\(\left(\dfrac{\sqrt{7}-1}{\sqrt{21}-\sqrt{3}}+\dfrac{\sqrt{3}-1}{\sqrt{21}-\sqrt{7}}\right)\times\dfrac{\sqrt{21}}{\sqrt{7}+\sqrt{3}}\)
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2 tháng 6 2021
`a)sqrt{28a^4}`
`=sqrt{7.4.a^4}`
`=2sqrt7a^2`
`b)A=((sqrt{21}-sqrt7)/(sqrt3-1)+(sqrt{10}-sqrt5)/(sqrt2-1)):1/(sqrt7-sqrt5)`
`=((sqrt7(sqrt3-1))/(sqrt3-1)+(sqrt5(sqrt2-1))/(sqrt2-1)).(sqrt7-sqrt5)`
`=(sqrt7+sqrt5)(sqrt7-sqrt5)`
`=7-5=2`
`c)` $\begin{cases}\dfrac{3}{2x}-y=6\\\dfrac{1}{x}+2y=-4\end{cases}$
`<=>` $\begin{cases}\dfrac{3}{x}-2y=12\\\dfrac{1}{x}+2y=-4\end{cases}$
`<=>` $\begin{cases}\dfrac{4}{x}=8\\2y+\dfrac{1}{x}=-4\end{cases}$
`<=>` $\begin{cases}x=\dfrac12\\2y=-4-2=-6\end{cases}$
`<=>` $\begin{cases}x=\dfrac12\\y=-3\end{cases}$
Vậy HPT có nghiệm `(x,y)=(1/2,-3)`.
H9
HT.Phong (9A5)
CTVHS
8 tháng 9 2023
\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)
\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)
\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(5-36\right)\)
\(B=-\left(-31\right)\)
\(B=31\)
_____________________________
\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)
\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)
\(=3\sqrt{3}-\sqrt{3}+1\)
\(=2\sqrt{3}+1\)
7 tháng 11 2017
a)\(\dfrac{\sqrt{21}\left(\sqrt{7}-\sqrt{3}\right)}{\sqrt{7}-\sqrt{3}}+\dfrac{4\left(5+\sqrt{21}\right)}{4}-\dfrac{\sqrt{3}.\sqrt{2}.\sqrt{7}}{\sqrt{3}}\)=\(5+2\sqrt{21}-\sqrt{14}\)
c) (\(\sqrt{2-\sqrt{3}}.\sqrt{2+\sqrt{3}}\))+\(\sqrt{2}\left(\sqrt{2-\sqrt{3}}\right)\)=1+\(\sqrt{2\sqrt{2}-\sqrt{6}}\)
24 tháng 7 2021
Bài 1:
Ta có: \(\left(3\sqrt{50}-5\sqrt{18}+3\sqrt{8}\right)\cdot\sqrt{2}\)
\(=\left(15\sqrt{2}-15\sqrt{2}+6\sqrt{2}\right)\cdot\sqrt{2}\)
\(=6\sqrt{2}\cdot\sqrt{2}\)
=12
24 tháng 7 2021
Bài 2:
1) ĐKXĐ: \(x\le0\)
2) ĐKXĐ: \(x\le2\)
3) ĐKXĐ: \(x>\dfrac{-3}{2}\)
4) ĐKXĐ: x>0
5) ĐKXĐ: x<3
24 tháng 12 2023
a: \(A=\dfrac{1-\dfrac{1}{\sqrt{49}}+\dfrac{1}{49}-\dfrac{1}{\left(7\sqrt{7}\right)^2}}{\dfrac{\sqrt{64}}{2}-\dfrac{4}{7}+\left(\dfrac{2}{7}\right)^2-\dfrac{4}{343}}\)
\(=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}\)
\(=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4\left(1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}\right)}=\dfrac{1}{4}\)
b: \(M=1-\dfrac{5}{\sqrt{196}}-\dfrac{5}{\left(2\sqrt{21}\right)^2}-\dfrac{\sqrt{25}}{204}-\dfrac{\left(\sqrt{5}\right)^2}{374}\)
\(=1-\dfrac{5}{14}-\dfrac{5}{84}-\dfrac{5}{204}-\dfrac{5}{374}\)
\(=1-5\left(\dfrac{1}{14}+\dfrac{1}{84}+\dfrac{1}{204}+\dfrac{1}{374}\right)\)
\(=1-5\left(\dfrac{1}{2\cdot7}+\dfrac{1}{7\cdot12}+\dfrac{1}{12\cdot17}+\dfrac{1}{17\cdot22}\right)\)
\(=1-\left(\dfrac{5}{2\cdot7}+\dfrac{5}{7\cdot12}+\dfrac{5}{12\cdot17}+\dfrac{5}{17\cdot22}\right)\)
\(=1-\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{22}\right)\)
\(=1-\left(\dfrac{1}{2}-\dfrac{1}{22}\right)\)
\(=1-\dfrac{11-1}{22}=1-\dfrac{10}{22}=\dfrac{12}{22}=\dfrac{6}{11}\)
12 tháng 11 2021
5: \(=\dfrac{1}{x-y}\cdot x^3\cdot\left(x-y\right)^2=x^3\left(x-y\right)\)
18 tháng 8 2021
Bài 2:
a: Ta có: \(\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{60}+6\right):2\sqrt{3}\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{12}\left(\sqrt{5}+\sqrt{3}\right):2\sqrt{3}\)
\(=2\sqrt{12}:2\sqrt{3}\)
=2
b: Ta có: \(\sqrt{5-\sqrt{21}}-\sqrt{\dfrac{7}{2}}\)
\(=\dfrac{\sqrt{10-2\sqrt{21}}-\sqrt{7}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}-\sqrt{3}-\sqrt{7}}{\sqrt{2}}\)
\(=-\dfrac{\sqrt{6}}{2}\)
8 tháng 7 2022
a: \(=\dfrac{1}{\sqrt{6}-1+1}-\dfrac{1}{\sqrt{6}+1-1}\)
\(=\dfrac{1}{\sqrt{6}}-\dfrac{1}{\sqrt{6}}\)
=0
b: \(=\dfrac{3+\sqrt{7}-3+\sqrt{7}}{2}=\dfrac{2\sqrt{7}}{2}=\sqrt{7}\)
c: \(=\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\)
\(=3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}=6\sqrt{2}\)
\(=\left[\dfrac{\sqrt{7}-1}{\sqrt{3}\left(\sqrt{7}-1\right)}+\dfrac{\sqrt{3}-1}{\sqrt{7}\left(\sqrt{3}-1\right)}\right]\cdot\dfrac{\sqrt{21}}{\sqrt{7}+\sqrt{3}}\\ =\left(\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{7}}\right)\cdot\dfrac{\sqrt{21}}{\sqrt{7}+\sqrt{3}}\\ =\dfrac{\sqrt{3}+\sqrt{7}}{\sqrt{21}}\cdot\dfrac{\sqrt{21}}{\sqrt{3}+\sqrt{7}}=1\)