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27 tháng 9 2017
\(B=\sqrt{12-6\sqrt{3}}+\sqrt{21-12\sqrt{3}}\)
\(B=\sqrt{\left(\sqrt{3}\right)^2-6\sqrt{3}+3^2}+\sqrt{\left(2\sqrt{3}\right)^2-12\sqrt{3}+3^2}\)
\(B=\sqrt{\left(\sqrt{3}-3\right)^2}+\sqrt{\left(2\sqrt{3}-3\right)^2}\)
\(B=\left|\sqrt{3}-3\right|+\left|2\sqrt{3}-3\right|\)
\(B=\left(3-\sqrt{3}\right)+\left(2\sqrt{3}-3\right)\)( vi \(\sqrt{3}-3< 0\); \(2\sqrt{3}-3>0\))
\(B=3-\sqrt{3}+2\sqrt{3}-3\)
\(B=\sqrt{3}\)
26 tháng 5 2022
a: \(A=\dfrac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}-\dfrac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)
\(=-\sqrt{x}+3-\sqrt{x}+3-6=-2\sqrt{x}\)
b: \(\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right):\left(\dfrac{2\sqrt{x}}{\sqrt{x}+1}-1\right)\)
\(=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x+1\right)}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{2\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}-x-1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{1}{x+1}\)
g: \(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\left(\dfrac{x-1}{\sqrt{x}+1}-2\right)\)
\(=\dfrac{\sqrt{x}+1+\sqrt{x}-1}{x-1}\cdot\left(\sqrt{x}-1-2\right)\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-1}\)
17 tháng 10 2020
1) Ta có: \(\left(\sqrt{12}-6\sqrt{3}+\sqrt{24}\right)\cdot\sqrt{6}-\left(\frac{5}{2}\sqrt{2}+12\right)\)
\(=\left(2\sqrt{3}-6\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}-\left(\sqrt{\frac{25}{4}\cdot2}+12\right)\)
\(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}-\left(\sqrt{\frac{50}{4}}+12\right)\)
\(=-12\sqrt{2}+12-\frac{5\sqrt{2}}{2}-12\)
\(=\frac{-24\sqrt{2}-5\sqrt{2}}{2}\)
\(=\frac{-29\sqrt{2}}{2}\)
2) Ta có: \(\frac{26}{2\sqrt{3}+5}-\frac{4}{\sqrt{3}-2}\)
\(=\frac{26\left(5-2\sqrt{3}\right)}{\left(5+2\sqrt{3}\right)\left(5-2\sqrt{3}\right)}+\frac{4}{2-\sqrt{3}}\)
\(=\frac{26\left(5-2\sqrt{3}\right)}{25-12}+\frac{4\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=2\left(5-2\sqrt{3}\right)+4\left(2+\sqrt{3}\right)\)
\(=10-4\sqrt{3}+8+4\sqrt{3}\)
\(=18\)
3) ĐK để phương trình có nghiệm là: x≥0
Ta có: \(\sqrt{x^2-6x+9}=2x\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x\)
\(\Leftrightarrow\left|x-3\right|=2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x\\x-3=-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3-2x=0\\x-3+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x-3=0\\3x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=3\\3x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
Vậy: S={1}
4) ĐK để phương trình có nghiệm là: \(x\ge\frac{1}{2}\)
Ta có: \(\sqrt{4x^2+1}=2x-1\)
\(\Leftrightarrow\left(\sqrt{4x^2+1}\right)^2=\left(2x-1\right)^2\)
\(\Leftrightarrow4x^2+1=4x^2-4x+1\)
\(\Leftrightarrow4x^2+1-4x^2+4x-1=0\)
\(\Leftrightarrow4x=0\)
hay x=0(loại)
Vậy: S=∅
PA
25 tháng 7 2017
\(x^2-4x-6=\sqrt{2x^2-8x+12}\)
\(\Leftrightarrow\left(x^2+2x\right)-\left(6x+6+\sqrt{2x^2-8x+12}\right)=0\)
\(\Leftrightarrow x\left(x+2\right)-\dfrac{36x^2+72x+36-\left(2x^2-8x+12\right)}{\left(6x+6\right)-\sqrt{2x^2-8x+12}}=0\)
\(\Leftrightarrow x\left(x+2\right)-\dfrac{2\left(17x+6\right)\left(x+2\right)}{\left(6x+6\right)-\sqrt{2x^2-8x+12}}=0\)
\(\Leftrightarrow\left(x+2\right)\left[x-\dfrac{2\left(17x+6\right)}{\left(6x+6\right)-\sqrt{2x^2-8x+12}}\right]=0\)
Pt \(x-\dfrac{2\left(17x+6\right)}{\left(6x+6\right)-\sqrt{2x^2-8x+12}}\) vô nghiệm
=> x + 2 = 0
<=> x = - 2 (nhận)
PA
25 tháng 7 2017
\(\sqrt{x+2-4\sqrt{x-2}}+\sqrt{x+7-6\sqrt{x-2}}=1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-2}-2\right)^2}+\sqrt{\left(\sqrt{x-2}-3\right)^2}=1\)
\(\Leftrightarrow\left|\sqrt{x-2}-2\right|+\left|\sqrt{x-2}-3\right|=1\)
Ta có:
\(VT=\left|\sqrt{x-2}-2\right|+\left|3-\sqrt{x-2}\right|\ge\left|\sqrt{x-2}-2+3-\sqrt{x-2}\right|=1\)
Dấu "=" xảy ra khi \(\left(\sqrt{x-2}-2\right)\left(3-\sqrt{x-2}\right)\ge0\)
Bảng xét dấu:
Vậy \(6\le x\le11\)
`\sqrt{x+12+6\sqrt{x+3}}-\sqrt{x+12-6\sqrt{x+3}}` `ĐK: x >= -3`
`=\sqrt{(\sqrt{x+3})^2+2.\sqrt{x+2}.3+3^2}-\sqrt{(\sqrt{x+3})^2-2.\sqrt{x+2}.3+3^2}`
`=\sqrt{(\sqrt{x+3}+3)^2}-\sqrt{(\sqrt{x+3}-3)^2}`
`=|\sqrt{x+3}+3|-|\sqrt{x+3}-3|`
`=\sqrt{x+3}+3-|\sqrt{x+3}-3|`
`@` Với `\sqrt{x+3}-3 >= 0<=>\sqrt{x+3} >= 3<=>x+3 >= 9<=>x >= 6` (t/m)
`=>\sqrt{x+3}+3-|\sqrt{x+3}-3|=\sqrt{x+3}+3-\sqrt{x+3}+3=6`
`@` Với `\sqrt{x+3}-3 < 0<=>\sqrt{x+3} < 3<=>x+3 < 9<=>x < 6`
Kết hợp đk `x >= -3 =>-3 <= x < 6`
`=>\sqrt{x+3}+3-|\sqrt{x+3}-3|=\sqrt{x+3}+3-3+\sqrt{x+3}=2\sqrt{x+3}`
\(\sqrt{x+12+6\sqrt{x+3}}-\sqrt{x+12-6\sqrt{x+3}}\) \(\left(ĐKXĐ:x\ge-3\right)\)
\(=\sqrt{\left(x+3\right)+2\sqrt{x+3}.3+9}-\sqrt{\left(x+3\right)-2\sqrt{x+3}.3+9}\)
\(=\sqrt{\left[\left(\sqrt{x}+3\right)+3\right]^2}-\sqrt{\left[\left(\sqrt{x}+3\right)-3\right]^2}\)
\(=|\left(\sqrt{x}+3\right)+3|-|\left(\sqrt{x}+3\right)-3|\)
\(=\left(\sqrt{x}+3\right)+3-\left(\sqrt{x}+3\right)+3=6\) ( Với \(x\ge-3\) )