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21 tháng 6 2017

MATH ERROR BN ƠI ! LÀM GÌ TÍNH ĐƯỢC HIHI ! ! !s2.jpg DarkEvil HK Huy

21 tháng 6 2017

lam gi ma tinh duoc 

25 tháng 9 2021

\(1,=20-7=13\\ b,=12-50=-38\\ c,=\sqrt{7}-2+\sqrt{7}+2=2\sqrt{7}\\ d,=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\\ e,=11+2\sqrt{30}\\ f,=8-2\sqrt{15}\\ g,=11+2\sqrt{6}\)

25 tháng 9 2021

1) \(=\left(2\sqrt{5}\right)^2-\left(\sqrt{7}\right)^2=20-7=13\)

2) \(=\left(2\sqrt{3}\right)^2-\left(5\sqrt{2}\right)^2=12-50=-38\)

3) \(=\sqrt{7}-2+\sqrt{7}+2=2\sqrt[]{7}\)

4) \(=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\)

5) \(=5+6-2\sqrt{5.6}=11-2\sqrt{30}\)

6) \(=3+5-2\sqrt{3.5}=8-4\sqrt{2}\)

7) \(=\left(2\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+2\sqrt{2\sqrt{2}.3}=11+2\sqrt{6\sqrt{2}}\)

14 tháng 10 2019

B = \(\frac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\frac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\)

=>  \(\frac{2}{\sqrt{2}}B=\frac{8+2\sqrt{7}}{6+\sqrt{8+2\sqrt{7}}}+\frac{8-2\sqrt{7}}{6-\sqrt{8-2\sqrt{7}}}\)

=> \(\frac{2}{\sqrt{2}}B=\frac{\left(\sqrt{7}+1\right)^2}{6+\sqrt{7}+1}+\frac{\left(\sqrt{7}-1\right)^2}{6-\sqrt{7}+1}\)

=> \(\frac{2}{\sqrt{2}}B=\frac{\left(\sqrt{7}+1\right)^2}{\sqrt{7}\left(\sqrt{7}+1\right)}+\frac{\left(\sqrt{7}-1\right)^2}{\sqrt{7}\left(\sqrt{7}-1\right)}\)

=> \(\frac{2}{\sqrt{2}}B=\frac{\sqrt{7}+1}{\sqrt{7}}+\frac{\sqrt{7}-1}{\sqrt{7}}=\frac{2\sqrt{7}}{\sqrt{7}}=2\)

=> B = \(\sqrt{2}\)

19 tháng 9 2023

\(A=\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)

\(A=\dfrac{\sqrt{2}\cdot\left(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\right)}{\sqrt{2}}\)

\(A=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{12+6\sqrt{7}}}{\sqrt{\text{2}}}\)

\(A=\dfrac{\sqrt{21-2\cdot\sqrt{21}\cdot\sqrt{3}+3}-\sqrt{21+2\cdot\sqrt{21}\cdot\sqrt{3}+3}}{\sqrt{2}}\)

\(A=\dfrac{\sqrt{\left(\sqrt{21}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{21}+\sqrt{3}\right)^2}}{\sqrt{2}}\)

\(A=\dfrac{\left|\sqrt{21}-\sqrt{3}\right|-\left|\sqrt{21}+\sqrt{3}\right|}{\sqrt{2}}\)

\(A=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{\text{2}}}\)

\(A=\dfrac{-\sqrt{6}}{\sqrt{2}}\)

\(A=-\sqrt{\dfrac{6}{2}}\)

\(A=-\sqrt{3}\)

19 tháng 9 2023

\(A=\sqrt[]{12-3\sqrt[]{7}}-\sqrt[]{12+3\sqrt[]{7}}\)

Giả sử \(\sqrt[]{12-3\sqrt[]{7}}-\sqrt[]{12+3\sqrt[]{7}}>0\)

\(\Leftrightarrow\sqrt[]{12-3\sqrt[]{7}}>\sqrt[]{12+3\sqrt[]{7}}\)

\(\Leftrightarrow12-3\sqrt[]{7}>12+3\sqrt[]{7}\)

\(\Leftrightarrow6\sqrt[]{7}< 0\left(sai\right)\)

Vậy \(\sqrt[]{12-3\sqrt[]{7}}-\sqrt[]{12+3\sqrt[]{7}}< 0\) hay \(A< 0\)

\(\Leftrightarrow A^2=12-3\sqrt[]{7}+12+3\sqrt[]{7}-2\sqrt[]{\left(12-3\sqrt[]{7}\right)\left(12+3\sqrt[]{7}\right)}\)

\(\Leftrightarrow A^2=24-2\sqrt[]{\left(144-63\right)}\)

\(\Leftrightarrow A^2=24-2\sqrt[]{81}\)

\(\Leftrightarrow A^2=24-18=6\)

\(\Leftrightarrow A=-\sqrt[]{6}\)

17 tháng 10 2023

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14 tháng 9 2021

\(\sqrt{7-2\sqrt{12}}=\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)

=> Chọn C

19 tháng 10 2016

a, =\(9\sqrt{2}\)

b, =21

21 tháng 9 2018

a) \(=9\sqrt{2}\)

b) \(=21\)

học tốt.

23 tháng 9 2023

\(a,\dfrac{3}{\sqrt{7}-4}+\dfrac{4+\sqrt{7}}{3}\)

\(=\dfrac{9}{3\left(\sqrt{7}-4\right)}+\dfrac{\left(\sqrt{7}-4\right)\left(\sqrt{7}+4\right)}{3\left(\sqrt{7}-4\right)}\)

\(=\dfrac{9+7-16}{3\left(\sqrt{7}-4\right)}\)

\(=0\)

\(b,\left(\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}\right):\dfrac{1}{2\sqrt{3}}\)

\(=\left[\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}\right]\cdot2\sqrt{3}\)

\(=\left(\sqrt{2}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}\right)\cdot2\sqrt{3}\)

\(=\left(\sqrt{2}+\sqrt{3}-\sqrt{2}\right)\cdot2\sqrt{3}\)

\(=\sqrt{3}\cdot2\sqrt{3}\)

\(=6\)

#\(Toru\)

a: \(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{3}+1+\sqrt{3}-1\right)=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

b: \(\dfrac{1}{\sqrt{7-\sqrt{24}}+1}-\dfrac{1}{\sqrt{7+\sqrt{24}}+1}\)

\(=\dfrac{1}{\sqrt{6}-1+1}-\dfrac{1}{\sqrt{6}+1+1}\)

\(=\dfrac{1}{\sqrt{6}}-\dfrac{1}{\sqrt{6}+2}=\dfrac{\sqrt{6}+2-\sqrt{6}}{\sqrt{6}\left(\sqrt{6}+2\right)}\)

\(=\dfrac{2}{\sqrt{6}\left(\sqrt{6}+2\right)}=\dfrac{2}{6+2\sqrt{6}}=\dfrac{1}{3+\sqrt{6}}=\dfrac{3-\sqrt{6}}{3}\)