Cho biểu thức A = \(\frac{1}{x-2}\) + \(\frac{1}{x+2}\) +\(\frac{x^2+1}{x^2-4}\)
a, Rút gọn
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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
\(A=\left(\dfrac{1}{\sqrt{x-1}}+\dfrac{1}{\sqrt{x+1}}\right)^2\cdot\dfrac{x^2-1}{2}-\sqrt{x^2-1}\\ \Rightarrow A=\left(\dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x^2-1}}\right)^2\cdot\dfrac{x^2-1}{2}-\sqrt{x^2-1}\\ \Rightarrow A=\dfrac{\left(\sqrt{x+1}+\sqrt{x-1}\right)^2}{2}-\sqrt{x^2-1}\\ \Rightarrow A=\dfrac{2x+2\sqrt{x^2-1}-2\sqrt{x^2-1}}{2}\\ \Rightarrow A=x\)
5 tháng 8 2016
\(A=\frac{1}{x-2}+\frac{1}{x+2}+\frac{x^2}{x^2-4}\)
\(ĐKXĐ:x\ne\pm2\)
\(a,A=\frac{x+2}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}+\frac{x^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x+2+x-2+x^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2x+x^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x\left(2+x\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x}{x-2}\)
\(b,A=\frac{x}{x-2}\)
\(=\frac{x-2+2}{x-2}\)
\(=\frac{x-2}{x-2}+\frac{2}{x-2}\)
\(=1+\frac{2}{x-2}\)
\(\text{Để A có giá trị nguyên thì:2⋮ x-2}\)
\(\text{hay }x-2\inƯ\left(2\right)=\left\{-1;1;-2;2\right\}\)
\(\Rightarrow x\in\left\{1;3;0;4\right\}\left(tm\right)\)
\(\text{Vậy }x\in\left\{1;3;0;4\right\}\) \(\text{thì A có giá trị nguyên.}\)
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.
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
\(A=\frac{x}{x+1}-\frac{3-3x}{x^2-x+1}+\frac{x+4}{x^3+1}\)
\(A=\frac{x}{x+1}-\frac{3-3x}{x^2-x+1}+\frac{x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(A=\frac{x\left(x^2-x+1\right)-\left(3+3x\right)\left(x+1\right)+\left(x+4\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(A=\frac{x^3-x^2+x-9x-3-3x^2+x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(A=\frac{x^3-x^2-3x^2+x-9x+x+3+4}{x^3+1}\)
\(A=\frac{x^3+2x^2-4x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)
30 tháng 1 2022
\(A=\left(\dfrac{1}{x-2}+\dfrac{2x}{\left(x-2\right)\left(x+2\right)}+\dfrac{1}{x+2}\right)\cdot\dfrac{2-x}{x}\)
\(=\dfrac{x+2+2x+x-2}{-\left(2-x\right)\left(x+2\right)}\cdot\dfrac{2-x}{x}\)
\(=\dfrac{4x}{-\left(x+2\right)\cdot x}=\dfrac{-4}{x+2}\)
20 tháng 1 2021
\(A=\left(\frac{x^2-16}{x-4}+1\right):\left(\frac{x-2}{x-3}+\frac{x+3}{x+1}+\frac{x+2-x^2}{x^2-2x-3}\right)\)
\(=\left(x+5\right):\left(\frac{\left(x-2\right)\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}+\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}+\frac{x+2-x^2}{\left(x-3\right)\left(x+1\right)}\right)\)
\(=\left(x+5\right):\left(\frac{x^2+x-2x-2+x^2-9+x+2-x^2}{\left(x-3\right)\left(x+1\right)}\right)\)
\(=\left(x+5\right):\left(\frac{x^2-9}{\left(x-3\right)\left(x+1\right)}\right)\)
\(=\left(x+5\right):\left(\frac{x+3}{x+1}\right)=\frac{x+3}{\left(x+5\right)\left(x+1\right)}\)
BT
9 tháng 11 2017
\(A=\left(\frac{x^2-1}{x^4-x^2+1}-\frac{1}{x^2+1}\right).\left(x^4+\frac{1-x^4}{1+x^2}\right)\)
\(=\left(\frac{\left(x^2-1\right)\left(x^2+1\right)-\left(x^4-x^2+1\right)}{\left(x^4-x^2+1\right)\left(x^2+1\right)}\right).\left(x^4+\frac{\left(1+x^2\right)\left(1-x^2\right)}{1+x^2}\right)\)
\(=\frac{x^4-1-x^4+x^2-1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\left(x^4+1-x^2\right)\)
\(=\frac{x^2-2}{x^2+1}\).
26 tháng 3 2020
\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{1+x+1-x}{\left(1+x\right)\left(1-x\right)}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2+2x^2+2-2x^2}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4+4x^4+4-4x^4}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{8+8x^8+8-8x^8}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{16}{1+x^{16}}\)
\(=\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}}\)
\(=\frac{16+16x^{16}+16-16x^{16}}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)
\(=\frac{32}{1-x^{32}}\)
Đào Thị Phương Duyên mk trả lời câu kia của pn rùi ak
nhẩm cx ra (x+1)2 / x2 -4