giúp mình giải bài toán rút gọn này ạ :\(\sqrt{11-4\sqrt{7}}-\sqrt{29-4\sqrt{7}}\)
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6 tháng 8 2021
Bài 20:
a) \(\sqrt{9-4\sqrt{5}}\cdot\sqrt{9+4\sqrt{5}}=\sqrt{81-80}=1\)
b) \(\left(2\sqrt{2}-6\right)\cdot\sqrt{11+6\sqrt{2}}=2\left(\sqrt{2}-3\right)\left(3+\sqrt{2}\right)\)
\(=2\left(2-9\right)=2\cdot\left(-7\right)=-14\)
c: \(\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
=2
d) \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)
\(=\left(4-2\sqrt{3}\right)\left(2+\sqrt{3}\right)\)
\(=8+4\sqrt{3}-4\sqrt{3}-6\)
=2
12 tháng 10 2023
a: \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)
\(=4-\sqrt{15}+\sqrt{15}=4\)
b: \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=2+\sqrt{3}-2+\sqrt{3}\)
\(=2\sqrt{3}\)
c: \(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(=\sqrt{\left(2\sqrt{5}+3\right)^2}-\sqrt{\left(2\sqrt{5}-3\right)^2}\)
\(=2\sqrt{5}+3-2\sqrt{5}+3=6\)
HT
1
29 tháng 7 2021
\(\sqrt{4-2\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{3}-1+2-\sqrt{3}\)
=1
HD
26 tháng 7 2021
Bài 2
b, `\sqrt{3x^2}=x+2` ĐKXĐ : `x>=0`
`=>(\sqrt{3x^2})^2=(x+2)^2`
`=>3x^2=x^2+4x+4`
`=>3x^2-x^2-4x-4=0`
`=>2x^2-4x-4=0`
`=>x^2-2x-2=0`
`=>(x^2-2x+1)-3=0`
`=>(x-1)^2=3`
`=>(x-1)^2=(\pm \sqrt{3})^2`
`=>` $\left[\begin{matrix} x-1=\sqrt{3}\\ x-1=-\sqrt{3}\end{matrix}\right.$
`=>` $\left[\begin{matrix} x=1+\sqrt{3}\\ x=1-\sqrt{3}\end{matrix}\right.$
Vậy `S={1+\sqrt{3};1-\sqrt{3}}`
H9
HT.Phong (9A5)
CTVHS
18 tháng 7 2023
\(\sqrt{7-\sqrt{24}}-\dfrac{\sqrt{50}-5}{\sqrt{10}-\sqrt{5}}+\sqrt{\left(11+\sqrt{120}\right)\left(11+2\sqrt{30}\right)^2}\)
\(=\sqrt{7-2\sqrt{6}}-\dfrac{5\left(\sqrt{2}-1\right)}{\sqrt{5}\left(\sqrt{2}-1\right)}+\left|11+2\sqrt{30}\right|\sqrt{11-2\sqrt{30}}\)
\(=\sqrt{1^2-2\sqrt{6}\cdot1+\left(\sqrt{6}\right)^2}-\dfrac{\sqrt{5}\cdot\sqrt{5}}{\sqrt{5}}+\left(11+2\sqrt{30}\right)\sqrt{\left(\sqrt{6}\right)^2-2\sqrt{5}\cdot\sqrt{6}+\left(\sqrt{5}\right)^2}\)
\(=\sqrt{\left(1-\sqrt{6}\right)^2}-\sqrt{5}+\left(11+2\sqrt{30}\right)\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}\)
\(=\left|1-\sqrt{6}\right|-\sqrt{5}+\left(11+2\sqrt{30}\right)\left|\sqrt{6}-\sqrt{5}\right|\)
\(=-1+6-\sqrt{5}+\left(\sqrt{6}+\sqrt{5}\right)^2\left(\sqrt{6}-\sqrt{5}\right)\)
\(=\sqrt{6}-1-\sqrt{5}+\left[\left(\sqrt{6}\right)^2-\left(\sqrt{5}\right)^2\right]\left(\sqrt{6}+\sqrt{5}\right)\)
\(=\sqrt{6}-1-\sqrt{5}+\left(6-5\right)\left(\sqrt{6}+\sqrt{5}\right)\)
\(=\sqrt{6}-1-\sqrt{5}+\sqrt{6}+\sqrt{5}\)
\(=2\sqrt{6}-1\)
18 tháng 7 2023
\(=\sqrt{6+1-2\sqrt{6}}-\dfrac{\sqrt{5}\left(\sqrt{10}-\sqrt{5}\right)}{\sqrt{10}-\sqrt{5}}+\sqrt{\left(11-\sqrt{120}\right)\left(11+\sqrt{120}\right)^2}\\ =\sqrt{\left(\sqrt{6}-\sqrt{1}\right)^2}-\sqrt{5}+\sqrt{\left(11^2-120\right)\left(11+2\sqrt{30}\right)}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{1\left(6+5+2\sqrt{6\cdot5}\right)}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{6}+\sqrt{5}=2\sqrt{6}-\sqrt{1}\)
24 tháng 9 2021
\(P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1:\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{2\sqrt{x}+7}{x-4}\right)\)
\(=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1:\left(\dfrac{x+2\sqrt{x}-x+\sqrt{x}+2-2\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)
\(=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-5}\)
\(=\dfrac{-x+8\sqrt{x}-15+\left(x-4\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{-x+8\sqrt{x}-15+x\sqrt{x}-2x-4\sqrt{x}+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{x\sqrt{x}-3x+4\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)
24 tháng 9 2021
\(ĐK:x\ge0;x\ne4\\ P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1:\dfrac{x+2\sqrt{x}-x+\sqrt{x}+2-2\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-5}\\ P=\dfrac{\left(3-\sqrt{x}\right)\left(\sqrt{x}-5\right)+\left(x-4\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\\ P=\dfrac{8\sqrt{x}-15-x+x\sqrt{x}-2x-4\sqrt{x}+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\\ P=\dfrac{x\sqrt{x}-3x+4\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)
5 tháng 8 2021
22,
1, Đặt √(3-√5) = A
=> √2A=√(6-2√5)
=> √2A=√(5-2√5+1)
=> √2A=|√5 -1|
=> A=\(\dfrac{\sqrt{5}-1}{\text{√2}}\)
=> A= \(\dfrac{\sqrt{10}-\sqrt{2}}{2}\)
2, Đặt √(7+3√5) = B
=> √2B=√(14+6√5)
=> √2B=√(9+2√45+5)
=> √2B=|3+√5|
=> B= \(\dfrac{3+\sqrt{5}}{\sqrt{2}}\)
=> B= \(\dfrac{3\sqrt{2}+\sqrt{10}}{2}\)
3,
Đặt √(9+√17) - √(9-√17) -\(\sqrt{2}\)=C
=> √2C=√(18+2√17) - √(18-2√17) -\(2\)
=> √2C=√(17+2√17+1) - √(17-2√17+1) -\(2\)
=> √2C=√17+1- √17+1 -\(2\)
=> √2C=0
=> C=0
26,
|3-2x|=2\(\sqrt{5}\)
TH1: 3-2x ≥ 0 ⇔ x≤\(\dfrac{-3}{2}\)
3-2x=2\(\sqrt{5}\)
-2x=2\(\sqrt{5}\) -3
x=\(\dfrac{3-2\sqrt{5}}{2}\) (KTMĐK)
TH2: 3-2x < 0 ⇔ x>\(\dfrac{-3}{2}\)
3-2x=-2\(\sqrt{5}\)
-2x=-2√5 -3
x=\(\dfrac{3+2\sqrt{5}}{2}\) (TMĐK)
Vậy x=\(\dfrac{3+2\sqrt{5}}{2}\)
6 tháng 8 2021
2, \(\sqrt{x^2}\)=12 ⇔ |x|=12 ⇔ x=12, -12
3, \(\sqrt{x^2-2x+1}\)=7
⇔ |x-1|=7
TH1: x-1≥0 ⇔ x≥1
x-1=7 ⇔ x=8 (TMĐK)
TH2: x-1<0 ⇔ x<1
x-1=-7 ⇔ x=-6 (TMĐK)
Vậy x=8, -6
4, \(\sqrt{\left(x-1\right)^2}\)=x+3
⇔ |x-1|=x+3
TH1: x-1≥0 ⇔ x≥1
x-1=x+3 ⇔ 0x=4 (KTM)
TH2: x-1<0 ⇔ x<1
x-1=-x-3 ⇔ 2x=-2 ⇔x=-1 (TMĐK)
Vậy x=-1
LT
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
23 tháng 8 2016
\(=\sqrt{2.5^2.3}+\sqrt{0.4^2}.\sqrt{2^2.3.5}+4,5.\frac{2}{\sqrt{3}}-\sqrt{2.3}.\)
\(=\sqrt{2.5^2.3^2}+\frac{2}{3}.\sqrt{2^2.3^2.5}+9-\sqrt{2.3}\)
\(=3.5.3.\sqrt{2}+2.2.3.\sqrt{5}+9-3.\sqrt{2.3}\)
\(=45\sqrt{2}+12\sqrt{5}+9-3\sqrt{6}\)
\(\sqrt{11-4\sqrt{7}}-\sqrt{29-4\sqrt{7}}\)
\(=\sqrt{7-2.\sqrt{7}.2+4}-\sqrt{28-2.2\sqrt{7}.1+1}\)
\(=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}.2+2^2}-\sqrt{\left(2\sqrt{7}\right)^2-2.2\sqrt{7}.1+1^2}\)
\(=\sqrt{\left(\sqrt{7}-2\right)^2}-\sqrt{\left(2\sqrt{7}-1\right)^2}\)
\(=\left|\sqrt{7}-2\right|-\left|2\sqrt{7}-1\right|\)
\(=\sqrt{7}-2-2\sqrt{7}+1\)
\(=-\sqrt{7}-1\)