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A
2
Y
24 tháng 6 2023
\(1,\sqrt{432}-\sqrt{363}+\sqrt{48}-\sqrt{75}+\sqrt{108}-\sqrt{147}\)
\(=\sqrt{12^2.3}-\sqrt{11^2.3}+\sqrt{4^2.3}-\sqrt{5^2.3}+\sqrt{6^2.3}-\sqrt{7^2.3}\)
\(=12\sqrt{3}-11\sqrt{3}+4\sqrt{3}-5\sqrt{3}+6\sqrt{3}-7\sqrt{3}\)
\(=\sqrt{3}.\left(12-11+4-5+6-7\right)\)
\(=-\sqrt{3}\)
\(2,6\sqrt{60}-5\sqrt{8}+3\sqrt{15}+4\sqrt{32}+3\sqrt{128}-2\sqrt{1250}\)
\(=6.2\sqrt{15}-5.2\sqrt{2}+3\sqrt{15}+4.4\sqrt{2}+3.8\sqrt{2}-2.25\sqrt{2}\)
\(=12\sqrt{15}+3\sqrt{15}-10\sqrt{2}+16\sqrt{2}+24\sqrt{2}-50\sqrt{2}\)
\(=\sqrt{15}.\left(12+3\right)+\sqrt{2}.\left(-10+16+24-50\right)\)
\(=15\sqrt{15}-20\sqrt{2}\)
H9
HT.Phong (9A5)
CTVHS
24 tháng 6 2023
1/ \(\sqrt{432}-\sqrt{363}+\sqrt{48}-\sqrt{75}+\sqrt{108}-\sqrt{147}\)
\(=12\sqrt{3}-11\sqrt{3}+4\sqrt{3}-5\sqrt{3}+6\sqrt{3}-7\sqrt{3}\)
\(=\left(12-11+4-5+6-7\right)\sqrt{3}\)
\(=-\sqrt{3}\)
2/ \(6\sqrt{60}-5\sqrt{8}+3\sqrt{15}+4\sqrt{32}+3\sqrt{128}-2\sqrt{1250}\)
\(=12\sqrt{15}-10\sqrt{2}+3\sqrt{15}+16\sqrt{2}+24\sqrt{2}-50\sqrt{2}\)
\(=\left(12+3\right)\sqrt{15}+\left(-10+16+24-50\right)\sqrt{2}\)
\(=15\sqrt{15}-20\sqrt{2}\)
2 tháng 10 2021
a) \(\sqrt{\left(2x-3\right)^2}=7\)
\(\Leftrightarrow\left|2x-3\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
b) \(\sqrt{64x+128}-\sqrt{25x+50}+\sqrt{4x+8}=20\left(đk:x\ge-2\right)\)
\(\Leftrightarrow8\sqrt{x+2}-5\sqrt{x+2}+2\sqrt{x+2}=20\)
\(\Leftrightarrow5\sqrt{x+2}=20\)
\(\Leftrightarrow\sqrt{x+2}=4\Leftrightarrow x+2=16\Leftrightarrow x=14\left(tm\right)\)
c) \(\sqrt{x^2-9}-3\sqrt{x-3}=0\left(đk:x\ge3\right)\)
\(\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\\sqrt{x+3}=3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x+3=9\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
HP
2 tháng 10 2021
a. \(\sqrt{\left(2x-3\right)^2}=7\)
<=> \(\left|2x-3\right|=7\)
<=> \(\left[{}\begin{matrix}2x-3=7\left(x\ge\dfrac{3}{2}\right)\\-2x+3=7\left(x< \dfrac{3}{2}\right)\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}2x=10\\-2x=4\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=5\left(TM\right)\\x=-2\left(TM\right)\end{matrix}\right.\)
b. \(\sqrt{64x+128}-\sqrt{25x+50}+\sqrt{4x+8}=20\) ĐK: \(x\ge-2\)
<=> \(\sqrt{64\left(x+2\right)}-\sqrt{25\left(x+2\right)}+\sqrt{4\left(x+2\right)}-20=0\)
<=> \(8\sqrt{x+2}-5\sqrt{x+2}+2\sqrt{x+2}-20=0\)
<=> \(\sqrt{x+2}.\left(8-5+2\right)-20=0\)
<=> \(5\sqrt{x+2}=20\)
<=> \(\sqrt{x+2}=4\)
<=> \(\left(\sqrt{x+2}\right)^2=4^2\)
<=> \(\left|x+2\right|=16\)
<=> \(\left[{}\begin{matrix}x+2=16\left(x\ge-2\right)\\x+2=-16\left(x< -2\right)\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=14\left(TM\right)\\x=-18\left(TM\right)\end{matrix}\right.\)
c. \(\sqrt{x^2-9}-3\sqrt{x-3}=0\) ĐK: \(x\ge3\)
<=> \(\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)
<=> \(\sqrt{x-3}.\sqrt{x+3}-3\sqrt{x-3}=0\)
<=> \(\left(\sqrt{x+3}-3\right).\sqrt{x-3}=0\)
<=> \(\left[{}\begin{matrix}\sqrt{x+3}-3=0\\\sqrt{x-3}=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=6\\x=3\end{matrix}\right.\)
5 tháng 7 2021
Bài 1 :
a, ĐKXĐ : \(\dfrac{2x+1}{x^2+1}\ge0\)
Mà \(x^2+1\ge1>0\)
\(\Rightarrow2x+1\ge0\)
\(\Rightarrow x\ge-\dfrac{1}{2}\)
Vậy ...
b, Ta có : \(\sqrt[3]{-27}+\sqrt[3]{64}-\sqrt[3]{-\dfrac{128}{2}}\)
\(=-3+4-\left(-4\right)=-3+4+4=5\)
5 tháng 7 2021
Bài 2 :
\(a,=2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}\)
\(=\sqrt{5}\left(2+6+5-12\right)=\sqrt{2}\)
\(b,=\sqrt{5}+\sqrt{5}+\left|\sqrt{5}-2\right|\)
\(=2\sqrt{5}+\sqrt{5}-2=3\sqrt{5}-2\)
\(c,=\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)
\(=\dfrac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\)
\(=3\)
NT
2
27 tháng 5 2017
\(A=2\sqrt{8}-3\sqrt{18}+4\sqrt{128}-5\sqrt{32}\)
\(A=2\sqrt{4.2}-3\sqrt{9.2}+4\sqrt{64.2}-5\sqrt{16.2}\)
\(A=4\sqrt{2}-9\sqrt{2}+32\sqrt{2}-20\sqrt{2}\)
\(A=7\sqrt{2}\)
NT
0
MN
2
H9
HT.Phong (9A5)
CTVHS
27 tháng 7 2023
a) \(\sqrt{128\left(x-y\right)^2}\)
\(=\sqrt{8^2\cdot2\left(x-y\right)^2}\)
\(=\left|8\left(x-y\right)\right|\sqrt{2}\)
\(=8\left|\left(x-y\right)\right|\sqrt{2}\)
b) \(\sqrt{150\left(4x^2-4x+1\right)}\)
\(=\sqrt{5^2\cdot6\left(2x-1\right)^2}\)
\(=\left|5\left(2x-1\right)\right|\sqrt{6}\)
\(=5\left|2x-1\right|\sqrt{6}\)
c) \(\sqrt{x^3-6x^2+12x-8}\)
\(=\sqrt{\left(x-2\right)^3}\)
\(=\sqrt{\left(x-2\right)^2\left(x-2\right)}\)
\(=\left|x-2\right|\sqrt{x-2}\)
27 tháng 7 2023
a: \(=\sqrt{64\cdot2\cdot\left(x-y\right)^2}=8\sqrt{2}\cdot\left|x-y\right|\)
b; \(=\sqrt{25\cdot6\left(2x-1\right)^2}=5\sqrt{6}\cdot\left|2x-1\right|\)
c: \(=\sqrt{\left(x-2\right)^3}=\left|x-2\right|\cdot\sqrt{x-2}\)
V
1
9 tháng 12 2018
\(5\sqrt{\dfrac{1}{5}}-\dfrac{8}{1+\sqrt{5}}-\dfrac{\sqrt{20}-5}{2-\sqrt{5}}=\dfrac{5\sqrt{5}}{5}-\dfrac{8\left(\sqrt{5}-1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}-\dfrac{\sqrt{5}\left(2-\sqrt{5}\right)}{2-\sqrt{5}}=\sqrt{5}-\dfrac{8\left(\sqrt{5}-1\right)}{4}-\sqrt{5}=-\dfrac{8\left(\sqrt{5}-1\right)}{4}=-2\left(\sqrt{5}-1\right)=2-2\sqrt{5}\)