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MD
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24 tháng 8 2023
\(M=\left(\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right)\left(\sqrt{6}+1\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+1\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+1\right)\)
\(=6+\sqrt{6}-11\sqrt{6}-11=-5-10\sqrt{6}\)
H9
HT.Phong (9A5)
CTVHS
24 tháng 8 2023
\(M=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+1\right)\)
\(M=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}+2\right)\left(\sqrt{6}-2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3+\sqrt{6}\right)\left(3-\sqrt{6}\right)}\right]\left(\sqrt{6}+1\right)\)
\(M=\left[\dfrac{15\left(\sqrt{6}-1\right)}{6-1}+\dfrac{4\left(\sqrt{6}+2\right)}{6-4}-\dfrac{12\left(3+\sqrt{6}\right)}{9-6}\right]\left(\sqrt{6}+1\right)\)
\(M=\left[3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+1\right)\)
\(M=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\cdot\left(\sqrt{6}+1\right)\)
\(M=\left(5\sqrt{6}-4\sqrt{6}+1-12\right)\left(\sqrt{6}+1\right)\)
\(M=\left(\sqrt{6}-11\right)\left(\sqrt{6}+1\right)\)
\(M=6+\sqrt{6}-11\sqrt{6}-11\)
\(M=-10\sqrt{6}-5\)
9 tháng 11 2023
Bài 1:
\(\sqrt{\left(4-\sqrt{5}\right)^2}+\sqrt{5+2\sqrt{5}+1}\)
\(=\left|4-\sqrt{5}\right|+\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=4-\sqrt{5}+\sqrt{5}+1=5\)
Bài 2:
a: ĐKXĐ: x>=3
\(\sqrt{x-3}=6\)
=>x-3=36
=>x=36+3=39(nhận)
b: ĐKXĐ: \(x\in R\)
\(\sqrt{\left(x-3\right)^2}=12\)
=>\(\left|x-3\right|=12\)
=>\(\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)
Bài 3:
a: \(P=\left(\dfrac{3-x\sqrt{x}}{3-\sqrt{x}}+\sqrt{x}\right)\cdot\left(\dfrac{3-\sqrt{x}}{3-x}\right)\)
\(=\dfrac{3-x\sqrt{x}+\sqrt{x}\left(3-\sqrt{x}\right)}{3-\sqrt{x}}\cdot\dfrac{3-\sqrt{x}}{3-x}\)
\(=\dfrac{3-x\sqrt{x}+3\sqrt{x}-x}{3-x}\)
\(=\dfrac{-\sqrt{x}\left(x-3\right)-\left(x-3\right)}{-\left(x-3\right)}=\dfrac{\left(x-3\right)\left(\sqrt{x}+1\right)}{x-3}=\sqrt{x}+1\)
b: \(P=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{x+\sqrt{x}}\right):\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)
\(=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\dfrac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
c: \(A=\sqrt{3x-1}+3\cdot\sqrt{12x-4}-\sqrt{6^2\left(3x-1\right)}+\sqrt{5}\)
\(=\sqrt{3x-1}+6\sqrt{3x-1}-6\sqrt{3x-1}+\sqrt{5}\)
\(=\sqrt{3x-1}+\sqrt{5}\)
d: \(A=\left(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}\)
\(=\left(\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{2\left(a-2\right)}{a+2}\)
25 tháng 3 2020
Câu 2:
ĐKXĐ \(\hept{\begin{cases}x\ge0\\x-1\ne0\\x+2\sqrt{x}+1\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne1\\\left(\sqrt{x}+1\right)^2\ne0\end{cases}}\)
\(Q=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right)\left(x+\sqrt{x}\right)\)
\(=\left[\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\sqrt{x}\left(\sqrt{x}+1\right)\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\sqrt{x}\left(\sqrt{x}+1\right)\)
\(=\frac{x+\sqrt{x}-2-\left(x-\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\sqrt{x}\)
\(=\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\sqrt{x}=\frac{2x}{x-1}\)
1 tháng 7 2016
1) \(x+5\sqrt{x}+6=x+2\sqrt{x}+3\sqrt{x}+6\)
\(=\sqrt{x}\left(\sqrt{x}+2\right)+3\left(\sqrt{x}+2\right)\)
\(=\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\)
2) \(=\left(4\sqrt{3}+3\sqrt{15}-12\sqrt{15}\right)\sqrt{3}\)
\(=\left(4\sqrt{3}-8\sqrt{15}\right)\sqrt{3}=12-24\sqrt{5}\)
13 tháng 8 2017
em hổng có biết đâu vì em chưa hc lp 9 mới lại đề bài dài kinh khủng
24 tháng 7 2017
a/ \(\frac{1}{2-\sqrt{3}}+\frac{3+\sqrt{3}}{\sqrt{3}}-\frac{4}{\sqrt{3}-1}\)
\(=2+\sqrt{3}+\sqrt{3}+1-2\sqrt{3}-2\)
\(=1\)
b/ \(\sqrt{3x+40}-4=x\)
\(\sqrt{3x+40}=x+4\)
Điều kiện: \(\hept{\begin{cases}3x+40\ge0\\x+4\ge0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ge-\frac{40}{3}\\x\ge-4\end{cases}}\)
\(\Leftrightarrow x\ge-\frac{40}{3}\)
Ta có: \(3x+40=x^2+8x+16\)
\(\Leftrightarrow x^2+5x-24=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-8\left(l\right)\\x=3\end{cases}}\)