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\(A=2\sqrt{5}-\sqrt{45}+2\sqrt{20}=2\sqrt{5}-\sqrt{3^2.5}+2\sqrt{2^2.5}=2\sqrt{5}-3\sqrt{5}+4\sqrt{5}=3\sqrt{5}\)
\(B=\left(\sqrt{18}-\frac{1}{2}\cdot\sqrt{32}+12\sqrt{2}\right):\sqrt{2}=\left(3\sqrt{2}-\frac{1}{2}\cdot4\sqrt{2}+12\sqrt{2}\right):\sqrt{2}\)
\(=13\sqrt{2}:\sqrt{2}=13\)
\(C=\left(\sqrt{12}+2\sqrt{27}-3\sqrt{3}\right)\cdot\sqrt{3}=\left(2\sqrt{3}+6\sqrt{3}-3\sqrt{3}\right)\cdot\sqrt{3}=5\sqrt{3}\cdot\sqrt{3}=15\)
\(D=\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}=-\sqrt{5}+15\sqrt{2}\)
Bài 2 xét x=0 => A =0
xét x>0 thì \(A=\frac{1}{x-2+\frac{2}{\sqrt{x}}}\)
để A nguyên thì \(x-2+\frac{2}{\sqrt{x}}\inƯ\left(1\right)\)
=>cho \(x-2+\frac{2}{\sqrt{x}}\)bằng 1 và -1 rồi giải ra =>x=?
1,Ta có \(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ac}\)
=> \(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=2\)
\(a+2=a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\)
\(b+2=\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{a}\right)\)
\(c+2=\left(\sqrt{c}+\sqrt{b}\right)\left(\sqrt{c}+\sqrt{a}\right)\)
=> \(\frac{\sqrt{a}}{a+2}+\frac{\sqrt{b}}{b+2}+\frac{\sqrt{c}}{c+2}=\frac{\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)}+\frac{\sqrt{b}}{\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{a}\right)}+...\)
=> \(\frac{\sqrt{a}}{a+2}+...=\frac{2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}=\frac{4}{\sqrt{\left(a+2\right)\left(b+2\right)\left(c+2\right)}}\)
=> M=0
Vậy M=0
a) ĐKXĐ: \(x\ne9\)
\(P=\frac{x\sqrt{x}+5\sqrt{x}-12-2\left(\sqrt{x}-3\right)^2-\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{x\sqrt{x}+5\sqrt{x}-12-2x+12\sqrt{x}-18-x-5\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{x\sqrt{x}-3x+12\sqrt{x}-36}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{\left(\sqrt{x}-3\right)\left(x+12\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{x+12}{\sqrt{x}+2}\)
b) Ta có: \(P=\frac{x+12}{\sqrt{x}+2}=\frac{x-4+16}{\sqrt{x}+2}=\sqrt{x}-2+\frac{16}{\sqrt{x}+2}\)
\(=\left(\sqrt{x}+2\right)+\frac{16}{\sqrt{x}+2}-4\)
\(\ge2\sqrt{\left(\sqrt{x}+2\right).\frac{16}{\sqrt{x}+2}}-4=4\)
P = 4 thì \(\left(\sqrt{x}+2\right)^2=16\Rightarrow\sqrt{x}=2\Rightarrow x=4\)
Vậy GTNN của P là 4 khi x = 4.
ai nay dung kinh nghiem la chinh
cau a)
ta thay \(10+6\sqrt{3}=\left(1+\sqrt{3}\right)^3\)
\(6+2\sqrt{5}=\left(1+\sqrt{5}\right)^2\)
khi do \(x=\frac{\sqrt[3]{\left(\sqrt{3}+1\right)^3}\left(\sqrt{3}-1\right)}{\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{5}}\)
\(x=\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{1+\sqrt{5}-\sqrt{5}}\)
\(x=\frac{3-1}{1}=2\)
suy ra
x^3-4x+1=1
A=1^2018
A=1
b)
ta thay
\(7+5\sqrt{2}=\left(1+\sqrt{2}\right)^3\)
khi do
\(x=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\frac{1}{\sqrt[3]{\left(1+\sqrt{2}\right)^3}}\)
\(x=1+\sqrt{2}-\frac{1}{1+\sqrt{2}}=\frac{\left(1+\sqrt{2}\right)^2-1}{1+\sqrt{2}}=\frac{2+2\sqrt{2}}{1+\sqrt{2}}\)
x=2
thay vao
x^3+3x-14=0
B=0^2018
B=0
a) \(2\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=2\sqrt{3}+2-\sqrt{3}\)
\(=\left(2\sqrt{3}-\sqrt{3}\right)+2\)
\(=\sqrt{3}+2\)
b) \(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\)
\(=\frac{1+\sqrt{5}}{\sqrt{5}-1}+\frac{\sqrt{5}-1}{1+\sqrt{5}}\)
\(=\frac{\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}-1\right)\left(1+\sqrt{5}\right)}+\frac{\left(\sqrt{5}-1\right)^2}{\left(\sqrt{5}-1\right)\left(1+\sqrt{5}\right)}\)
\(=\frac{\left(\sqrt{5}+1\right)^2+\left(\sqrt{5}-1\right)^2}{\left(\sqrt{5}-1\right)\left(1+\sqrt{5}\right)}\)
\(=\frac{12}{4}=3\)
c) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}\)
\(=\frac{7-4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}+\frac{7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=\frac{14}{1}=14\)
\(=\frac{7}{\sqrt{3}-\sqrt{2}}-7\sqrt{3}-6\sqrt{2}=\frac{7-\left(7\sqrt{3}+6\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}\)
\(=\frac{7-21+7\sqrt{6}-6\sqrt{6}+12}{\sqrt{3}-\sqrt{2}}=\frac{-2+\sqrt{6}}{\sqrt{3}-\sqrt{2}}=\frac{\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}=\sqrt{2}\)