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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
\(a,2\sqrt{18}-4\sqrt{50}+3\sqrt{32}\\ =6\sqrt{2}-20\sqrt{2}+12\sqrt{2}=-2\sqrt{2}\\ b,\sqrt{\left(\sqrt{8}-4\right)^2}+\sqrt{8}\\ =4-\sqrt{8}+\sqrt{8}\\ =4\\ c,\sqrt{14-6\sqrt{5}}+\sqrt{6+2\sqrt{5}}\\ =\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}=3+\sqrt{5}+\sqrt{5}+1\\ =4+2\sqrt{5}\)
1, \(\sqrt{8}-3\sqrt{32}+\sqrt{72}=2\sqrt{2}-12\sqrt{2}+6\sqrt{2}=-4\sqrt{2}\)
2,\(6\sqrt{12}-2\sqrt{48}+5\sqrt{75}-7\sqrt{108}=12\sqrt{3}-8\sqrt{3}+25\sqrt{3}-42\sqrt{3}=-13\sqrt{3}\)
3, \(\sqrt{20}+3\sqrt{45}-6\sqrt{80}-\dfrac{1}{3}\sqrt{125}=2\sqrt{5}+9\sqrt{5}-24\sqrt{5}-\dfrac{5}{3}.\sqrt{5}=-\dfrac{44}{3}.\sqrt{5}\)
4, \(2\sqrt{5}-\sqrt{125}-\sqrt{80}=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}=-7\sqrt{5}\)
5, \(3\sqrt{2}-\sqrt{8}+\sqrt{50}-4\sqrt{32}=3\sqrt{2}-2\sqrt{2}+5\sqrt{2}-16\sqrt{2}=-10\sqrt{2}\)
1.
\(\sqrt{50}-3\sqrt{8}+\sqrt{32}=5\sqrt{2}-6\sqrt{2}+4\sqrt{2}=3\sqrt{2}\)
2.
a, ĐK: \(x\in R\)
\(pt\Leftrightarrow\sqrt{\left(x-2\right)^2}=1\)
\(\Leftrightarrow\left|x-2\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
b, ĐK: \(x\ge3\)
\(pt\Leftrightarrow\sqrt{x-3}\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=1\left(l\right)\end{matrix}\right.\)
\(=5\sqrt{2}-\dfrac{3}{2}\cdot4\sqrt{2}-\dfrac{1}{3}\cdot6\sqrt{2}+8=-3\sqrt{2}+8\)
a: \(\dfrac{\sqrt{50}-\sqrt{32}+\sqrt{8}}{\sqrt{2}}\)
\(=\dfrac{5\sqrt{2}-4\sqrt{2}+2\sqrt{2}}{\sqrt{2}}\)
\(=\dfrac{3\sqrt{2}}{\sqrt{2}}=3\)
b: \(\dfrac{4}{\sqrt{5}-1}-5\sqrt{\dfrac{1}{5}}\)
\(=\dfrac{4\left(\sqrt{5}+1\right)}{5-1}-\sqrt{5}\)
\(=\sqrt{5}+1-\sqrt{5}\)
=1
a) \(A=2\sqrt{8}-3\sqrt{32}+\sqrt{50}\)
\(A=2\sqrt{4.2}-3\sqrt{16.2}+\sqrt{25.2}\)
\(A=2.2\sqrt{2}-3.4\sqrt{2}+5\sqrt{2}\)
\(A=4\sqrt{2}-12\sqrt{2}+5\sqrt{2}\)
\(A=\left(4-12+5\right)\sqrt{2}\)
\(A=-3\sqrt{2}\)
b) \(B=\sqrt{12}+4\sqrt{27}-3\sqrt{48}\)
\(B=\sqrt{4.3}+4\sqrt{9.3}-3\sqrt{16.3}\)
\(B=2\sqrt{3}+4.3\sqrt{3}-3.4\sqrt{3}\)
\(B=2\sqrt{3}\)
c) \(C=\sqrt{20a}+4\sqrt{45a}-2\sqrt{125a}\left(a\ge0\right)\)
\(C=\sqrt{4.5a}+4\sqrt{9.5a}-2\sqrt{25.5a}\)
\(C=2\sqrt{5a}+4.3\sqrt{5a}-2.5\sqrt{5a}\)
\(C=2\sqrt{5a}+12\sqrt{5a}-10\sqrt{5a}\)
\(C=\left(2+12-10\right)\sqrt{5a}\)
\(C=4\sqrt{5a}\)
a) ta có \(2\sqrt{8}=2\sqrt{4.2}=4\sqrt{2},3\sqrt{32}=3\sqrt{16.2}=12\sqrt{2},\sqrt{50}=\sqrt{25.2}=5\sqrt{2}\) \(\Rightarrow A=4\sqrt{2}-12\sqrt{2}+5\sqrt{2}=-3\sqrt{2}\) b) ta có \(\sqrt{12}=\sqrt{4.3}=2\sqrt{3},4\sqrt{27}=4\sqrt{9.3}=12\sqrt{3},3\sqrt{48}=3\sqrt{16.3}=12\sqrt{3}\Rightarrow B=2\sqrt{3}+12\sqrt{3}-12\sqrt{3}=26\sqrt{3}\)c) ta có \(\sqrt{20a}=\sqrt{4.5a}=2\sqrt{5a},4\sqrt{45a}=4\sqrt{9.5a}=12\sqrt{5a},2\sqrt{125a}=2\sqrt{25.5a}=10\sqrt{5a}\Rightarrow C=2\sqrt{5a}+12\sqrt{5a}-10\sqrt{5a}=4\sqrt{5a}\)
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,
\(2\sqrt{18}-3\sqrt{80}-5\sqrt{147}+5\sqrt{245}-3\sqrt{98}\\ =2\sqrt{9\cdot2}-3\sqrt{16\cdot5}-5\sqrt{49\cdot3}+5\sqrt{49\cdot5}-3\sqrt{49\cdot2}\\ =6\sqrt{2}-12\sqrt{5}-35\sqrt{3}+35\sqrt{5}-21\sqrt{2}\\ =-15\sqrt{2}-35\sqrt{3}+23\sqrt{5}\)
MK CHỈ VIẾT KQ THUI NHA ! VÌ DÀI QUÁ ...
A= 4,236067977 B = 2,414213562 C= 0,8218544151
D= 3,968118785 E= \(-\)\(10\sqrt{2}\) F=17,10050878 (\(3\sqrt{5}+6\sqrt{3}\))
G=\(-7\sqrt{5}\) H= \(-10\sqrt{2}\)
K VÀ KB NHOA ! Dương Nguyễn Ngọc Khánh !
\(3\sqrt{2}-\sqrt{8}+\sqrt{50}-4\sqrt{32}=3\sqrt{2}-2\sqrt{2}+5\sqrt{2}-16\sqrt{2}=-10\sqrt{2}.\)