Rút gọn biểu thức:
\(A=\dfrac{\sqrt{2+\sqrt{4-x^2}}\left[\sqrt{\left(2+x\right)^3}-\sqrt{\left(2-x\right)^3}\right]}{4+\sqrt{4-x^2}}\)với \(-2\le x\le2\)
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S
21 tháng 8 2018
\(A=\dfrac{\sqrt{2+\sqrt{4-x^2}}\left(\sqrt{\left(2+x\right)^3}-\sqrt{\left(2-x\right)^3}\right)}{4+\sqrt{4-x^2}}\)
\(\Rightarrow A=\sqrt{\left(2+x\right)^{^{ }3}}-\sqrt{\left(2-x\right)^3}=\left(\sqrt{2+x}-\sqrt{2-x}\right)\left(4+\sqrt{4-x^2}\right)\)
\(\Rightarrow A=\dfrac{\sqrt{4+2\sqrt{4-x^2}}\left(\sqrt{2+x}-\sqrt{2-x}\right)\left(4+\sqrt{4-x^2}\right)}{\sqrt{2}\left(4+\sqrt{4-x^2}\right)}\)
\(\Rightarrow A=\dfrac{\left(\sqrt{2+x}+\sqrt{2-x}\right)\left(\sqrt{2+x}-\sqrt{2-x}\right)}{\sqrt{2}}=2\sqrt{2}\)
HP
3 tháng 9 2021
a, \(A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
b, \(A\in Z\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}\in Z\)
\(\Leftrightarrow\sqrt{x}+3\inƯ_3=\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow\sqrt{x}=0\)
\(\Leftrightarrow x=0\)
3 tháng 9 2021
\(a,A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\left(x\ge0;x\ne9\right)\\ A=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\\ A=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)
\(b,A\in Z\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}\in Z\Leftrightarrow-3⋮\sqrt{x}+3\\ \Leftrightarrow\sqrt{x}+3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{-6;-4;-2;0\right\}\)
Mà \(\sqrt{x}\ge0\)
\(\Leftrightarrow x\in\left\{0\right\}\)
Vậy \(x=0\) thì A nguyên
21 tháng 6 2021
`a)(2sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)/(3-sqrtx)(x>=0,x ne 4,x ne 9)`
`=(2sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)/(sqrtx-3)`
`=(2sqrtx-9+(sqrtx-3)(sqrtx+3)+(2sqrtx+1)(sqrtx-2))/(x-5sqrtx+6)`
`=(2sqrtx-9+x-9+2x-3sqrtx-2)/(x-5sqrtx+6)`
`=(3x-sqrtx-20)/
10 tháng 8 2015
\(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\left(\frac{2x-6\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\frac{3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+9\right)}\right).\frac{\sqrt{x}+3}{2\left(\sqrt{x}-1\right)}\)
\(=\frac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}+3}{2\sqrt{x}-2}=\frac{-3\sqrt{x}-3}{2x-8\sqrt{x}+6}\)
Nếu đề ko sai thì đấy là kết quả
28 tháng 8 2022
\(=\dfrac{8-x}{2+\sqrt[3]{x}}:\dfrac{4+2\sqrt[3]{x}+\sqrt[3]{x^2}}{2+\sqrt[3]{x}}+\dfrac{\sqrt[3]{x^2}-2\sqrt[3]{x}+2\sqrt[3]{x}}{\sqrt[3]{x}-2}\cdot\dfrac{\sqrt[3]{x^2}-1}{\sqrt[3]{x}\left(\sqrt[3]{x}+1\right)}\)
\(=2-\sqrt[3]{x}+\dfrac{\sqrt[3]{x}-1}{\sqrt[3]{x}-2}\)
\(=\dfrac{4-4\sqrt[3]{x}+\sqrt[3]{x^2}-\sqrt[3]{x}+1}{2-\sqrt[3]{x}}\)
\(=\dfrac{\sqrt[3]{x^2}-5\sqrt[3]{x}+5}{2-\sqrt[3]{x}}\)
LM
3
TM
27 tháng 7 2016
\(\sqrt{\left(x-4\right)^2}+\frac{x-4}{\sqrt{x^2-8x+16}}\)
\(=x-4+\frac{x-4}{\sqrt{\left(x-4\right)^2}}\)
\(=x-4+\frac{x-4}{x-4}\)
\(=x-4+1\)
\(=x-3\)
27 tháng 7 2016
\(\sqrt{\left(x-4\right)^2}+\frac{x-4}{\sqrt{x^2-8x+16}}\)
\(=x-4+\frac{x-4}{\sqrt{\left(x+4\right)^2}}\)
\(=x-4+\frac{x-4}{x-4}\)
\(=x-4+1\)
= x - 3
28 tháng 8 2022
Sửa đề: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\cdot\dfrac{x-4}{4-\sqrt{x}}\)
a: \(P=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}\cdot\dfrac{x-4}{4-\sqrt{x}}=\dfrac{2x}{4-\sqrt{x}}\)
b: Để P>3 thì P-3>0
\(\Leftrightarrow-\dfrac{2x}{\sqrt{x}-4}-3>0\)
\(\Leftrightarrow\dfrac{-2x-3\sqrt{x}+12}{\sqrt{x}-4}>0\)
\(\Leftrightarrow\dfrac{5\sqrt{x}-12}{\sqrt{x}-4}< 0\)
=>12/5<căn x<4
=>144/25<x<16
