HC
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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
28 tháng 6 2021
`a)((sqrt(14)-sqrt7)/(1-sqrt2)+(sqrt{15}-sqrt5)/(1-sqrt3)):1/(sqrt7-sqrt5)`
`=((sqrt7(sqrt2-1))/(1-sqrt2)+(sqrt5(sqrt3-1))/(1-sqrt3)).(sqrt7-sqrt5)`
`=(-sqrt7-sqrt5)*(sqrt7-sqrt5)`
`=-(sqrt7+sqrt5)(sqrt7+sqrt5)`
`=-(7-5)=-2`
`b)sqrt2+1/sqrt{5+2sqrt6}+2/sqrt{8+2sqrt{15}}`
`=sqrt2+1/sqrt{3+2sqrt{3}.sqrt2+2}+2/sqrt{5+2sqrt{5}.sqrt3+3}`
`=sqrt2+1/sqrt{(sqrt3+sqrt2)^2}+2/sqrt{(sqrt5+sqrt3)^2}`
`=sqrt2+1/(sqrt3+sqrt2)+2/(sqrt5+sqrt3)`
`=sqrt2+((sqrt3+sqrt2)(sqrt3-sqrt2))/(sqrt3+sqrt2)+((sqrt5+sqrt3)(sqrt5-sqrt3))/(sqrt5+sqrt3)`
`=sqrt2+sqrt3-sqrt2+sqrt5-sqrt3=sqrt5`
28 tháng 6 2021
a) Ta có: \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
\(=\left(-\dfrac{\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\dfrac{\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
\(=-2\)
b) Ta có: \(\sqrt{2}+\dfrac{1}{\sqrt{5+2\sqrt{6}}}+\dfrac{2}{\sqrt{8+2\sqrt{15}}}\)
\(=\sqrt{2}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{2}{\sqrt{5}+\sqrt{3}}\)
\(=\sqrt{2}+\sqrt{3}-\sqrt{2}+\sqrt{5}-\sqrt{3}\)
\(=\sqrt{5}\)
NP
3
27 tháng 6 2021
`a)sqrt{8-2sqrt7}+sqrt{16-6sqrt7}`
`=sqrt{(sqrt7-1)^2}+sqrt{(3-sqrt7)^2}`
`=sqrt7-1+3-sqrt7=2`
`b)sqrt{(sqrt7-1)^2}-sqrt{11+4sqrt7}`
`=sqrt7-1-sqrt{(2+sqrt7)^2}`
`=sqrt7-1-2-sqrt7=-3`
27 tháng 6 2021
a, \(=\sqrt{7-2\sqrt{7}+1}+\sqrt{7-2.3\sqrt{7}+9}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}+\sqrt{\left(3-\sqrt{7}\right)^2}=\left|\sqrt{7}-1\right|+\left|3-\sqrt{7}\right|\)
\(=\sqrt{7}-1+3-\sqrt{7}=2\)
\(b,=\left|\sqrt{7}-1\right|-\sqrt{7+2.2\sqrt{7}+4}\)
\(=\left|\sqrt{7}-1\right|-\sqrt{\left(\sqrt{7}+2\right)^2}=\left|\sqrt{7}-1\right|-\left|\sqrt{7}+2\right|\)
\(=\sqrt{7}-1-\sqrt{7}-2=-3\)
30 tháng 7 2023
3:
a: \(\sqrt{\dfrac{2}{3}}=\sqrt{\dfrac{6}{9}}=\dfrac{\sqrt{6}}{3}\)
b: \(\dfrac{x}{y}\cdot\sqrt{\dfrac{y}{x}}=\sqrt{\dfrac{x^2}{y^2}\cdot\dfrac{y}{x}}=\sqrt{\dfrac{x}{y}}=\dfrac{\sqrt{xy}}{y}\)
2:
a: 2căn 7=căn 28
3căn 2=căn 18
mà 28>18
nên 2*căn 7>3*căn 2
b: 5=2+3
mà 3>căn 2
nên 2+3>2+căn 2
=>5>2+căn 2
31 tháng 7 2023
1) a) \(\sqrt{98}-\sqrt{72}+0,5\sqrt{8}\)
\(=\sqrt{49.2}-\sqrt{36.2}+0,5\sqrt{4.2}\)
\(=7\sqrt{2}-6\sqrt{2}+0,5.2\sqrt{2}\)
\(=7\sqrt{2}-6\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
b) \(\sqrt{9a}-\sqrt{16a}+\sqrt{49}\)
\(=3\sqrt{a}-4\sqrt{a}+7=7-\sqrt{a}\)
2. a) \(2\sqrt{7}=\sqrt{4.7}=\sqrt{28}\)
\(3\sqrt{2}=\sqrt{9.2}=\sqrt{18}\)
Mà \(\sqrt{28}>\sqrt{18}\Rightarrow2\sqrt{7}>3\sqrt{2}\)
b) \(5=2+3=2+\sqrt{9}\)
Vì \(\sqrt{9}>\sqrt{2}\Rightarrow2+\sqrt{9}>2+\sqrt{2}\Rightarrow5>2+\sqrt{2}\)
3. a) \(\sqrt{\dfrac{2}{3}}=\sqrt{\dfrac{6}{9}}=\dfrac{\sqrt{6}}{3}\)
b) \(\dfrac{x}{y}.\sqrt{\dfrac{y}{x}}=\sqrt{\dfrac{x^2}{y^2}.\dfrac{y}{x}}=\sqrt{\dfrac{x}{y}}=\dfrac{\sqrt{xy}}{y}\)
28 tháng 6 2021
`a)(5sqrt2-2sqrt5)/(sqrt5-sqrt2)+6/(2-sqrt{10})`
`=(sqrt{10}(sqrt5-sqrt2))/(sqrt5-sqrt2)+(6(2+sqrt{10}))/(4-10)`
`=sqrt{10}-(2+sqrt{10})`
`=-2`
`b)6/(sqrt5-1)+7/(1-sqrt3)-2/(sqrt3-sqrt5)`
`=(6(sqrt5+1))/(5-1)+(7(1+sqrt3))/(1-3)-(2(sqrt3+sqrt5))/(3-5)`
`=(6(sqrt5+1))/4-(7+7sqrt3)/2+sqrt3+sqrt5`
`=(3sqrt5+3)/2-(7+7sqrt3)/2+sqrt3+sqrt5`
`=(3sqrt5+3-7-7sqrt3+2sqrt3+2sqrt5)/2`
`=(5sqrt5-5sqrt3-4)/2`
AT
4 tháng 7 2021
a) \(A=\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2}-\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1^2}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}=\left|\sqrt{3}-1\right|-\left|\sqrt{3}+1\right|\)
\(=\sqrt{3}-1+-\sqrt{3}-1=-2\)
b) \(B=\sqrt{11-6\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{3^2-2.3.\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}\)
\(=\sqrt{\left(3-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}=\left|3-\sqrt{2}\right|-\left|\sqrt{2}-1\right|\)
\(=3-\sqrt{2}-\sqrt{2}+1=4-2\sqrt{2}\)
c) \(C=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{7-2\sqrt{10}}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\sqrt{\left(\sqrt{5}\right)^2-2.\sqrt{5}.\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\left(\sqrt{5}+\sqrt{3}\right)\left|\sqrt{5}-\sqrt{2}\right|\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{2}\right)=5-\sqrt{10}+\sqrt{15}-\sqrt{6}\)
VH
0
\(1,\)
\(a,\sqrt{6-2\sqrt{5}}=\sqrt{\sqrt{5^2}-2.\sqrt{5}.1+1}=\sqrt{\left(\sqrt{5}-1\right)^2}=\left|\sqrt{5}-1\right|=\sqrt{5}-1\)
\(b,\sqrt{8+2\sqrt{7}}=\sqrt{\sqrt{7^2}+2.\sqrt{7}.1+1}=\sqrt{\left(\sqrt{7}+1\right)^2}=\left|\sqrt{7}+1\right|=\sqrt{7}+1\)
\(2,\)
\(a,\sqrt{\left(\sqrt{10}-3\right)^2}-\sqrt{10}\)
\(=\left|\sqrt{10}-3\right|-\sqrt{10}\)
\(=\sqrt{10}-\sqrt{10}-3\)
\(=-3\)
\(b,\sqrt{\left(5+\sqrt{7}\right)^2}-\sqrt{8-2\sqrt{7}}\)
\(=\left|5+\sqrt{7}\right|-\sqrt{\left(\sqrt{7}-1\right)^2}\)
\(=5+\sqrt{7}-\left|\sqrt{7}-1\right|\)
\(=5+\sqrt{7}-\sqrt{7}+1\)
\(=6\)