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26 tháng 6 2018

\(\sqrt{4\left(a-3\right)^2}\)

\(=\sqrt{4\left(a^2-6a+9\right)}\)

\(=\sqrt{4a^2-24a+36}\)

\(=\sqrt{\left(2a-6\right)^2}\)

\(=\left|2a-6\right|\)

\(=2a-6\)

\(\sqrt{4\left(a-3\right)^2}\)

\(=\sqrt{2^2\left(a-3\right)^2}\)

\(=2\left(a-3\right)\)

\(=2a-6\)

15 tháng 7 2016

\(\sqrt{4\left(a-3\right)^2}=\sqrt{\left[2\left(a-3\right)\right]^2}=2\left(a-3\right)\)3)

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 2019

a) \(\sqrt{\left(3-6a\right)^2}=6a-3\)

( vì \(a\ge\frac{1}{2}\)\(\Rightarrow3-6a< 0\))

31 tháng 10 2018

Rút gọn bt:

Câu 1: a, \(\left(\sqrt{50}+\sqrt{48}-\sqrt{72}\right)2\sqrt{3}\)

b, \(\sqrt{25a}+2\sqrt{45a}-3\sqrt{80a}+2\sqrt{16a}\left(a\ge0\right)\)ư

Câu 2: Cho bt: P =\(\left(1+\frac{\sqrt{a}}{a+1}\right):\left(\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\right)\)

a, Tìm ĐKXĐ . Rút gọn P 

B, Tìm x nguyên để P có gt nguyên

c, Tìm GTNN của P với a >1

Câu 3: Giair các pt 

a, \(\sqrt{\left(2x-1\right)^2}=4\)

b, \(\sqrt{4x+4}+\sqrt{9x+9}-8\sqrt{\frac{x+1}{16}}=5\)

23 tháng 8 2019

\(P=\frac{x\sqrt{x}-8}{x+2\sqrt{x}+4}+3\left(1-\sqrt{x}\right).\)

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

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

\(=\sqrt{x}-2+3-3\sqrt{x}=-2\sqrt{x}+1\)

\(Q=\frac{2P}{1-P}=\frac{2\left(-2\sqrt{x}+1\right)}{1-\left(-2\sqrt{x}+1\right)}\)

\(=\frac{-4\sqrt{x}+2}{1+2\sqrt{x}-1}=\frac{-2\sqrt{x}+1}{\sqrt{x}}\)

\(=\frac{-2\sqrt{x}}{\sqrt{x}}+\frac{1}{\sqrt{x}}=-2+\frac{1}{\sqrt{x}}\)

\(Q\in Z\Leftrightarrow-2+\frac{1}{\sqrt{x}}\in Z\Rightarrow\frac{1}{\sqrt{x}}\in Z\)

\(\Rightarrow1\)\(⋮\)\(\sqrt{x}\)\(\Rightarrow\sqrt{x}\inƯ_1\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=1\\\sqrt{x}=-1\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x\in\varnothing\end{cases}}}\)

Vậy \(Q\in Z\Leftrightarrow x=1\)

31 tháng 10 2018

\(a,\left(\sqrt{50}+\sqrt{48}-\sqrt{72}\right)2\sqrt{3}\)

\(=\left(5\sqrt{2}+4\sqrt{3}-6\sqrt{2}\right)2\sqrt{3}\)

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

\(=24-2\sqrt{6}\)

21 tháng 8 2018

\(A=\dfrac{\sqrt{2+\sqrt{4-x^2}}\left(\sqrt{\left(2+x\right)^3}-\sqrt{\left(2-x\right)^3}\right)}{4+\sqrt{4-x^2}}\)

\(\Rightarrow A=\sqrt{\left(2+x\right)^{^{ }3}}-\sqrt{\left(2-x\right)^3}=\left(\sqrt{2+x}-\sqrt{2-x}\right)\left(4+\sqrt{4-x^2}\right)\)

\(\Rightarrow A=\dfrac{\sqrt{4+2\sqrt{4-x^2}}\left(\sqrt{2+x}-\sqrt{2-x}\right)\left(4+\sqrt{4-x^2}\right)}{\sqrt{2}\left(4+\sqrt{4-x^2}\right)}\)

\(\Rightarrow A=\dfrac{\left(\sqrt{2+x}+\sqrt{2-x}\right)\left(\sqrt{2+x}-\sqrt{2-x}\right)}{\sqrt{2}}=2\sqrt{2}\)

21 tháng 8 2018

\(2\sqrt{2}\)