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a: \(=\sqrt{\dfrac{25}{16}\cdot\dfrac{49}{9}\cdot\dfrac{1}{100}}=\dfrac{5}{4}\cdot\dfrac{7}{3}\cdot\dfrac{1}{10}=\dfrac{35}{120}=\dfrac{7}{24}\)
b: \(=\sqrt{1.44\cdot0.81}=1.2\cdot0.9=1.08\)
c: \(=\sqrt{\dfrac{\left(165-124\right)\left(165+124\right)}{164}}=\sqrt{\dfrac{1}{4}\cdot289}=\dfrac{17}{2}\)
d: \(=\sqrt{\dfrac{\left(149-76\right)\left(149+76\right)}{\left(457-384\right)\left(457+384\right)}}=\sqrt{\dfrac{225}{841}}=\dfrac{15}{29}\)
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:
\(VT=\dfrac{1}{\sqrt{a}-\sqrt{b}}:\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}:\dfrac{\sqrt{ab}\cdot\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\cdot\dfrac{1}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{\left(a-b\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{1}{a-b}=VP\left(dpcm\right)\)
1) \(\frac{\sqrt{165^2-124^2}}{164}=\frac{\sqrt{\left(165-124\right)\left(165+124\right)}}{164}=\frac{\sqrt{41}\cdot\sqrt{289}}{164}=\frac{\sqrt{41}\cdot17}{164}=\frac{17}{4\sqrt{41}}\)
2) \(\frac{\sqrt{149^2-76^2}}{\sqrt{457^2-384^2}}=\frac{\sqrt{\left(149+76\right)\left(149-76\right)}}{\sqrt{\left(457+384\right)\left(457-384\right)}}=\frac{\sqrt{225}\cdot\sqrt{73}}{\sqrt{841}\cdot\sqrt{73}}=\frac{25}{29}\)
a) \(\sqrt{\frac{165^2-124^2}{164}}=\sqrt{\frac{\left(165-124\right)\left(165+124\right)}{164}}=\sqrt{\frac{41.289}{164}}\)
\(=\sqrt{\frac{11849}{164}}=\sqrt{72,25}=8,5\)
b)\(\sqrt{\frac{149^2-76^2}{457^2-384^2}}=\sqrt{\frac{\left(149-76\right)\left(149+76\right)}{\left(457-384\right)\left(457+384\right)}}\) \(=\sqrt{\frac{73.225}{73.841}}=\sqrt{\frac{225}{841}}=\sqrt{\frac{15^2}{29^2}}=\frac{15}{29}\)
c)\(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\) \(=\sqrt{2^2+3+2.2.\sqrt{3}}-\sqrt{2^2+3-2.2.\sqrt{3}}\)
\(=\sqrt{2^2+2.2.\sqrt{3}+\sqrt{3}^2}-\sqrt{2^2-2.2.\sqrt{3}+\sqrt{3}^2}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}=\left(2+\sqrt{3}\right)-\left(2-\sqrt{3}\right)\)
\(=2+\sqrt{3}-2+\sqrt{3}=2\sqrt{3}\)
a) \(\sqrt{\frac{\left(165-124\right)\left(165+124\right)}{164}}=\sqrt{\frac{41.289}{164}}=\sqrt{\frac{289}{4}}=\frac{17}{2}\)
b) tương tự ý a
c) \(\left(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\right)^2=7+4\sqrt{3}+7-4\sqrt{3}-2.\sqrt{7+4\sqrt{3}}.\sqrt{7-4\sqrt{3}}\)
\(=14-2\sqrt{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=14-2\sqrt{49-48}\)
\(=14-2.1=12\)
\(\Rightarrow\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}=\sqrt{12}=2\sqrt{3}\)
\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)
\(ĐK:x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)
\(\Leftrightarrow4x^2-9=4x+12\)
\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)
\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(ĐK:x\ge5\)
\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)
ĐK:x>=1
\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)
\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)
\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
\(ĐK:x\ge3\)
\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)
\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)
\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}=0\) (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)
\(\sqrt{1\dfrac{9}{16}.5\dfrac{4}{9}.0,01}=\sqrt{\dfrac{25}{16}.\dfrac{49}{9}.\dfrac{1}{100}}=\sqrt{\dfrac{25}{16}}.\sqrt{\dfrac{49}{9}}.\sqrt{\dfrac{1}{100}}=\dfrac{5}{4}.\dfrac{7}{3}.\dfrac{1}{10}=\dfrac{5.7.1}{4.3.10}=\dfrac{35}{120}=\dfrac{7}{24}\)
\(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+\sqrt{3}\right)=\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\sqrt{2}-\sqrt{3}=\sqrt{3}+2+\sqrt{2}-\sqrt{2}-\sqrt{3}=2\)
a) HD: Đổi hỗn số và số thập phân thành phân số.
ĐS:
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b)
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d) ĐS:
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