\(\sqrt{(1 -\sqrt{5})^2} - \sqrt{16 - 4\sqrt{5}}\)
Những câu hỏi liên quan
tính1.sqrt{147}+sqrt{54}-4sqrt{27}2.sqrt{28}-4sqrt{63}+7sqrt{112}3.sqrt{49}-5sqrt{28}+dfrac{1}{2}sqrt{63}4.left(2sqrt{6}-4sqrt{3}-dfrac{1}{4}sqrt{8}right).3sqrt{6}5.(2sqrt{1dfrac{9}{16}}-5sqrt{5dfrac{1}{16}}):sqrt{16}6.left(sqrt{48}-3sqrt{27}-sqrt{147}right):sqrt{3}7.left(sqrt{50}-3sqrt{49}right):sqrt{2}-sqrt{162}:sqrt{2}8.left(2sqrt{1dfrac{9}{10}}-sqrt{5dfrac{1}{10}}right):sqrt{10}9.2sqrt{dfrac{16}{3}}-3sqrt{dfrac{1}{27}}-6sqrt{dfrac{4}{75}}10.2sqrt{27}-6sqrt{dfrac{4}{3}}+dfrac{3}{5}sqrt{75}11....
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tính
1.\(\sqrt{147}+\sqrt{54}-4\sqrt{27}\)
2.\(\sqrt{28}-4\sqrt{63}+7\sqrt{112}\)
3.\(\sqrt{49}-5\sqrt{28}+\dfrac{1}{2}\sqrt{63}\)
4.\(\left(2\sqrt{6}-4\sqrt{3}-\dfrac{1}{4}\sqrt{8}\right).3\sqrt{6}\)
5.(\(2\sqrt{1\dfrac{9}{16}}-5\sqrt{5\dfrac{1}{16}}\)):\(\sqrt{16}\)
6.\(\left(\sqrt{48}-3\sqrt{27}-\sqrt{147}\right):\sqrt{3}\)
7.\(\left(\sqrt{50}-3\sqrt{49}\right):\sqrt{2}-\sqrt{162}:\sqrt{2}\)
8.\(\left(2\sqrt{1\dfrac{9}{10}}-\sqrt{5\dfrac{1}{10}}\right):\sqrt{10}\)
9.\(2\sqrt{\dfrac{16}{3}}-3\sqrt{\dfrac{1}{27}}-6\sqrt{\dfrac{4}{75}}\)
10.\(2\sqrt{27}-6\sqrt{\dfrac{4}{3}}+\dfrac{3}{5}\sqrt{75}\)
11.\(\dfrac{\sqrt{18}}{\sqrt{2}}-\dfrac{\sqrt{12}}{\sqrt{3}}\)
12.\(\dfrac{\sqrt{27}}{\sqrt{3}}+\dfrac{\sqrt{98}}{\sqrt{2}}-\sqrt{175}:\sqrt{7}\)
13.\(\left(\dfrac{\sqrt{8}}{\sqrt{2}}-\dfrac{\sqrt{180}}{\sqrt{5}}\right).\sqrt{5}-\sqrt{\dfrac{81}{11}}.\sqrt{11}\)
14.\(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
15.\(\left(3\sqrt{2}-2\sqrt{3}\right)\left(3\sqrt{2}+2\sqrt{3}\right)\)
16.\(\left(1+\sqrt{5}-\sqrt{3}\right)\left(1+\sqrt{5}+\sqrt{3}\right)\)
rút gọn các biểu thức sau:
a \(\sqrt[3]{8\sqrt{5}-16}.\sqrt[3]{8\sqrt{5}+16}\)
b \(\sqrt[3]{7-5\sqrt{2}}-\sqrt[6]{8}\)
c \(\sqrt[3]{4}.\sqrt[3]{1-\sqrt{3}}.\sqrt[6]{4+2\sqrt{3}}\)
d \(\dfrac{2}{\sqrt[3]{3}-1}-\dfrac{4}{\sqrt[3]{9}-\sqrt[3]{3}+1}\)
`c)root{3}{4}.root{3}{1-sqrt3}.root{6}{(sqrt3+1)^2}`
`=root{3}{4(1-sqrt3)}.root{3}{1+sqrt3}`
`=root{3}{4(1-sqrt3)(1+sqrt3)}`
`=root{3}{4(1-3)}=-2`
`d)2/(root{3}{3}-1)-4/(root{9}-root{3}{3}+1)`
`=(2(root{3}{9}+root{3}{3}+1))/(3-1)-(4(root{3}{3}+1))/(3+1)`
`=root{3}{9}+root{3}{3}+1-root{3}{3}-1`
`=root{3}{9}`
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`a)root{3}{8sqrt5-16}.root{3}{8sqrt5+16}`
`=root{3}{(8sqrt5-16)(8sqrt5+16)}`
`=root{3}{320-256}`
`=root{3}{64}=4`
`b)root{3}{7-5sqrt2}-root{6}{8}`
`=root{3}{1-3.sqrt{2}+3.2.1-2sqrt2}-root{6}{(2)^3}`
`=root{3}{(1-sqrt2)^3}-sqrt2`
`=1-sqrt2-sqrt2=1-2sqrt2`
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THỰC HIỆN PHÉP TÍNH1,sqrt{3+sqrt{5}}.sqrt{2}2,sqrt{3-sqrt{5}.sqrt{8}}3,(sqrt{dfrac{3}{4}}-sqrt{3}+5sqrt{dfrac{4}{3})}.sqrt{12}4,(sqrt{dfrac{1}{7}}-sqrt{dfrac{16}{7}}+sqrt{7}):sqrt{7}5, sqrt{36-12sqrt{5}}:sqrt{6}6,sqrt{3-sqrt{5}:}sqrt{2}
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THỰC HIỆN PHÉP TÍNH
1,\(\sqrt{3+\sqrt{5}}.\sqrt{2}\)
2,\(\sqrt{3-\sqrt{5}.\sqrt{8}}\)
3,\((\sqrt{\dfrac{3}{4}}-\sqrt{3}+5\sqrt{\dfrac{4}{3})}.\sqrt{12}\)
4,\((\sqrt{\dfrac{1}{7}}-\sqrt{\dfrac{16}{7}}+\sqrt{7}):\sqrt{7}\)
5, \(\sqrt{36-12\sqrt{5}}:\sqrt{6}\)
6,\(\sqrt{3-\sqrt{5}:}\sqrt{2}\)
1: \(\sqrt{3+\sqrt{5}}\cdot\sqrt{2}=\sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)
3) \(\left(\sqrt{\dfrac{3}{4}}-\sqrt{3}+5\cdot\sqrt{\dfrac{4}{3}}\right)\cdot\sqrt{12}\)
\(=\left(\dfrac{\sqrt{3}}{2}-\dfrac{2\sqrt{3}}{2}+5\cdot\dfrac{2}{\sqrt{3}}\right)\cdot\sqrt{12}\)
\(=\dfrac{17\sqrt{3}}{6}\cdot2\sqrt{3}\)
\(=\dfrac{34\cdot3}{6}=\dfrac{102}{6}=17\)
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16 tháng 11 2021 lúc 15:47
rút gọn!!!
\(\sqrt{(1-\sqrt{5})^2} - \sqrt{16 - 4\sqrt{5}}\)
Tính:
1) \(\sqrt{14-2\sqrt{33}}\)
2) \(\sqrt{12-2\sqrt{35}}\)
3) \(\sqrt{16-2\sqrt{55}}\)
4) \(\sqrt{14-6\sqrt{5}}\)
5) \(\sqrt{17-12\sqrt{2}}\)
6) \(\sqrt{27-12\sqrt{5}}\)
7) \(\sqrt{4+\sqrt{15}}\)
LÀM CHI TIẾT GIÚP MK NHÉ!
1)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}=\sqrt{11}-\sqrt{3}\)
2)
\(=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}=\sqrt{7}-\sqrt{5}\)
3)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{11}-\sqrt{5}\right)}=\sqrt{11}-\sqrt{5}\)
4)
\(=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)
5)
\(=\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}=\sqrt{\left(3-2\sqrt{2}\right)^2}=3-2\sqrt{2}\)
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a : \(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
b : \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
c : \(\sqrt{\left(2\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
d : \(\sqrt{52-16\sqrt{3}}+\sqrt{\left(4\sqrt{3}-7\right)^2}\)
a.
$A=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}$
$A\sqrt{2}=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}$
$A\sqrt{2}=\sqrt{(\sqrt{3}-1)^2}+\sqrt{(\sqrt{3}+1)^2}$
$=|\sqrt{3}-1|+|\sqrt{3}+1|=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}$
$\Rightarrow A=2\sqrt{3}: \sqrt{2}=\sqrt{6}$
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$B=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}$
$B\sqrt{2}=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}$
$B\sqrt{2}=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}$
$=|\sqrt{7}-1|-|\sqrt{7}+1|=\sqrt{7}-1-(\sqrt{7}+1)=-2$
$\Rightarrow B=-2:\sqrt{2}=-\sqrt{2}$
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\(a,\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(A-\sqrt{2}=\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)\cdot\sqrt{2}\\ =\sqrt{2-\sqrt{3}}\cdot\sqrt{2}-\sqrt{2+\sqrt{3}}\cdot\sqrt{2}\\ =\sqrt{\left(2-\sqrt{3}\right)\cdot2}-\sqrt{\left(2+\sqrt{3}\right)\cdot2}\\ =\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\\ =\sqrt{3-2\sqrt{3}+1}-\sqrt{3+2\sqrt{3}+1}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\\ =\left|\sqrt{3}-1\right|-\left|\sqrt{3}+1\right|\\ =\sqrt{3}-1-\sqrt{3}-1\\ =-2\)
Ta có :
\(A-\sqrt{2}=-2\\ \Leftrightarrow A=\dfrac{-2}{\sqrt{2}}=\dfrac{-\left(\sqrt{2}\right)^2}{\sqrt{2}}=-\sqrt{2}\)
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C làm giống câu a, nhé.
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\(\sqrt{\left(2\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\\ =\left|2\sqrt{5}+1\right|-\left|\sqrt{5}-2\right|\\ =2\sqrt{5}+1-\sqrt{5}+2\\ =3+\sqrt{5}\)
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\(\sqrt{52-16\sqrt{3}}+\sqrt{\left(4\sqrt{3}-7\right)^2}\\ =\sqrt{48-2\cdot4\cdot\sqrt{3}\cdot2+4}+\left|4\sqrt{3}-7\right|\\ =\sqrt{\left(4\sqrt{3}\right)^2-2\cdot4\cdot\sqrt{3}\cdot2+2^2}+4\sqrt{3}-7\\ =\sqrt{\left(4\sqrt{3}-2\right)^2}+4\sqrt{3}-7\\ =4\sqrt{3}-2+4\sqrt{3}-7\\ =8\sqrt{3}-9\)
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c.
$C=\sqrt{(2\sqrt{5}+1)^2}-\sqrt{(\sqrt{5}-2)^2}$
$=|2\sqrt{5}+1|-|\sqrt{5}-2|=2\sqrt{5}+1-(\sqrt{5}-2)=\sqrt{5}+3$
d.
$D=\sqrt{52-16\sqrt{3}}+\sqrt{4\sqrt{3}-7)^2}$
$=\sqrt{(4\sqrt{3})^2-2.4\sqrt{3}.2+2^2}+|4\sqrt{3}-7|$
$=\sqrt{(4\sqrt{3}-2)^2}+|4\sqrt{3}-7|$
$=|4\sqrt{3}-2|+|4\sqrt{3}-7|$
$=4\sqrt{3}-2+7-4\sqrt{3}=5$
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a) \(\dfrac{2}{5}\sqrt{25}\) -\(\dfrac{1}{2}\sqrt{4}\) b)0,5\(\sqrt{0,09}\) +5\(\sqrt{0,81}\) c)\(\dfrac{2}{5}\sqrt{\dfrac{25}{36}}\) -\(\dfrac{5}{2}\sqrt{\dfrac{4}{25}}\)
d)-2\(\sqrt{\dfrac{-36}{-16}}\) + 5\(\sqrt{\dfrac{-81}{-25}}\)
`#3107.101107`
a)
`2/5 \sqrt{25} - 1/2 \sqrt{4}`
`= 2/5 * \sqrt{5^2} - 1/2 * \sqrt{2^2}`
`= 2/5*5 - 1/2*2`
`= 2 - 1`
`= 1`
b)
`0,5*\sqrt{0,09} + 5*\sqrt{0,81}`
`= 0,5*\sqrt{(0,3)^2} + 5*\sqrt{(0,9)^2}`
`= 0,5*0,3 + 5*0,9`
`= 0,15 + 4,5`
`= 4,65`
c)
`2/5\sqrt{25/36} - 5/2\sqrt{4/25}`
`= 2/5*\sqrt{(5^2)/(6^2)} - 5/2*\sqrt{(2^2)/(5^2)}`
`= 2/5*5/6 - 5/2*2/5`
`= 1/3 - 1`
`= -2/3`
d)
`-2 \sqrt{(-36)/(-16)} + 5 \sqrt{(-81)/(-25)}`
`= -2*\sqrt{36/16} + 5*\sqrt{81/25}`
`= -2*\sqrt{(6^2)/(4^2)} + 5*\sqrt{(9^2)/(5^2)}`
`= -2*6/4 + 5*9/5`
`= -3 + 9`
`= 6`
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Bài 1: Tìm x, biết
a)\(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)
b) \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
c)\(\sqrt{16x-16}-\sqrt{9x-9}+\sqrt{4x-4}+\sqrt{x-1}=8\)
d) \(\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}=2\)
a) Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)
\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)
\(\Leftrightarrow4\sqrt{x-3}=20\)
\(\Leftrightarrow x-3=25\)
hay x=28
b) Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)
\(\Leftrightarrow2\sqrt{x+2}=6\)
\(\Leftrightarrow x+2=9\)
hay x=7
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giải pt
a.\(\sqrt{x^2-4x+4}=5\)
b.\(\sqrt{16x+16}-3\sqrt{x+1}+\sqrt{4x+4}=16-\sqrt{x+1}\)
Lời giải:
a. ĐKXĐ: $x\in\mathbb{R}$
PT $\Leftrightarrow \sqrt{(x-2)^2}=5$
$\Leftrightarrow |x-2|=5$
$\Leftrightarrow x-2=5$ hoặc $x-2=-5$
$\Leftrightarrow x=7$ hoặc $x=-3$ (đều tm)
b. ĐKXĐ: $x\geq -1$
PT $\Leftrightarrow \sqrt{16}.\sqrt{x+1}-3\sqrt{x+1}+\sqrt{4}.\sqrt{x+1}=16-\sqrt{x+1}$
$\Leftrightarrow 4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}$
$\Leftrightarrow 4\sqrt{x+1}=16$
$\Leftrightarrow \sqrt{x+1}=4$
$\Leftrightarrow x+1=16$
$\Leftrightarrow x=15$ (tm)
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