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2
20 tháng 8 2020
1. \(\left(\sqrt{5}-\sqrt{6}\right)=\left(\sqrt{5}\right)^2-2\sqrt{5}\sqrt{6}+\left(\sqrt{6}\right)^2=5-2\sqrt{30}+6\)
2. \(\left(\sqrt{3}-\sqrt{5}\right)^2=\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot\sqrt{5}+\left(\sqrt{5}\right)^2=3-2\sqrt{15}+5\)
3. \(\left(2\sqrt{2}+\sqrt{3}\right)^2=\left(2\sqrt{2}\right)^2+2\cdot2\sqrt{2}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2=8+4\sqrt{6}+3\)
4. \(\left(\sqrt{4}-\sqrt{17}\right)^2=\left(\sqrt{4}\right)^2-2\cdot\sqrt{4}\cdot\sqrt{17}+\left(\sqrt{17}\right)^2=4-4\sqrt{47}+17\)
5. \(\sqrt{\left(\sqrt{5}-3\right)^2}=\left|\sqrt{5}-3\right|=\left|-3+\sqrt{5}\right|=3-\sqrt{5}\)
6. \(\left(2\sqrt{5}-\sqrt{7}\right)\left(2\sqrt{5}+\sqrt{7}\right)=\left(2\sqrt{5}\right)^2-\left(\sqrt{7}\right)^2=4\cdot5-7=13\)
7. \(\left(5\sqrt{2}+2\sqrt{3}\right)\left(2\sqrt{3}-5\sqrt{2}\right)=\left(2\sqrt{3}\right)^2-\left(5\sqrt{2}\right)^2=12-50=-38\)
8. \(\sqrt{\left(5+2\sqrt{6}\right)^2}-\sqrt{\left(5-2\sqrt{6}\right)^2}=\left|5+2\sqrt{6}\right|-\left|5-2\sqrt{6}\right|=5+2\sqrt{6}-\left(5-2\sqrt{6}\right)=4\sqrt{6}\)9. \(\sqrt{\left(\sqrt{7}-2\right)^2}+\sqrt{\left(\sqrt{7}+2\right)^2}=\left|\sqrt{7}-2\right|+\left|\sqrt{7}+2\right|=-2+\sqrt{7}+2+\sqrt{7}=2\sqrt{7}\)
10. \(\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\left|\sqrt{3}+\sqrt{2}\right|+\left|\sqrt{3}-\sqrt{2}\right|=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\)
#em mới lớp 8 nên không chắc lắm ạ :((
DK
26 tháng 6 2016
\(\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left(5-2\sqrt{2}-\sqrt{3}\right)\)
\(=\left(\sqrt{6}+\sqrt{3}+\sqrt{2}+1\right)\left(5\sqrt{6}-4\sqrt{3}-3\sqrt{2}+5-2\sqrt{2}-\sqrt{3}\right)\)
\(=\left(\sqrt{6}+\sqrt{3}+\sqrt{2}+1\right)\left(5\sqrt{6}-5\sqrt{3}-5\sqrt{2}+5\right)\)
\(=5\left(\sqrt{6}+\sqrt{3}+\sqrt{2}+1\right)\left(\sqrt{6}-\sqrt{3}-\sqrt{2}+1\right)\)
\(=5\left[\left(\sqrt{6}+1\right)^2-\left(\sqrt{3}+\sqrt{2}\right)^2\right]\)
\(=5.\left(6+1+2\sqrt{6}-3-2\sqrt{6}\right)\)
\(=5.2=10\)
Chúc bạn học tốt và nhớ click cho mình với nhá!
7 tháng 8 2018
a) \(A=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\)
\(A=\sqrt{\left(2+\sqrt{3}\right)\left(\sqrt{2+\sqrt{3}}+2\right)\left(-\sqrt{2+\sqrt{3}}+2\right)}\)
\(A=\sqrt{1}\)
\(A=1\)
b)\(B=\left(\frac{\sqrt{x}}{\sqrt{xy}-y}-\frac{\sqrt{y}}{\sqrt{xy}-x}\right).\left(x\sqrt{y}-y\sqrt{x}\right)\)
\(B=\frac{\sqrt{xy}}{\sqrt{xy}-y}x\sqrt{y}+\frac{\sqrt{x}}{\sqrt{xy}-y}y\sqrt{x}+\left(-\frac{\sqrt{y}}{\sqrt{xy}-x}\right)^2x\sqrt{y}+y\sqrt{x}\)
\(B=x\frac{\sqrt{x}}{\sqrt{xy}-y}\sqrt{y}+y\frac{\sqrt{x}}{\sqrt{xy}-y}\sqrt{x}+x\frac{\sqrt{x}}{\sqrt{xy}-x}\sqrt{y}-y\sqrt{x}\frac{\sqrt{y}}{\sqrt{xy}-y}\)
\(B=\frac{-x^{\frac{5}{2}}\sqrt{y}+\sqrt{x}.y^{\frac{5}{2}}}{\left(\sqrt{xy}-y\right)\left(\sqrt{xy}-x\right)}\)
\(B=\frac{\left(\sqrt{x}.y^{\frac{5}{2}}-x^{\frac{5}{2}}\sqrt{y}\right)\left(y+\sqrt{xy}\right)\left(x+\sqrt{xy}\right)}{\left(-y^2+xy\right)\left(-x^2+xy\right)}\)
c) \(C=\sqrt{\left(3-\sqrt{5}\right)^2+\sqrt{6}-2\sqrt{5}}\)
\(C=14-6\sqrt{5}+\sqrt{6}-2\sqrt{5}\)
\(C=14-8\sqrt{5}+\sqrt{6}\)
\(C=\sqrt{14-8\sqrt{5}+\sqrt{6}}\)
25 tháng 1 2021
a) Ta có: \(\left(x-\sqrt{2}\right)+3\left(x^2-2\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right)+3\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right)\left(1+3x+3\sqrt{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{2}=0\\3x+3\sqrt{2}+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\3x=-3\sqrt{2}-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=\dfrac{-3\sqrt{2}-1}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{\sqrt{2};\dfrac{-3\sqrt{2}-1}{3}\right\}\)
b) Ta có: \(x^2-5=\left(2x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)
\(\Leftrightarrow\left(x+\sqrt{5}\right)\left(x-\sqrt{5}\right)-\left(2x-\sqrt{5}\right)\left(x+\sqrt{5}\right)=0\)
\(\Leftrightarrow\left(x+\sqrt{5}\right)\left(x-\sqrt{5}-2x+\sqrt{5}\right)=0\)
\(\Leftrightarrow-x\left(x+\sqrt{5}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\x+\sqrt{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\sqrt{5}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-\sqrt{5}\right\}\)