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24 tháng 12 2021

Câu 12: Ko có câu nào đúng

a: \(=\left(2\sqrt{3}-12\sqrt{3}+15\sqrt{3}\right)\cdot\sqrt{3}=5\sqrt{3}\cdot\sqrt{3}=15\)

b: \(=\left(6\sqrt{2}-16\sqrt{2}+15\sqrt{2}\right):5=\sqrt{2}\)

c: \(=\dfrac{\left(2\sqrt{5}-6\sqrt{5}+15\sqrt{5}\right)}{\sqrt{5}}=17-6=11\)

6 tháng 7 2021

a) \(\left(\sqrt{125}+\sqrt{45}-2\sqrt{80}\right).\sqrt{5}=\left(5\sqrt{5}+3\sqrt{5}-8\sqrt{5}\right).\sqrt{5}\)

\(=0.\sqrt{5}=0\)

b) \(\frac{5-2\sqrt{6}}{\sqrt{2}-\sqrt{3}}=\frac{\left(5-2\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}=\frac{\left(5\sqrt{2}+5\sqrt{3}-4\sqrt{3}-6\sqrt{2}\right)}{-1}\)

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

6 tháng 7 2021

a,\(\left(\sqrt{125}+\sqrt{45}-2\sqrt{80}\right).\sqrt{5}\)

\(=\left(5\sqrt{5}+3\sqrt{5}-8\sqrt{5}\right).\sqrt{5}\)

\(=0.\sqrt{5}\)

\(=0\)

b,\(\frac{5-2\sqrt{6}}{\sqrt{2}-\sqrt{3}}\)

\(=\frac{\left(5-2\sqrt{6}\right).\left(\sqrt{2}+\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right).\left(\sqrt{2}+\sqrt{3}\right)}\)

\(=\frac{\sqrt{3}-\sqrt{2}}{-1}\)

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

\(A=5\sqrt{\dfrac{1}{5}}+\dfrac{5}{2}\cdot\sqrt{20}-\sqrt{80}\)

\(=\dfrac{5}{\sqrt{5}}+\dfrac{5}{2}\cdot2\sqrt{5}-4\sqrt{5}\)

\(=\sqrt{5}+5\sqrt{5}-4\sqrt{5}=2\sqrt{5}\)

10 tháng 8 2018

*\(A=2\sqrt{80\sqrt{7}}-2\sqrt{45\sqrt{7}}-5\sqrt{20\sqrt{7}}\)

\(A=16\sqrt{5\sqrt{7}}-6\sqrt{5\sqrt{7}}-10\sqrt{5\sqrt{7}}\)

\(A=\left(16-6-10\right)\sqrt{5\sqrt{7}}=0\)

* \(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)

\(B^3=5+2\sqrt{13}+5-2\sqrt{13}+3\left(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\right).\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}\)

\(B^3=10-9B\)

\(\Rightarrow B^3+9B-10=0\)

\(\Rightarrow B^3-B^2+B^2-B+10B-10=0\)

\(\Rightarrow B^2\left(B-1\right)+B\left(B-1\right)+10\left(B-1\right)=0\)

\(\Rightarrow\left(B-1\right)\left(B^2+B+10\right)=0\)

\(\Rightarrow B=1\)

30 tháng 6 2021

30 tháng 6 2021

T = \(\dfrac{\sqrt{5}\left(\sqrt{16}-\sqrt{9}\right)}{4-5}-5\sqrt{5}+\dfrac{1}{\sqrt{5}-2}+2\sqrt{5}\)

   = \(-\sqrt{5}-5\sqrt{5}+2\sqrt{5}+\dfrac{1}{\sqrt{5}-2}\)

   = \(-4\sqrt{5}+\dfrac{1}{\sqrt{5}-2}\)

   = \(\dfrac{-4\sqrt{5}\left(\sqrt{5}-2\right)+1}{\sqrt{5}-2}\)

   = \(\dfrac{-20+8\sqrt{5}+1}{\sqrt{5}-2}\)

   = \(\dfrac{-19+8\sqrt{5}}{\sqrt{5}-2}\)

   = \(\dfrac{19-8\sqrt{5}}{2-\sqrt{5}}\)

   = \(\dfrac{\left(-2+3\sqrt{5}\right)\left(\sqrt{5}-2\right)}{-\left(\sqrt{5}-2\right)}=2-3\sqrt{5}\)

17 tháng 5 2021

`A)đk:x>=0,x ne 25`

`A=9=>A=(3+2)/(3-5)=-5/2`

`B)B=(3sqrtx-15+20-2sqrtx)/(x-25)`

`=(sqrtx+5)/(x-25)`

`=1/(sqrtx-5)`

`A=B.|x-4|`

`<=>A/B=|x-4|`

`<=>\sqrtx+2=|x-4|`

`<=>\sqrtx+2=(sqrtx+2)|sqrtx-2|`

`<=>|sqrtx-2|=1`

`+)sqrtx-2=1<=>x=9(tm)`

`+)sqrtx-2=-1<=>x=1(tm)`

Vậy `S={1,9}`

17 tháng 5 2021

a, Thay x=9 vào biểu thức A ta có

\(A=\dfrac{\sqrt{9}+2}{\sqrt{9}-5}\)

\(A=\dfrac{3+2}{3-5}=\dfrac{5}{-2}=-2,5\)

Vậy A =-2,5 khi x=9

17 tháng 1 2018

Chọn đáp án A

22 tháng 6 2023

\(I=\left(2\sqrt{3}-5\sqrt{27}+4\sqrt{12}\right):\sqrt{3}\)

\(=\left(2\sqrt{3}-5\sqrt{3}.\sqrt{3^2}+2\sqrt{2^2}.\sqrt{3}\right):\sqrt{3}\)

\(=\left(2\sqrt{3}-15\sqrt{3}+8\sqrt{3}\right):\sqrt{3}\)

\(=-5\sqrt{3}.\dfrac{1}{\sqrt{3}}\)

\(=-5\)

\(K=\sqrt{125}-4\sqrt{45}+3\sqrt{20}-\sqrt{80}\)

\(=\sqrt{5^2.5}-4\sqrt{3^2.5}+3\sqrt{2^2.5}-\sqrt{4^2.5}\)

\(=5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}\)

\(=\sqrt{5}.\left(5-12+6-4\right)\)

\(=-5\sqrt{5}\)

\(L=2\sqrt{9}+\sqrt{25}-5\sqrt{4}\)

\(=2\sqrt{3^2}+\sqrt{5^2}-5\sqrt{2^2}\)

\(=2.3+5-5.2\)

\(=1\)

\(N=2\sqrt{32}-5\sqrt{27}-4\sqrt{8}+3\sqrt{75}\)

\(=2.4\sqrt{2}-5.3\sqrt{3}-4.2\sqrt{2}+3.5\sqrt{3}\)

\(=8\sqrt{2}-8\sqrt{2}-15\sqrt{3}+15\sqrt{3}\)

\(=0\)

\(O=2\sqrt{3.5^2}-3\sqrt{3.2^2}+\sqrt{3.3^2}\)

\(=2.5\sqrt{3}-3.2\sqrt{3}+3\sqrt{3}\)

\(=10\sqrt{3}-6\sqrt{3}+3\sqrt{3}\)

\(=7\sqrt{3}\)

\(L=\dfrac{2\sqrt{3}-15\sqrt{3}+8\sqrt{3}}{\sqrt{3}}=2-15+8=-5\)

\(K=5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}=-5\sqrt{5}\)

L=2*3+5-5*2=5-4=1

N=8căn 2-8căn2-15căn3+15căn 3=0

O=10căn 3-6căn3+3căn3=7căn 3

4 tháng 11 2021

\(a,=3\sqrt{5}-2\sqrt{5}-\sqrt{5}+5\sqrt{5}=5\sqrt{5}\\ b,=9\sqrt{a}-6\sqrt{a}-\sqrt{a}=2\sqrt{a}\\ c,Sửa:3\sqrt[3]{27}-3\sqrt[3]{-8}-3\sqrt[3]{-125}\\ =3\cdot3-3\left(-2\right)-3\left(-5\right)\\ =9+6+15=30\)