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tìm giá trị nhỏ nhất của biểu thức:
D= x+2y -√2x−y- 5√4y−3+ 13 ( x≥12 ; y≥ 34 )
\(\frac{\left(5\sqrt{7}+7\sqrt{5}\right)}{\sqrt{35}}\)
= \(\frac{\sqrt{5}.\left(\sqrt{35}+7\right)}{\sqrt{35}}\)
= \(\frac{\sqrt{35}+7}{\sqrt{7}}\)
= \(\sqrt{5}+\sqrt{7}\)
\(\frac{5\sqrt{7}+7\sqrt{5}}{\sqrt{35}}=\frac{\sqrt{5}.\sqrt{5}.\sqrt{7}+\sqrt{7}.\sqrt{7}.\sqrt{5}}{\sqrt{35}}.\)
\(=\frac{\sqrt{5}.\sqrt{35}+\sqrt{7}.\sqrt{35}}{\sqrt{35}}\)
\(=\frac{\sqrt{35}\left(\sqrt{5}+\sqrt{7}\right)}{\sqrt{35}}=\sqrt{5}+\sqrt{7}\)
a) \(A=\frac{1}{2}\sqrt{32}+\sqrt{98}-\frac{1}{6}\sqrt{18}=\frac{1}{2}\sqrt{4^2.2}+\sqrt{7^2.2}-\frac{1}{6}.\sqrt{3^2.2}\)
\(=\frac{1}{2}\sqrt{4^2}.\sqrt{2}+\sqrt{7^2}.\sqrt{2}-\frac{1}{6}.\sqrt{3^2}.\sqrt{2}\)\(=\frac{1}{2}.4\sqrt{2}+7\sqrt{2}-\frac{1}{6}.3.\sqrt{2}\)\(=2.\sqrt{2}+7\sqrt{2}-\frac{1}{2}\sqrt{2}=\left(2+7-\frac{1}{2}\right)\sqrt{2}=\frac{17}{2}\sqrt{2}\)
a) \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{1+2\sqrt{5}+\left(\sqrt{5}\right)^2}+\sqrt{1-2\sqrt{5}+\left(\sqrt{5}\right)^2}\)\(=\sqrt{\left(1+\sqrt{5}\right)^2}+\sqrt{\left(1-\sqrt{5}\right)^2}=1+\sqrt{5}-\left(1-\sqrt{5}\right)=1+\sqrt{5}-1+\sqrt{5}=2\sqrt{5}\)
a) \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
b) \(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}\)
\(=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
\(=2\sqrt{2}+\sqrt{5}+2\sqrt{2}-\sqrt{5}=4\sqrt{2}\)
c) \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
\(=2\sqrt{2\sqrt{3}}-10\sqrt{2\sqrt{3}}+8\sqrt{2\sqrt{3}}=0\)
a) \(\sqrt{x}+\sqrt{\frac{x}{9}}-\frac{1}{3}\sqrt{4x}=5\)
ĐK : x ≥ 0
<=>\(\sqrt{x}+\sqrt{x\times\frac{1}{9}}-\frac{1}{3}\sqrt{2^2x}=5\)
<=> \(\sqrt{x}+\sqrt{x\times\left(\frac{1}{3}\right)^2}-\left(\frac{1}{3}\times\left|2\right|\right)\sqrt{x}=5\)
<=> \(\sqrt{x}+\left|\frac{1}{3}\right|\sqrt{x}-\left(\frac{1}{3}\times2\right)\sqrt{x}=5\)
<=> \(\sqrt{x}+\frac{1}{3}\sqrt{x}-\frac{2}{3}\sqrt{x}=5\)
<=> \(\sqrt{x}\left(1+\frac{1}{3}-\frac{2}{3}\right)=5\)
<=> \(\sqrt{x}\times\frac{2}{3}=5\)
<=> \(\sqrt{x}=\frac{15}{2}\)
<=> \(x=\frac{225}{4}\)( tm )
Đề khó quá!!!!!!!!!!!!!!
Nhưng mik vẫn giải được kìa chứng tỏ khả năng của mình thuộc đẳng cấp nào rồi đấy ^-^
Cho tích thôi nạo
M=\(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
M3=\(\text{ }\text{ }\text{ }\text{ }\left(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\right)^3\)
M3=\(5+2\sqrt{13}+3\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}\)\(\left(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\right)\)\(+5-2\sqrt{13}\)
M3=\(10+3\sqrt[3]{25-4\cdot13}\cdot M\)
M3=\(10+3\sqrt[3]{-27}\cdot M\)
M3=\(10-9M\)
\(M^3+9M-10=0\)
\(M=1\)
\(=>\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}=1\)