`1a)A=(sqrt{99}-sqrt{18}-sqrt{11}).sqrt{11}+3sqrt{22}`
`=(3sqrt{11}-sqrt{11}-3sqrt2).sqrt{11}+3sqrt{22}`
`=(2sqrt{11}-3sqrt{2}).sqrt{11}+3sqrt{22}`
`=2.11-3sqrt{22}+3sqrt{22}`
`=22`
`b)B=sqrt{4+2sqrt3}+sqrt{4-2sqrt3}`
`=sqrt{3+2sqrt+1}+sqrt{3-2sqrt3+1}`
`=sqrt{(sqrt3+1)^2}+sqrt{(sqrt3-1)^2}`
`=sqrt3+1+sqrt3-1`
`=2sqrt3`
`c)C=5/(sqrt7+sqrt2)-(7-sqrt7)/(sqrt7-1)+6sqrt{1/2}`
`=((sqrt7-sqrt2)(sqrt7+sqrt2))/(sqrt7+sqrt2)-(sqrt7(sqrt7-1))/(sqrt7-1)+3.2.sqrt{1/2}`
`=sqrt7-sqrt2-sqrt7+3sqrt2`
`=2sqrt2`
`2)a)sqrt{2x-1}=sqrt{x-1}`
`đk:x>=1`
`pt<=>2x-1=x-1`
`<=>x=0(l)`
Vậy pt vô nghiệm
`b)sqrt{4-x^2}-x+2=0`
`đk:-2<=x<=2`
`pt<=>\sqrt{(2-x)(2+x)}=x-2`
`<=>4-x^2=x^2-4x+4`
`<=>2x^2-4x=0`
`<=>x(2x-4)=0`
`<=>` \(\left[ \begin{array}{l}x=2\\x=0\end{array} \right.\)
Vậy `S={0,2}`
Bài \(1\)
\(a\)) \(A=\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right).\sqrt{11}+3\sqrt{22}\)
\(A=\left(3\sqrt{11}-3\sqrt{2}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)
\(A=3.11-3\sqrt{22}-11+3\sqrt{22}\)
\(A=22\)
\(b\))\(B=\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)
\(B=\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}+1}+\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}+1}\)
\(B=\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(B=\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|\)
\(B=2\sqrt{3}\)
\(c\))\(C=\dfrac{5}{\sqrt{7}+\sqrt{2}}-\dfrac{7-\sqrt{7}}{\sqrt{7}-1}+6\sqrt{\dfrac{1}{2}}\)
\(C=\dfrac{5\left(\sqrt{7}-\sqrt{2}\right)}{5}-\dfrac{\sqrt{7}\left(\sqrt{7}-1\right)}{\left(\sqrt{7}-1\right)}+\sqrt{\dfrac{36}{2}}\)
\(C=\sqrt{7}-\sqrt{2}-\sqrt{7}+3\sqrt{2}\)
\(C=2\sqrt{2}\)
\(a\)) ĐK \(x>-1\)
\(\sqrt{2x-1}=\sqrt{x+1}\)
<=>\(2x-1=x+1\)
<=>\(x=2\)\(\left(TM\right)\)
Vậy \(x=2\)
\(b\)) ĐK \(x\) ≤ 2
\(\sqrt{4-x^2}-x+2=0\)
<=> \(\sqrt{4-x^2}=x-2\)
<=>\(4-x^2=x^2-4x+4\)
<=>\(2x^2-4x=0\)
<=>\(2x\left(x-2\right)=0\)
<=>\(\left[{}\begin{matrix}2x=0\\x-2=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy \(S=\left\{0;2\right\}\)
a, \(A=\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)
\(=\left(3\sqrt{11}-3\sqrt{2}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)
\(=\left(2\sqrt{11}-3\sqrt{2}\right)\sqrt{11}+3\sqrt{22}\)
\(=22-3\sqrt{22}+3\sqrt{22}=22\)
b, \(B=\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)
\(B=\sqrt{3+2\sqrt{3}+1}+\sqrt{3-2\sqrt{3}+1}\)
\(B=\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(B=\sqrt{3}+1+\sqrt{3}-1=2\sqrt{3}\)
c,\(C=\dfrac{5}{\sqrt{7}+\sqrt{2}}-\dfrac{7-\sqrt{7}}{\sqrt{7}-1}+6\sqrt{\dfrac{1}{2}}\)
\(C=\dfrac{5\left(\sqrt{7}-\sqrt{2}\right)}{7-2}-\dfrac{\sqrt{7}\left(\sqrt{7}-1\right)}{\sqrt{7}-1}+3\sqrt{2}\)
\(C=\sqrt{7}-\sqrt{2}-\sqrt{7}+3\sqrt{2}=2\sqrt{2}\)
bài 2:
a, \(\sqrt{2x-1}=\sqrt{x+1}\left(x\ge\dfrac{1}{2}\right)\)
\(=>2x-1=x+1< =>x=2\left(TM\right)\)
vậy x=2
b, \(\sqrt{4-x^2}-x+2=0\left(-2\le x\le2,\right)\)
\(=>\sqrt{4-x^2}=x-2=>4-x^2=x^2-4x+4\)
\(< =>-2x^2+4x=0< =>-2x\left(x-2\right)=0\)
\(=>\left[{}\begin{matrix}x=0\left(loai\right)\\x=2\left(TM\right)\end{matrix}\right.\) vậy...
