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Bài 1:
a) \(\sqrt{1,44\cdot1,21-1,44\cdot0,4}\)
\(=\sqrt{1,44\cdot\left(1,21-0,4\right)}\)
\(=\sqrt{1,44\cdot0,81}\)
\(=\sqrt{1,44}\cdot\sqrt{0,81}\)
\(=1,2\cdot0,9\)
\(=1,08\)
b) \(\dfrac{\sqrt{5}-2}{\sqrt{5}+2}+\sqrt{80}\)
\(=\dfrac{\left(\sqrt{5}-2\right)^2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}+4\sqrt{5}\)
\(=\dfrac{5-4\sqrt{5}+4}{1}+4\sqrt{5}\)
\(=9-4\sqrt{5}+4\sqrt{5}\)
\(=9\)
c) \(\sqrt[3]{16}+\sqrt[3]{2}\left(\sqrt[3]{4}-\sqrt[3]{2}\right)\)
\(=\sqrt[3]{2^3\cdot2}+\sqrt[3]{2\cdot4}-\sqrt[3]{2\cdot2}\)
\(=2\sqrt[3]{2}+\sqrt[3]{8}-\sqrt[3]{4}\)
\(=2\sqrt[3]{2}+2-\sqrt[3]{4}\)
Bài 2: Ta có:
\(VT=\dfrac{1}{\sqrt{a}-\sqrt{b}}:\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}:\dfrac{\sqrt{ab}\cdot\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\cdot\dfrac{1}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{\left(a-b\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{1}{a-b}=VP\left(dpcm\right)\)
a) Ta có: \(\sqrt{2}\left(\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}\right)\)
\(=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
\(=\sqrt{5}-1-\sqrt{5}-1=-2\)
b) Ta có: \(\sqrt{13+30\sqrt{2}+\sqrt{9+4\sqrt{2}}}\)
\(=\sqrt{13+30\sqrt{2}+2\sqrt{2}+1}\)
\(=\sqrt{14+32\sqrt{2}}\)
c) Ta có: \(\sqrt{6+2\sqrt{5}-\sqrt{13+\sqrt{48}}}\)
\(=\sqrt{6+2\sqrt{5}-2\sqrt{3}-1}\)
\(=\sqrt{5+2\sqrt{5}-2\sqrt{3}}\)
Không làm mất tính tổng quát của bài toán, giả sử \(a\ge b\ge c\)(1)
Có \(\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{a+c}{ac}}+\sqrt{\frac{b+c}{bc}}=\sqrt{\frac{1}{b}+\frac{1}{a}}+\sqrt{\frac{1}{c}+\frac{1}{a}}+\sqrt{\frac{1}{c}+\frac{1}{b}}\)
Từ (1) => \(\hept{\begin{cases}\frac{2}{a}\le\frac{1}{a}+\frac{1}{b}\\\frac{2}{b}\le\frac{1}{b}+\frac{1}{c}\\\frac{2}{c}\le\frac{1}{a}+\frac{1}{c}\end{cases}}\Rightarrow\hept{\begin{cases}\sqrt{\frac{2}{a}}\le\sqrt{\frac{1}{a}+\frac{1}{b}}\\\sqrt{\frac{2}{b}}\le\sqrt{\frac{1}{b}+\frac{1}{c}}\\\sqrt{\frac{2}{c}}\le\sqrt{\frac{1}{a}+\frac{1}{c}}\end{cases}}\)
=>\(\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{1}{b}+\frac{1}{a}}+\sqrt{\frac{1}{c}+\frac{1}{a}}+\sqrt{\frac{1}{c}+\frac{1}{b}}\)
=>\(\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{a+c}{ac}}+\sqrt{\frac{b+c}{bc}}\)
Ta có đpcm
a) Ta có: \(\left(\sqrt{7}-\sqrt{2}\right)\cdot\sqrt{9+2\sqrt{14}}\)
\(=\left(\sqrt{7}-\sqrt{2}\right)\cdot\left(\sqrt{7}+\sqrt{2}\right)\)
=7-2
=5
d) Ta có: \(\dfrac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-\dfrac{6\sqrt{2}-4}{3-\sqrt{2}}\)
\(=2\sqrt{2}-\sqrt{7}+5\sqrt{7}-\dfrac{2\sqrt{2}\left(3-\sqrt{2}\right)}{3-\sqrt{2}}\)
\(=2\sqrt{2}+4\sqrt{7}-2\sqrt{2}\)
\(=4\sqrt{7}\)
14:
\(A=\sqrt{-4x^2+4x+7}\)
\(=\sqrt{-\left(4x^2-4x-7\right)}\)
\(=\sqrt{-\left(4x^2-4x+1-8\right)}\)
\(=\sqrt{-\left(2x-1\right)^2+8}< =\sqrt{8}=2\sqrt{2}\)
Dấu = xảy ra khi 2x-1=0
=>\(x=\dfrac{1}{2}\)
13:
\(a+b+c>=\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)
=>\(2a+2b+2c-2\sqrt{ab}-2\sqrt{bc}-2\sqrt{ac}>=0\)
=>\(\left(a-2\sqrt{ab}+b\right)+\left(b-2\sqrt{bc}+c\right)+\left(a-2\sqrt{ac}+c\right)>=0\)
=>\(\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2+\left(\sqrt{a}-\sqrt{c}\right)^2>=0\)(luôn đúng)
a) áp dụng bất đẳng thức CÔ SI => dpcm