Cho a= \(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\) Hãy tính giá trị của biểu thức sau :
M = \(\sqrt{15a^2-8a\sqrt{15}+16}\)
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MA
9 tháng 9 2016
\(C=\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}\)
\(C^2=\left(\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}\right)^2\)
\(C^2=x^2+2\sqrt{x^2-1}-2\sqrt{\left(x^2+2\sqrt{x^2-1}\right)\left(x^2-2\sqrt{x^2-1}\right)}+x^2-2\sqrt{x^2-1}\)
\(C^2=2x^2-2\sqrt{x^4-2x^2\sqrt{x^2-1}+2x^2\sqrt{x^2-1}-\left(2\sqrt{x^2-1}\right)^2}\)
\(C^2=2x^2-2\sqrt{x^4-4\left(x^2-1\right)}\)
\(C^2=2x^2-2\sqrt{x^4-4x^2+4}\)
\(C=\sqrt{2x^2-2\sqrt{x^4-4x^2+4}}\)
Thay: \(x=\sqrt{5}\) vào C, ta có:
\(C=\sqrt{2\sqrt{5}^2-2\sqrt{\sqrt{5}^4-4\sqrt{5}^2+4}}\)
\(C=\sqrt{10-2\sqrt{25-20+4}}\)
\(C=\sqrt{10-2\sqrt{9}}\)
\(C=\sqrt{10-6}\)
\(C=\orbr{\begin{cases}-2\\2\end{cases}}\)
Mà theo bài ra: \(\sqrt{x^2+2\sqrt{x^2-1}}>\sqrt{x^2-2\sqrt{x^2-1}}\)
\(\Rightarrow\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}>0\)
\(\Rightarrow C=2\)
26 tháng 7 2017
Giải
Ta có: \(\sqrt{\dfrac{5}{3}}+\sqrt{\dfrac{3}{5}}=\dfrac{\sqrt{5}}{\sqrt{3}}+\dfrac{\sqrt{3}}{\sqrt{5}}=\dfrac{8}{\sqrt{15}}\)
Vậy M = \(\sqrt{15\left(\dfrac{8}{15}\right)^2-8.\dfrac{8}{\sqrt{15}}.\sqrt{15}+16}\)
= \(\sqrt{8^2-8^2+16}=\sqrt{16}=4\)
CD
1
20 tháng 6 2017
\(M=\sqrt{15a^2-8a\sqrt{15}+16}=\sqrt{\left(\sqrt{15}a-4\right)^2}\)
\(=\sqrt{15}a-4=\sqrt{15}\left(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\right)-4\)
\(=\left(3+5\right)-4=4\)
23 tháng 7 2019
ĐK: \(x-9\ne0\Rightarrow x\ne9\)
\(\sqrt{x}\ge0\Rightarrow x\ge0\)
\(x+\sqrt{x}-6\ne0\Rightarrow x+3\sqrt{x}-2\sqrt{x}-6\ne0\Rightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)\ne0\)
\(\Rightarrow\sqrt{x}-2\ne0\Rightarrow\sqrt{x}\ne2\Rightarrow x\ne4\)
ĐKXĐ: \(x\ge0;x\ne4;x\ne9\)
\(A=\left(\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{1}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\left(\frac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\left(\frac{1+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\frac{1+x-9-x+4\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{4\sqrt{x}-12}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-3\right)}\)
2, Với \(x=\frac{25}{16}\)\(\Rightarrow\sqrt{x}=\sqrt{\frac{25}{16}}=\frac{5}{4}\)
\(A=\frac{\frac{5}{4}\left(\frac{5}{4}-2\right)}{4\left(\frac{5}{4}-3\right)}=\frac{5}{4}.\left(-\frac{3}{4}\right):4\left(-\frac{7}{4}\right)=-\frac{15}{16}:-7=\frac{15}{112}\)
\(\orbr{\begin{cases}\orbr{\begin{cases}\\\end{cases}}\\\end{cases}}\)\(\orbr{\begin{cases}\orbr{\begin{cases}\sqrt{x}-2< 0\\\sqrt{x}-3>0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}< 2\\\sqrt{x}>3\end{cases}}\Rightarrow\orbr{\begin{cases}x< 4\\x>9\end{cases}}}\\\orbr{\begin{cases}\sqrt{x}-2>0\\\sqrt{x}-3< 0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}>2\\\sqrt{x}< 3\end{cases}\Rightarrow\orbr{\begin{cases}x>4\\x< 9\end{cases}}}}\end{cases}}\)
NN
20 tháng 7 2018
\(A=\frac{2}{\sqrt{5}-3}-\frac{2}{\sqrt{5}+3}=\frac{2\left(\sqrt{5}+3\right)-2\left(\sqrt{5}-3\right)}{-4}=\frac{2\sqrt{5}+6-2\sqrt{5}+6}{-4}=\frac{12}{-4}=-3\)
Vay ........
24 tháng 7 2016
2) \(A=\sqrt{15a^2-8a\sqrt{15}+16}\\ =\sqrt{\left(a\sqrt{15}-4\right)^2}\)
b) Khi a=\(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\) thì
\(A=\sqrt{\left[\left(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\right)\sqrt{15}-4\right]^2}\)
\(=\sqrt{\left[\left(3+5\right)-4\right]^2}\)
\(=\sqrt{4^2}\)
\(=4\)
26 tháng 10 2016
\(M=4\frac{1}{3}-\sqrt{16}+5\sqrt{\frac{4}{9}}-\frac{25}{\left(\sqrt{6}\right)^2}\)
\(=\frac{13}{3}-4+5\cdot\frac{2}{3}-\frac{25}{6}\)
\(=\frac{1}{3}+\frac{10}{3}-\frac{25}{6}\)
\(=\frac{11}{3}-\frac{25}{6}\)
\(=-\frac{1}{2}\)
Ta có: \(a=\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}=\frac{\sqrt{3}}{\sqrt{5}}+\frac{\sqrt{5}}{\sqrt{3}}=\frac{8\sqrt{15}}{15}\)
=> \(a^2=\frac{64}{15}\)
=> \(M=\sqrt{15a^2-8a\sqrt{15}+16}=\sqrt{15.\frac{64}{15}-8.\frac{8\sqrt{15}}{15}.\sqrt{15}+16}\)
\(M=\sqrt{64-64+16}=4\)