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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,
\(2\sqrt{5}-\sqrt{125}-\sqrt{80}\\ =2\sqrt{5}-\sqrt{25\cdot5}-\sqrt{16\cdot5}\\ =2\sqrt{5}-5\sqrt{5}-4\sqrt{5}\\ =-7\sqrt{5}\)
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,
\(2\sqrt{18}-3\sqrt{80}-5\sqrt{147}+5\sqrt{245}-3\sqrt{98}\\ =2\sqrt{9\cdot2}-3\sqrt{16\cdot5}-5\sqrt{49\cdot3}+5\sqrt{49\cdot5}-3\sqrt{49\cdot2}\\ =6\sqrt{2}-12\sqrt{5}-35\sqrt{3}+35\sqrt{5}-21\sqrt{2}\\ =-15\sqrt{2}-35\sqrt{3}+23\sqrt{5}\)
\(a,\left(\sqrt{32}-\sqrt{50}+\sqrt{8}\right):2\)
\(=\left(4\sqrt{2}-5\sqrt{2}+2\sqrt{2}\right):2\)
\(=\sqrt{2}:2\)
1. \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{2}\)
\(=2\sqrt{2}\)
a/ \(=5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}=-5\sqrt{5}\)
Mấy câu kia bấm máy tính là xong hết
B2:
a/ \(=\sqrt{-\left(x^2+5\right)}\)
Có \(x^2+5>0\forall x\Rightarrow-\left(x^2+5\right)< 0\forall x\)
Vậy biểu thức luôn ko đc xđ
b/ x-4\(\ge0\) \(\Rightarrow x\ge4\)
c/ Có -3<0
Để căn thức xđ\(\Leftrightarrow x+1< 0\Leftrightarrow x< -1\)
d/ Có -(x2+1)<0\(\forall\) x
Để căn thức có nghĩa \(\Leftrightarrow x-3< 0\Leftrightarrow x< 3\)
bạn tự làm đi,,dễ mà,,,cứ cố tách bình phương là ok mà
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 !