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21 tháng 9 2015

Phân tích ra như vừa nãy 

21 tháng 9 2015

Như sau :

\(D=\frac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}=\frac{\sqrt{x+3}\left[\left(x+2\right)\sqrt{x+3}+x\sqrt{3-x}\right]}{\sqrt{3-x}\left(x\sqrt{3-x}+\left(x+2\right)\sqrt{3+x}\right)}=\frac{\sqrt{x+3}}{\sqrt{3-x}}\)

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

\(B=\frac{x^2-5x+6+3\sqrt{x^2-6x+8}}{3x-12+\left(x-3\right)\sqrt{x^2-6x+8}}\)

\(=\frac{\left(x-3\right)\left(x-2\right)+3\sqrt{\left(x-4\right)\left(x-2\right)}}{3\left(x-4\right)+\left(x-3\right)\sqrt{\left(x-4\right)\left(x-2\right)}}\)

\(=\frac{\sqrt{x-2}\left(\left(x-3\right)\sqrt{x-2}+3\sqrt{x-4}\right)}{\sqrt{x-4}\left(3\sqrt{x-4}+\left(x-3\right)\sqrt{x-2}\right)}\)

\(=\frac{\sqrt{x-2}}{\sqrt{x-4}}\)

28 tháng 5 2017

\(=\frac{\left(x^2+5x+6\right)+x\sqrt{9-x^2}}{\left(3x-x^2\right)+\left(2+x\right)\sqrt{9-x^2}}\)

\(=\frac{\left(x+2\right)\left(3+x\right)+x\sqrt{\left(3+x\right)\left(3-x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3+x\right)\left(3-x\right)}}\) nhóm nhân tử chung

\(=\frac{\sqrt{3+x}\left(\left(x+2\right)\sqrt{3+x}+x\sqrt{3-x}\right)}{\sqrt{3-x}\left(x\sqrt{3-x}+\left(x+2\right)\sqrt{3+x}\right)}\)rồi rút gọn được

\(=\frac{\sqrt{3+x}}{\sqrt{3-x}}\)

10 tháng 8 2017

Hỏi đáp Toán

10 tháng 8 2017

cái gì vậy bạn???????nhonhung

1 tháng 6 2016

\(\frac{x^2+5x+x\sqrt{9-x^2}+6}{3x-x^2+\left(x+2\right)\sqrt{9-x^2}}\left(DK:-3\le x< 3\right)\)

\(=\frac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}\)

\(=\frac{\sqrt{x+3}\left[\sqrt{x+3}\left(x+2\right)+x\sqrt{3-x}\right]}{\sqrt{3-x}\left[x\sqrt{3-x}+\left(x+2\right)\sqrt{x+3}\right]}=\frac{\sqrt{x+3}\left(x\sqrt{x+3}+2\sqrt{x+3}+x\sqrt{3-x}\right)}{\sqrt{3-x}\left(x\sqrt{x+3}+2\sqrt{x+3}+x\sqrt{3-x}\right)}=\frac{\sqrt{x+3}}{\sqrt{3-x}}=\sqrt{\frac{x+3}{3-x}}\)

13 tháng 11 2021

\(M=\dfrac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}:2\sqrt{\dfrac{3-x+2x}{3-x}}\left(-3\le x< 3;x\ne-1\right)\\ M=\dfrac{\sqrt{x+3}\left(x+2+x\sqrt{3-x}\right)}{\sqrt{3-x}\left[x+\left(x+2\right)\sqrt{3+x}\right]}:2\sqrt{\dfrac{x+3}{3-x}}\\ M=\dfrac{\sqrt{x+3}\left(x+2+x\sqrt{3-x}\right)}{\sqrt{3-x}\left[x+\left(x+2\right)\sqrt{3+x}\right]}\cdot\dfrac{3-x}{2\sqrt{\left(3-x\right)}\sqrt{\left(x+3\right)}}\)

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