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24 tháng 8 2021

\(x-3-\sqrt{x^2-6x+9}\left(1\right)=x-3-\sqrt{\left(x-3\right)^2}=x-3-\left|x-3\right|\)

TH1: \(x< 3\)

\(\left(1\right)=x-3+x-3=2x-6\)

TH2: \(x\ge3\)

\(\left(1\right)=x-3-x+3=0\)

 

\(x-3-\sqrt{x^2-6x+9}\)

\(=x-3-\left|x-3\right|\)

\(=\left[{}\begin{matrix}x-3-x+3=0\left(x\ge3\right)\\x-3+x-3=2x-6\left(x< 3\right)\end{matrix}\right.\)

Ta có: \(T=\sqrt{x^2-6x+9}+\sqrt{x^2+6x+9}\)

\(=\left|x-3\right|+\left|x+3\right|\)

\(=3-x+x+3\)

\(=6\)

27 tháng 9 2021

\(B=\left(2x-3\right)\left(4x^2+6x+9\right)-\left(x+1\right)^2-\left(x-2\right)^3\)

\(=8x^3-27-x^2-2x-1-x^3+6x^2-12x+8\)

\(=7x^3+5x^2-14x-20\)

14 tháng 12 2023

9(x + 1)² - 16(y + 3)²

= 9(x² + 2x + 1) - 16(y² + 6y + 9)

= 9x² + 18x + 9 - 16y² - 96y - 144

= 9x² - 16y² + 18x - 96y - 135

23 tháng 6 2019

\(a,\)\(đkxđ\Leftrightarrow x\ge0\)và \(x-9\ne0\Rightarrow x\ne9\)

\(A=\frac{6\sqrt{x}}{x-9}-\frac{5\sqrt{x}}{3-\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+3}\)

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

\(=\frac{6\sqrt{x}+5x+15\sqrt{x}+x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{18\sqrt{x}+6x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{6\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{6\sqrt{x}}{\sqrt{x}-3}\)

23 tháng 6 2019

\(b,\)Để \(A>2\)\(\Rightarrow\frac{6\sqrt{x}}{\sqrt{x}-3}>2\)

\(\Rightarrow\frac{6\sqrt{x}}{\sqrt{x}-3}>\frac{12\sqrt{x}}{x-3}\)

\(\Rightarrow\frac{6\sqrt{x}-12\sqrt{x}}{\sqrt{x}-3}>0\)

\(\Rightarrow\frac{6\sqrt{x}}{\sqrt{x}-3}< 0\)

Vì \(\sqrt{x}\ge0;\)\(6>0\)\(\Rightarrow6\sqrt{x}\ge0\)

\(\Rightarrow\frac{6\sqrt{x}}{\sqrt{x}-3}>0\Leftrightarrow\sqrt{x}-3< 0\)

\(\Rightarrow\sqrt{x}< 3\Rightarrow\sqrt{x}< \sqrt{9}\)\(\Leftrightarrow x< 9\)

Mà \(x\ge0\left(đkxđ\right)\)\(\Rightarrow0\le x< 9\)

19 tháng 10 2021

\(1,\\ a,ĐK:x\ge-\dfrac{1}{2}\\ PT\Leftrightarrow\sqrt{2x+1}=\dfrac{2}{3}\Leftrightarrow2x+1=\dfrac{4}{9}\Leftrightarrow x=-\dfrac{5}{18}\left(tm\right)\\ b,PT\Leftrightarrow\left|x-3\right|=2\Leftrightarrow\left[{}\begin{matrix}x-3=2\\3-x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\\ 2,\\ a,=\left|5-x\right|=x-5\\ b,=\sqrt{4a\cdot44a}=\sqrt{176a^2}=4\left|a\right|\sqrt{11}=4a\sqrt{11}\\ c,=\sqrt{\left(2x-1\right)^2}=\left|2x-1\right|=2x-1\)