5\left(\sqrt{7}-\sqrt{2}\right)}{\left(\sqrt{7}-\sqrt{2}\right)\left(\sqrt{7}+\sqrt{2}\right)}+\frac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-\frac{7\sqrt{7}}{7}\)
\(A=\frac{5\left(\sqrt{7}-\sqrt{2}\right)}{7-2}+\frac{\left(\sqrt{2}+1\right)}{2-1}-\sqrt{7}\)
\(A=\sqrt{7}-\sqrt{2}+\sqrt{2}+1-\sqrt{7}=1\)
\(P=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
3 tháng 3 2022
Bài 1:
a: \(A=\dfrac{x+1+x}{x+1}:\dfrac{3x^2+x^2-1}{x^2-1}\)
\(=\dfrac{2x+1}{x+1}\cdot\dfrac{\left(x+1\right)\left(x-1\right)}{\left(2x+1\right)\left(2x-1\right)}=\dfrac{x-1}{2x-1}\)
b: Thay x=1/3 vào A, ta được:
\(A=\left(\dfrac{1}{3}-1\right):\left(\dfrac{2}{3}-1\right)=\dfrac{-2}{3}:\dfrac{-1}{3}=2\)
\(\frac{x^2-1}{x+1}=\frac{\left(x-1\right)\left(x+1\right)}{x+1}=x-1\)