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\(\frac{2}{x-3}\)\(\sqrt{\frac{x^2-6x+9}{4y^4}}\) (x < 3 ; y\(\ne\)0)
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10 tháng 8 2017
\(A=\frac{x^2+5x+6+x\sqrt{9-x^2}}{3x-x^2+\left(x+2\right)\sqrt{9-x^2}}\)
\(=\frac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}\)
\(=\frac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}\)
\(=\frac{\sqrt{3+x}\left(\left(x+2\right)\sqrt{x+3}+x\sqrt{3-x}\right)}{\sqrt{3-x}\left(\left(x+2\right)\sqrt{x+3}+x\sqrt{3-x}\right)}\)
\(=\frac{\sqrt{3+x}}{\sqrt{3-x}}\)
10 tháng 8 2017
\(B=\frac{x^2-5x+6+3\sqrt{x^2-6x+8}}{3x-12+\left(x-3\right)\sqrt{x^2-6x+8}}\)
\(=\frac{\left(x-3\right)\left(x-2\right)+3\sqrt{\left(x-4\right)\left(x-2\right)}}{3\left(x-4\right)+\left(x-3\right)\sqrt{\left(x-4\right)\left(x-2\right)}}\)
\(=\frac{\sqrt{x-2}\left(\left(x-3\right)\sqrt{x-2}+3\sqrt{x-4}\right)}{\sqrt{x-4}\left(3\sqrt{x-4}+\left(x-3\right)\sqrt{x-2}\right)}\)
\(=\frac{\sqrt{x-2}}{\sqrt{x-4}}\)
6 tháng 7 2016
\(=\frac{2x\left(x-2y\right)}{\left(x+2y\right)^2}:\frac{\left(2y-x\right)\left(2y+x\right)}{\left(x-2y\right)^2}:\frac{5xy\left(x-2y\right)}{\left(x+2y\right)^3}\)
Điều kiện: \(x\ne2y;x\ne-2y;x\ne0;y\ne0\)
\(=\frac{2x\left(x-2y\right)}{\left(x+2y\right)^2}:\frac{\left(2y+x\right)}{\left(x-2y\right)}:\frac{5xy\left(x-2y\right)}{\left(x+2y\right)^3}\)
\(=\frac{2x\left(x-2y\right)}{\left(x+2y\right)^2}\times\frac{x-2y}{x+2y}\times\frac{\left(x+2y\right)^3}{5xy\left(x-2y\right)}=\frac{2\left(x-2y\right)}{5y}\)
CN
2
10 tháng 10 2020
\(=\frac{2\sqrt{x}}{x-3}.\frac{\sqrt{\left(x-3\right)^2}}{\sqrt{x}}=\frac{2\left(x-3\right)}{x-3}=-2\)
AN
2
2 tháng 9 2017
Ta có : \(A=\sqrt{\left(x+3\right)^2}-\frac{\sqrt{x^2-6x+9}}{x+3}\)
\(\Rightarrow A=\left(x+3\right)-\frac{\left(x-3\right)^2}{\left(x+3\right)}\)
\(\Rightarrow A=\frac{\left(x+3\right)^2}{\left(x+3\right)}-\frac{\left(x-3\right)^2}{\left(x+3\right)}\)
\(\Rightarrow A=\frac{\left(x+3\right)^2-\left(x-3\right)^2}{\left(x+3\right)}\)
\(\Rightarrow A=\frac{\left(x+3+x-3\right)\left(x+3-x+3\right)}{\left(x+3\right)}\)
\(\Rightarrow A=\frac{18x}{\left(x+3\right)}\)
AN
3 tháng 9 2017
Mình quên ghi điều kiện mất là x<-3 nhé
Mấy bạn làm lại giúp mình
Mình cảm ơn :D
AN
0
\(\frac{2}{x-3}\sqrt{\frac{x^2-6x+9}{4y^4}}=\frac{2}{x-3}.\frac{\sqrt{x^2-6x+9}}{\sqrt{4y^4}}=\frac{2}{x-3}.\frac{\sqrt{\left(x-3\right)^2}}{\sqrt{\left(2y^2\right)^2}}\)
\(=\frac{2}{x-3}.\frac{x-3}{2y^2}=\frac{1}{y^2}\)
\(\frac{2}{x-3}\sqrt{\frac{x^2-6x+9}{4y^4}}\)
\(=\frac{2}{x-3}\sqrt{\frac{\left(x-3\right)^2}{\left(2y^2\right)^2}}\)
\(=\frac{2}{x-3}.\left|\frac{x-3}{2y^2}\right|\)
\(=\frac{2}{x-3}.\frac{3-x}{2y^2}\)( vi \(x< 3;y\ne0\))
\(=\frac{-2}{2y^2}\)
\(=\frac{-1}{y^2}\)