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a: Ta có: \(A=\sqrt{8}-2\sqrt{18}+3\sqrt{50}\)
\(=2\sqrt{2}-6\sqrt{2}+15\sqrt{2}\)
\(=11\sqrt{2}\)
b: Ta có: \(B=\sqrt{125}-10\sqrt{\dfrac{1}{20}}+\dfrac{5-\sqrt{5}}{\sqrt{5}}\)
\(=5\sqrt{5}-\sqrt{5}+\sqrt{5}-1\)
\(=5\sqrt{5}-1\)
\(a,=\dfrac{\sqrt{5}+1+\sqrt{5}-1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\dfrac{2\sqrt{5}}{4}=\dfrac{\sqrt{5}}{2}\\ b,=\sqrt{\left(3-\sqrt{5}\right)^2}+\left|2-\sqrt{5}\right|=3-\sqrt{5}+\sqrt{5}-2=1\\ c,=\dfrac{2\left(\sqrt{5}-\sqrt{3}\right)}{2}-\dfrac{-\sqrt{3}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}=\sqrt{5}-\sqrt{3}+\sqrt{3}=\sqrt{5}\)
a) 2√18 - 4√50 + 3√32
= 6√2 - 20√2 + 12√2
= -2√2
b) √(√8 - 4)² + √8
= 4 - √8 + √8
= 4
c) √(14 - 6√5) + √(6 + 2√5)
= √(3 - √5)² + √(√5 + 1)²
= 3 - √5 + √5 + 1
= 4
\(a,2\sqrt{18}-4\sqrt{50}+3\sqrt{32}\\ =6\sqrt{2}-20\sqrt{2}+12\sqrt{2}=-2\sqrt{2}\\ b,\sqrt{\left(\sqrt{8}-4\right)^2}+\sqrt{8}\\ =4-\sqrt{8}+\sqrt{8}\\ =4\\ c,\sqrt{14-6\sqrt{5}}+\sqrt{6+2\sqrt{5}}\\ =\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}=3+\sqrt{5}+\sqrt{5}+1\\ =4+2\sqrt{5}\)
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Bài 1:
a) \(A=\sqrt{8}+\sqrt{18}-\sqrt{32}\)
\(=2\sqrt{2}+3\sqrt{2}-4\sqrt{2}\)
\(=\sqrt{2}\)
b) \(B=\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{4-4\sqrt{5}+5}-\sqrt{5}\)
\(=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{5}\)
\(=\left|2-\sqrt{5}\right|-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}\)
\(=-2\)
Bài 2:
a) \(\left\{{}\begin{matrix}2x-3y=4\\x+3y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x=6\\x+3y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2+3y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
Vậy phương trình có nghiệm là: \(\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
b) ĐKXĐ: \(x\ne\pm2\)
Với \(x\ne\pm2\), ta có:
\(\dfrac{10}{x^2-4}+\dfrac{1}{2-x}=1\)
\(\Leftrightarrow\dfrac{10}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}=1\)
\(\Leftrightarrow\dfrac{10-x-2}{x^2-4}=1\)
\(\Leftrightarrow\dfrac{8-x}{x^2-4}=1\)
\(\Rightarrow x^2-4=8-x\)
\(\Leftrightarrow x^2+x-12=0\)
\(\Leftrightarrow x^2-3x+4x-12=0\)
\(\Leftrightarrow x\left(x-3\right)+4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\) (TM)
Vậy phương trình có tập nghiệm là: S ={3; -4}
\(a,=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}\)
\(=\sqrt{2}\left(3-12+8-5\right)=-6\sqrt{2}\)
\(b,=\left|\sqrt{2}-\sqrt{3}\right|+3\sqrt{2}=\sqrt{3}-\sqrt{2}+3\sqrt{2}=\sqrt{3}+2\sqrt{2}\)
\(c,=\sqrt{5}+\sqrt{5}+\dfrac{5}{\sqrt{5}}-1=3\sqrt{5}-1\)
\(d,=\sqrt{3-2.2\sqrt{3}+4}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+1+\sqrt{3}=2\)
a) \(3\sqrt{2}-4\sqrt{18}+2\sqrt{32}-\sqrt{50}=3\sqrt{2}-4\sqrt{9.2}+2\sqrt{16.2}-\sqrt{25.2}\)
\(=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}=-6\sqrt{2}\)
b) \(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}=\left|\sqrt{2}-\sqrt{3}\right|+\sqrt{9.2}=\sqrt{3}-\sqrt{2}+3\sqrt{2}\)
\(=2\sqrt{2}+\sqrt{3}\)
c) \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}=\sqrt{25.\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{9.5}+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}}\)
\(=\sqrt{5}+\sqrt{5}+\sqrt{5}-1=3\sqrt{5}-1\)
d) \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}=\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}+\left|\sqrt{3}+1\right|\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{3}+1=\left|2-\sqrt{3}\right|+\sqrt{3}+1=2-\sqrt{3}+\sqrt{3}+1=3\)
\(1,B=9-5=4\\ 2,\dfrac{\sqrt{5}+1}{3-2\sqrt{2}}-\dfrac{\sqrt{10}}{\sqrt{5}-2}+3\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(\sqrt{5}+1\right)\left(3+2\sqrt{2}\right)-\sqrt{10}\left(\sqrt{5}+2\right)+3\sqrt{2}-3\sqrt{5}\\ =3\sqrt{5}+2\sqrt{10}+3+2\sqrt{2}-5\sqrt{2}-2\sqrt{10}+3\sqrt{2}-3\sqrt{5}=3\)
\(3,\\ a,\left(\dfrac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\dfrac{\sqrt{x}+\sqrt{y}}{1+\sqrt{xy}}\right):\left(1+\dfrac{x+y+2xy}{1-xy}\right)\left(x,y\ge0;xy\ne1\right)\\ =\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)+\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{1-xy}:\dfrac{1-xy+x+y+2xy}{1-xy}\\ =\dfrac{\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}+\sqrt{x}-x\sqrt{y}+\sqrt{y}-y\sqrt{x}}{1-xy}\cdot\dfrac{1-xy}{1+x+y+xy}\\ =\dfrac{2\left(\sqrt{x}+\sqrt{y}\right)}{\left(1+x\right)+y\left(1+x\right)}=\dfrac{2\left(\sqrt{x}+\sqrt{y}\right)}{\left(1+y\right)\left(1+x\right)}\)
\(b,x=\dfrac{2}{2+\sqrt{3}}=\dfrac{2\left(2-\sqrt{3}\right)}{1}=4-2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}-1\)
Thay vào BT
\(=\dfrac{2\left(\sqrt{3}-1+\sqrt{y}\right)}{\left(1+y\right)\left(1+4-2\sqrt{3}\right)}=\dfrac{2\sqrt{3}-2+2\sqrt{y}}{\left(1+y\right)\left(3-2\sqrt{3}\right)}\\ =\dfrac{2\sqrt{3}-2+2\sqrt{y}}{3-2\sqrt{3}+3y-2y\sqrt{3}}\)
1:
\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)
2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)
\(=\dfrac{20-6}{2}=7\)
1.
a, \(2\sqrt{18}-4\sqrt{50}-3\sqrt{32}=6\sqrt{2}-20\sqrt{2}-12\sqrt{2}=-2\sqrt{2}\)
b, \(\sqrt{14-6\sqrt{5}}+\sqrt{14+6\sqrt{5}}=\sqrt{\left(\sqrt{5}-3\right)^2}+\sqrt{\left(\sqrt{5}+3\right)^2}\)
\(=\left|\sqrt{5}-3\right|+\left|\sqrt{5}+3\right|\)
\(=-\sqrt{5}+3+\sqrt{5}+3=6\)
c, \(\dfrac{\sqrt{10}+10}{1+\sqrt{10}}-\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}=\dfrac{\sqrt{10}\left(1+\sqrt{10}\right)}{1+\sqrt{10}}-\dfrac{\sqrt{10}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{5}-\sqrt{2}}\)
\(=\sqrt{10}-\sqrt{10}=0\)
2.
ĐK: \(x\in R\)
\(\sqrt{9x^2-30x+25}=5\)
\(\Leftrightarrow\sqrt{\left(3x-5\right)^2}=5\)
\(\Leftrightarrow\left|3x-5\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=5\\3x-5=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10}{3}\\x=0\end{matrix}\right.\)
Vậy ...
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