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\(a,ĐK:x\le\dfrac{5}{3}\\ PT\Leftrightarrow-3x+5=49\\ \Leftrightarrow x=-\dfrac{44}{3}\left(tm\right)\\ b,ĐK:x\ge-12\\ PT\Leftrightarrow\dfrac{1}{2}x+6=2\\ \Leftrightarrow\dfrac{1}{2}x=-4\\ \Leftrightarrow x=-8\left(tm\right)\\ c,ĐK:x\ge-\dfrac{1}{2}\\ PT\Leftrightarrow2x+1=13+4\sqrt{3}\\ \Leftrightarrow x=\dfrac{12+4\sqrt{3}}{2}=6+2\sqrt{3}\left(tm\right)\\ d,PT\Leftrightarrow\left|3x-1\right|=8\Leftrightarrow\left[{}\begin{matrix}3x-1=8\\1-3x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{7}{3}\end{matrix}\right.\)
\(a.\dfrac{2}{4-3\sqrt{2}}-\dfrac{2}{4+3\sqrt{2}}=\dfrac{2\left(4+3\sqrt{2}\right)-2\left(4-3\sqrt{2}\right)}{\left(4-3\sqrt{2}\right)\left(4+3\sqrt{2}\right)}=\dfrac{8+6\sqrt{2}-8+6\sqrt{2}}{16-18}=\dfrac{12\sqrt{2}}{-2}=-6\sqrt{2}\)\(b.\dfrac{2}{1+\sqrt{2}}+\dfrac{2}{1-\sqrt{2}}=\dfrac{2\left(1-\sqrt{2}\right)+2\left(1+\sqrt{2}\right)}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}=\dfrac{2-2\sqrt{2}+2+2\sqrt{2}}{1-2}=-4\)\(c.\left(\sqrt{14}-3\sqrt{2}\right)^2+6\sqrt{28}=14-12\sqrt{7}+18+12\sqrt{7}=14+18=32\)\(d.\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}=14-14\sqrt{2}+7+14\sqrt{2}=14+7=21\)\(e.\left(\sqrt{6}-\sqrt{5}\right)^2-2\sqrt{120}=6-2\sqrt{30}+5-4\sqrt{30}=11-6\sqrt{30}\)
a: \(=\left(2\sqrt{7}+\sqrt{7}+2\sqrt{14}\right)\cdot\sqrt{7}-\left(51+14\sqrt{2}\right)\)
\(=3\sqrt{7}\cdot\sqrt{7}+2\sqrt{14}\cdot\sqrt{7}-51-14\sqrt{2}\)
\(=21-51=-30\)
b: \(=\dfrac{\sqrt{10}}{2}+\dfrac{\sqrt{10}-\sqrt{6}}{2}=\dfrac{2\sqrt{10}-\sqrt{6}}{2}\)
c: \(=\dfrac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\sqrt{5}+\sqrt{3}}+\dfrac{\left(\sqrt{5}-\sqrt{2}\right)^2}{\sqrt{5}-\sqrt{2}}\)
\(=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{2}\)
\(=2\sqrt{5}+\sqrt{3}-\sqrt{2}\)
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}}`
a: Ta có: \(A=\left(\dfrac{6+\sqrt{20}}{3+\sqrt{5}}+\dfrac{\sqrt{14}-\sqrt{2}}{\sqrt{7}-1}\right):\left(2+\sqrt{2}\right)\)
\(=\left(2+\sqrt{2}\right):\left(2+\sqrt{2}\right)\)
=1
b: Ta có: \(B=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-\dfrac{11}{2\sqrt{3}+1}\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}-2\sqrt{3}+1\)
=1
`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`
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: \(\sqrt{2}+\dfrac{1}{\sqrt{5+2\sqrt{6}}}+\dfrac{2}{\sqrt{8+2\sqrt{15}}}\)
\(=\sqrt{2}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{2}{\sqrt{5}+\sqrt{3}}\)
\(=\sqrt{2}+\sqrt{3}-\sqrt{2}+\sqrt{5}-\sqrt{3}\)
\(=\sqrt{5}\)
1) \(\sqrt{12}\)+\(5\sqrt{3}-\sqrt{48}\)
= \(2\sqrt{3}+5\sqrt{3}-4\sqrt{3}\)
= (2+5-4).\(\sqrt{3}\)
= \(3\sqrt{3}\)
2)\(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)
= \(5\sqrt{5}+2\sqrt{5}-3.3\sqrt{5}\)
= \(5\sqrt{5}+2\sqrt{5}-9\sqrt{5}\)
= \(\left(5+2-9\right).\sqrt{5}\)
= -2\(\sqrt{2}\)
3)\(3\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)
= \(3.4\sqrt{2}+4.2\sqrt{2}-5.3\sqrt{2}
\)
= 12\(\sqrt{2}\) \(+8\sqrt{2}\) \(-15\sqrt{2}\)
= \(\left(12+8-15\right).\sqrt{2}\)
= \(5\sqrt{2}\)
4)\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
= \(3.2\sqrt{3}-4.3\sqrt{3}+5.4\sqrt{3}\)
= \(6\sqrt{3}-12\sqrt{3}+20\sqrt{3}\)
= \(\left(6-12+20\right).\sqrt{3}\)
= \(14\sqrt{3}\)
5)\(\sqrt{12}+\sqrt{75}-\sqrt{27}\)
= \(2\sqrt{3}+5\sqrt{3}-3\sqrt{3}\)
= \(\left(2+5-3\right).\sqrt{3}\)
= \(4\sqrt{3}\)
6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)
= \(2.3\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= 6\(\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= \(\left(6-7+9\right).\sqrt{2}\)
= 8\(\sqrt{2}\)
7)\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)
= \(3.2\sqrt{5}-2.3\sqrt{5}+4\sqrt{5}\)
= \(6\sqrt{5}-6\sqrt{5}+4\sqrt{5}\)
= \(4\sqrt{5}\)
8)\(\left(\sqrt{2}+2\right).\sqrt{2}-2\sqrt{2}\)
= \(\left(\sqrt{2}\right)^2+2\sqrt{2}-2\sqrt{2}\)
= 2
Bài 1:
a. Ta có \(\sqrt{\dfrac{2}{x^2}}=\dfrac{\sqrt{2}}{\left|x\right|}=\dfrac{\sqrt{2}}{x}\) ,để biểu thức có nghĩa thì \(x>0\)
b. Để biểu thức \(\sqrt{\dfrac{-3}{3x+5}}\) có nghĩa thì \(\dfrac{-3}{3x+5}\ge0\)
mà \(-3< 0\Rightarrow3x+5< 0\) \(\Rightarrow x< \dfrac{-5}{3}\)
Bài 2:
a. \(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\dfrac{\left(2+\sqrt{2}\right)\left(1-\sqrt{2}\right)}{1-2}=\dfrac{-\sqrt{2}}{-1}=\sqrt{2}\)
b. \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)
\(=14-14\sqrt{2}+7+14\sqrt{2}\)
\(=21\)
c. \(\left(\sqrt{14}-3\sqrt{2}\right)^2+6\sqrt{28}\)
\(=14-6\sqrt{28}+18+6\sqrt{28}\)
\(=32\)