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`c)(15/(sqrt6+1)+4/(sqrt6-2)-12/(3-sqrt6))*(sqrt6+11)`
`=((15(sqrt6-1))/(6-1)+(4(sqrt6+2))/(6-4)-(12(3+sqrt6))/(9-6))*(sqrt6+11)`
`=(3(sqrt6-1)+2(sqrt6+2)-4(3+sqrt6))*(sqrt6+11)`
`=(3sqrt6-3+2sqrt6+4-12-4sqrt6)*(sqrt6+11)`
`=(sqrt6-11)(sqrt6+11)`
`=6-121=-115`
c) Ta có: \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(=\left[3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
=6-121=-115
Bài 1:
a: Ta có: \(\sqrt{3x^2}=\sqrt{12}\)
\(\Leftrightarrow3x^2=12\)
\(\Leftrightarrow x^2=4\)
hay \(x\in\left\{2;-2\right\}\)
b: Ta có: \(\sqrt{\left(x-2\right)^2}=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
a)√x−2+12√4x−8=√9x−18−2
=>√x−2+12√4(x−2)=√9(x−2)−2
=>√x−2+12√22(x−2)=√32(x−2)−2
=>√x−2+12.2√(x−2)=3√(x−2)−2
=>√x−2+24√(x−2)=3√(x−2)−2
=>√x−2+24√(x−2)-3√(x−2)=-2
=>√x−2(1+24-3)=-2
=>22√x−2=-2
=>√x−2=-2/22
=>√x−2=-1/11
=>x−2=1/121
=>x=1/121+2=243/121
b)√(3x−1)2=5
=>|3x−1|=5
=>3x−1=5 hoặc 3x−1=-5
=>3x=6 hoặc 3x=-4
=>x=2 hoặc x=-4/3
\(B=\frac{2\sqrt{6}+\sqrt{3}+4\sqrt{2}+3}{\sqrt{6+3+2+2\sqrt{6}+2\sqrt{12}+2\sqrt{18}}}=\frac{2\sqrt{6}+\sqrt{3}+4\sqrt{2}+3}{\sqrt{\left(\sqrt{6}+\sqrt{3}+\sqrt{2}\right)^2}}=\frac{2\sqrt{6}+\sqrt{3}+4\sqrt{2}+3}{\sqrt{6}+\sqrt{3}+\sqrt{2}}\)
\(=\frac{\sqrt{6}+\sqrt{3}+\sqrt{2}+\sqrt{6}+3\sqrt{2}+3}{\sqrt{6}+\sqrt{3}+\sqrt{2}}=\frac{\sqrt{6}+\sqrt{3}+\sqrt{2}+\sqrt{3}\left(\sqrt{6}+\sqrt{3}+\sqrt{2}\right)}{\sqrt{6}+\sqrt{3}+\sqrt{2}}\)
\(=\frac{\left(\sqrt{6}+\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}+1\right)}{\sqrt{6}+\sqrt{3}+\sqrt{2}}=\sqrt{3}+1\)
Nhìn nó có dạng \(\sqrt{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}=\sqrt{a+b+c+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}\)
Thấy \(2\left(\sqrt{6}+\sqrt{12}+\sqrt{18}\right)=2\left(\sqrt{2.3}+\sqrt{2.6}+\sqrt{3.6}\right)\)
nên:
\(\sqrt{11+2\left(\sqrt{6}+\sqrt{12}+\sqrt{18}\right)}=\sqrt{\sqrt{2}^2+\sqrt{6}^2+\sqrt{3}^2+2\left(\sqrt{2.3}+\sqrt{2.6}+\sqrt{3.6}\right)}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)^2}=\sqrt{2}+\sqrt{3}+\sqrt{6}\)
Nguyễn Việt Lâm Anh ơi! Anh làm cho em câu này ạ!