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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=\(\dfrac{3+2\sqrt{5}}{2}\)
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
\(\dfrac{2+\sqrt{2}}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}\)
\(=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}\)
\(=\sqrt{2}+\sqrt{3}-\sqrt{2}=\sqrt{3}\)
`a)\sqrt{16x+48}+\sqrt{x+3}=15` `ĐK: x >= -3`
`<=>4\sqrt{x+3}+\sqrt{x+3}=15`
`<=>5\sqrt{x+3}=15`
`<=>\sqrt{x+3}=3`
`<=>x+3=9<=>x=6` (t/m).
`b)\sqrt{x^2-4}-3\sqrt{x-2}=0` `ĐK: x >= 2`
`<=>\sqrt{x-2}(\sqrt{x+2}-3)=0`
`<=>[(\sqrt{x-2}=0),(\sqrt{x+2}=3):}`
`<=>[(x-2=0),(x+2=9):}<=>[(x=2(t//m)),(x=7(t//m)):}`
a) Ta có: \(\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(=\frac{2+\sqrt{4+2\sqrt{3}}}{2-\sqrt{4-2\sqrt{3}}}\)
\(=\frac{2+\sqrt{\left(\sqrt{3}+1\right)^2}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\frac{2+\left|\sqrt{3}+1\right|}{2-\left|\sqrt{3}-1\right|}\)
\(=\frac{2+\sqrt{3}+1}{2-\sqrt{3}+1}\)(Vì \(\sqrt{3}>1>0\))
\(=\frac{3+\sqrt{3}}{3-\sqrt{3}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}\)
\(a=\frac{2+\sqrt{4+2\sqrt{3}}}{2-\sqrt{4-2\sqrt{3}}}=\frac{2+\sqrt{\left(\sqrt{3}+1\right)^2}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}=\frac{2+\sqrt{3}+1}{2-\sqrt{3}+1}=\frac{3+\sqrt{3}}{3-\sqrt{3}}=\frac{\left(3+\sqrt{3}\right)^2}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}\)
\(=\frac{12+6\sqrt{3}}{6}=2+\sqrt{3}\)
Xét \(A=\sqrt{3+\sqrt{7}}+\sqrt{3-\sqrt{7}}>0\)
\(A^2=6+2\sqrt{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}=6+2\sqrt{2}\)
\(\Rightarrow A=\sqrt{6+2\sqrt{2}}\)
\(\Rightarrow\sqrt{3+\sqrt{7}}+\sqrt{3-\sqrt{7}}-\sqrt{6+2\sqrt{2}}=\sqrt{6+2\sqrt{2}}-\sqrt{6+2\sqrt{2}}=0\)