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\(2\sqrt{20}-\sqrt{45}+3\sqrt{18}+3\sqrt{32}-\sqrt{50}\\ =4\sqrt{5}-3\sqrt{5}+9\sqrt{2}+12\sqrt{2}-5\sqrt{2}\\ =\sqrt{5}+16\sqrt{2}\)
\(\sqrt{50}-3\sqrt{98}+2\sqrt{8}+3\sqrt{32}-5\sqrt{18}\)
\(=5\sqrt{2}-21\sqrt{2}+4\sqrt{2}+12\sqrt{2}-15\sqrt{12}\)
\(=-15\sqrt{2}\)
\(\sqrt{16x-32}+\sqrt{25x-50}=187\sqrt{9x-18}\)
\(\Leftrightarrow\sqrt{16\left(x-2\right)}+\sqrt{25\left(x-2\right)}=187\sqrt{9\left(x-2\right)}\)
\(\Leftrightarrow4\sqrt{x-2}+5\sqrt{x-2}=561\sqrt{x-2}\)
\(\Leftrightarrow552\sqrt{x-2}=0\)
\(\Leftrightarrow\sqrt{x-2}=0\)
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
a) \(\sqrt{1-8x+16x^2}=\dfrac{1}{3}\)
\(\Leftrightarrow\sqrt{1^2-2\cdot4x\cdot1+\left(4x\right)^2}=\dfrac{1}{3}\)
\(\Leftrightarrow\sqrt{\left(4x-1\right)^2}=\dfrac{1}{3}\)
\(\Leftrightarrow\left|4x-1\right|=\dfrac{1}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-1=\dfrac{1}{3}\left(ĐK:x\ge\dfrac{1}{4}\right)\\4x-1=\dfrac{1}{3}\left(ĐK:x< \dfrac{1}{4}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{4}{3}\\4x=\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\left(tm\right)\\x=\dfrac{1}{6}\left(tm\right)\end{matrix}\right.\)
b) \(\sqrt{16x-32}+\sqrt{25x-50}=18+\sqrt{9x-18}\) (ĐK: \(x\ge2\))
\(\Leftrightarrow\sqrt{16\left(x-2\right)}+\sqrt{25\left(x-2\right)}=18+\sqrt{9\left(x-2\right)}\)
\(\Leftrightarrow4\sqrt{x-2}+5\sqrt{x-2}=18+3\sqrt{x-2}\)
\(\Leftrightarrow6\sqrt{x-2}=18\)
\(\Leftrightarrow\sqrt{x-2}=3\)
\(\Leftrightarrow x-2=9\)
\(\Leftrightarrow x=9+2\)
\(\Leftrightarrow x=11\left(tm\right)\)
\(\sqrt{32}-\sqrt{50}+\sqrt{18}\)
\(=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}\)
\(=2\sqrt{2}\)
\(a,=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}\)
\(=\sqrt{2}\left(3-12+8-5\right)=-6\sqrt{2}\)
\(b,=\left|\sqrt{2}-\sqrt{3}\right|+3\sqrt{2}=\sqrt{3}-\sqrt{2}+3\sqrt{2}=\sqrt{3}+2\sqrt{2}\)
\(c,=\sqrt{5}+\sqrt{5}+\dfrac{5}{\sqrt{5}}-1=3\sqrt{5}-1\)
\(d,=\sqrt{3-2.2\sqrt{3}+4}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+1+\sqrt{3}=2\)
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) \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}=\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}+\left|\sqrt{3}+1\right|\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{3}+1=\left|2-\sqrt{3}\right|+\sqrt{3}+1=2-\sqrt{3}+\sqrt{3}+1=3\)
a: \(A=\left(1-\sqrt{7}\right)\cdot\left(1+\sqrt{7}\right)=1-7=-6\)
b: \(B=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}=-4\sqrt{3}\)
c: \(C=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)
1,
\(2\sqrt{5}-\sqrt{125}-\sqrt{80}\\ =2\sqrt{5}-\sqrt{25\cdot5}-\sqrt{16\cdot5}\\ =2\sqrt{5}-5\sqrt{5}-4\sqrt{5}\\ =-7\sqrt{5}\)
2,
\(3\sqrt{2}-\sqrt{8}+\sqrt{50}-4\sqrt{32}\\ =3\sqrt{2}-\sqrt{4\cdot2}+\sqrt{25\cdot2}-4\sqrt{16\cdot2}\\ =3\sqrt{2}-2\sqrt{2}+5\sqrt{2}-16\sqrt{2}\\=-10\sqrt{2}\)
3,
\(\sqrt{18}-3\sqrt{80}-2\sqrt{50}+2\sqrt{45}\\ =\sqrt{9\cdot2}-3\sqrt{16\cdot5}-2\sqrt{25\cdot2}+2\sqrt{9\cdot5}\\ =3\sqrt{2}-12\sqrt{5}-10\sqrt{2}+6\sqrt{5}\\ =-7\sqrt{2}-6\sqrt{5}\)
4,
\(\sqrt{27}-2\sqrt{3}+2\sqrt{48}-3\sqrt{75}\\ =\sqrt{9\cdot3}-2\sqrt{3}+2\sqrt{16\cdot3}-3\sqrt{25\cdot2}\\ =3\sqrt{3}-2\sqrt{3}+8\sqrt{3}-15\sqrt{3}\\ =-6\sqrt{3}\)
5,
\(3\sqrt{2}-4\sqrt{18}+\sqrt{32}-\sqrt{50}\\ =3\sqrt{2}-4\sqrt{9\cdot2}+\sqrt{16\cdot2}-\sqrt{25\cdot2}\\ =3\sqrt{2}-12\sqrt{2}+4\sqrt{2}-5\sqrt{2}\\ =-10\sqrt{2}\)
6,
\(2\sqrt{3}-\sqrt{75}+2\sqrt{12}-\sqrt{147}\\ =2\sqrt{3}-\sqrt{25\cdot3}+2\sqrt{4\cdot3}-\sqrt{49\cdot3}\\ =2\sqrt{3}-5\sqrt{3}+4\sqrt{3}-7\sqrt{3}\\ =-6\sqrt{3}\)
7,
\(\sqrt{20}-2\sqrt{45}-3\sqrt{80}+\sqrt{125}\\ =\sqrt{4\cdot5}-2\sqrt{9\cdot5}-3\sqrt{16\cdot5}+\sqrt{25\cdot5}\\ =2\sqrt{5}-6\sqrt{5}-12\sqrt{5}+5\sqrt{5}\\ =-11\sqrt{5}\)
8,
\(6\sqrt{12}-\sqrt{20}-2\sqrt{27}+\sqrt{125}\\ =6\sqrt{4\cdot3}-\sqrt{4\cdot5}-2\sqrt{9\cdot3}+\sqrt{25\cdot5}\\ =12\sqrt{3}-2\sqrt{5}-6\sqrt{3}+5\sqrt{5}\\ =6\sqrt{3}+3\sqrt{5}\\ =3\left(2\sqrt{3}+\sqrt{5}\right)\)
9,
\(4\sqrt{24}-2\sqrt{54}+3\sqrt{6}-\sqrt{150}\\ =4\sqrt{4\cdot6}-2\sqrt{9\cdot6}+3\sqrt{6}-\sqrt{25\cdot6}\\ =8\sqrt{6}-6\sqrt{6}+3\sqrt{6}-5\sqrt{6}=0\)
10,
\(2\sqrt{18}-3\sqrt{80}-5\sqrt{147}+5\sqrt{245}-3\sqrt{98}\\ =2\sqrt{9\cdot2}-3\sqrt{16\cdot5}-5\sqrt{49\cdot3}+5\sqrt{49\cdot5}-3\sqrt{49\cdot2}\\ =6\sqrt{2}-12\sqrt{5}-35\sqrt{3}+35\sqrt{5}-21\sqrt{2}\\ =-15\sqrt{2}-35\sqrt{3}+23\sqrt{5}\)
\(\sqrt{32}+\sqrt{50}+3\sqrt{8}-\sqrt{18}\)
\(=4\sqrt{2}+5\sqrt{2}+6\sqrt{2}-3\sqrt{2}\)
\(=\left(4+5+6-3\right)\sqrt{2}\)
\(=12\sqrt{2}\)
=4căn 2+5căn 2+6căn 2-3căn 2
=12căn 2