1) Giải
\(\sqrt{7+\sqrt{33}}+\sqrt{7-\sqrt{33}}\)
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NH
1
20 tháng 1 2024
\(\sqrt{3x^2+33}+3\sqrt{x}=2x+7\)(ĐKXĐ: x>=0)
=>\(\sqrt{3x^2+33}-6+3\sqrt{x}-3=2x-2\)
=>\(\dfrac{3x^2+33-36}{\sqrt{3x^2+33}+6}+3\left(\sqrt{x}-1\right)=2\left(x-1\right)\)
=>\(\dfrac{3x^2-3}{\sqrt{3x^2+33}+6}+3\left(\sqrt{x}-1\right)-2\left(x-1\right)=0\)
=>\(\dfrac{3\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x+1\right)}{\sqrt{3x^2+33}+6}+3\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\)
=>\(\left(\sqrt{x}-1\right)\left(\dfrac{3\left(\sqrt{x}+1\right)\left(x+1\right)}{\sqrt{3x^2+33}+6}+3-2\left(\sqrt{x}+1\right)\right)=0\)
=>\(\sqrt{x}-1=0\)
=>x=1(nhận)
KT
26 tháng 7 2018
\(A=\sqrt{\frac{5+\sqrt{21}}{5-\sqrt{21}}}+\sqrt{\frac{5-\sqrt{21}}{5+\sqrt{21}}}\)
\(=\sqrt{\frac{\left(5+\sqrt{21}\right)^2}{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}}+\sqrt{\frac{\left(5-\sqrt{21}\right)^2}{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}}\)
\(=\sqrt{\frac{\left(5+\sqrt{21}\right)^2}{4}}+\sqrt{\frac{\left(5-\sqrt{21}\right)^2}{4}}\)
\(=\frac{5+\sqrt{21}}{2}+\frac{5-\sqrt{21}}{2}=5\)
\(B=\sqrt{7+\sqrt{33}}+\sqrt{7-\sqrt{33}}\)
\(\Rightarrow\)\(\sqrt{2}B=\sqrt{14+2\sqrt{33}}+\sqrt{14-2\sqrt{33}}\)
\(=\sqrt{\left(\sqrt{11}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}\)
\(=\sqrt{11}+\sqrt{3}+\sqrt{11}-\sqrt{3}=2\sqrt{11}\)
\(\Rightarrow\)\(B=\sqrt{22}\)
13 tháng 5 2022
a: \(A=\left(1-\sqrt{7}\right)\cdot\left(1+\sqrt{7}\right)=1-7=-6\)
b: \(B=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}=-4\sqrt{3}\)
c: \(C=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)
VT
7 tháng 11 2017
\(A=\sqrt{5+\sqrt{21}}+\sqrt{5-\sqrt{21}}\)
\(A^2=\left(\sqrt{5+\sqrt{21}}^2+2\sqrt{\left(5+\sqrt{21}\right)\left(5-\sqrt{21}\right)}+\sqrt{5-\sqrt{21}}^2\right)\)
\(A^2=5+\sqrt{21}+\sqrt{4\left(5+\sqrt{21}\right)\left(5-\sqrt{21}\right)}+5-\sqrt{21}\)
\(A^2=10+\sqrt{4.\left(25-21\right)}\)
\(A^2=10+\sqrt{4.4}=10+4=14\)
\(A=\sqrt{14}\)
Tương tự
24 tháng 6 2023
g: \(=\left|\sqrt{6}-1\right|=\sqrt{6}-1\)
h: \(=\left|2\sqrt{3}-1\right|=2\sqrt{3}-1\)
l: \(=\left|2-\sqrt{3}\right|-2=2-\sqrt{3}-2=-\sqrt{3}\)
j: \(=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|\)
\(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
6 tháng 11 2021
ĐKXĐ: \(x\ge3\)
\(pt\Leftrightarrow5\sqrt{x-3}+3\sqrt{x-3}-\sqrt{x-3}=7\)
\(\Leftrightarrow7\sqrt{x-3}=7\Leftrightarrow\sqrt{x-3}=1\)
\(\Leftrightarrow x-3=1\Leftrightarrow x=4\left(tm\right)\)
24 tháng 8 2021
a: \(\dfrac{1}{3}\cdot\sqrt{18}-\sqrt{192}-\dfrac{\sqrt{33}}{\sqrt{11}}+3\cdot\sqrt{5\dfrac{1}{3}}\)
\(=\dfrac{1}{3}\cdot3\sqrt{2}-8\sqrt{3}-\sqrt{3}+3\cdot\dfrac{4}{\sqrt{3}}\)
\(=\sqrt{2}-7\sqrt{3}+4\sqrt{3}\)
\(=\sqrt{2}+3\sqrt{3}\)
b: Ta có: \(\sqrt{\left(2\sqrt{3}-5\right)^2}-2\cdot\sqrt{7+4\sqrt{3}}\)
\(=5-2\sqrt{3}-2\cdot\left(2+\sqrt{3}\right)\)
\(=5-2\sqrt{3}-4-2\sqrt{3}\)
\(=-4\sqrt{3}+1\)
1 tháng 6 2023
i: =-12*căn 3/2căn 3=-6
h: =72căn 2/12căn 2=6
g: =25căn 12/5căn 6=5căn 2
f: =(15:5)*căn 6:3=3căn 2
d: =-1/2*6*căn 10=-3căn 10
\(A=\sqrt{7+\sqrt{33}}+\sqrt{7-\sqrt{33}}>0\)
\(\Rightarrow A^2=14+2\sqrt{7^2-33}\)
\(\Rightarrow A^2=14+2\sqrt{16}=22\)
\(\Rightarrow A=\sqrt{22}\)