Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
TRẢ LỜI :
\(=\sqrt{5}+\sqrt{5}+\sqrt{5}=3\sqrt{5}\)
c) √20 - √45 + 3√18 + √72
= √4.5 - √9.5 + 3√9.2 + √36.2
= 2√5 - 3√5 + 9√2 + 6√2
= -√5 + 15√2
a, \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}\sqrt{20}+\sqrt{5}\)
\(=\sqrt{5}+\dfrac{1}{2}.2\sqrt{5}+\sqrt{5}\)
\(=3\sqrt{5}\)
b, \(\sqrt{\dfrac{1}{2}}+\sqrt{4,5}+\sqrt{12,5}\)
\(=\sqrt{0,5}+3\sqrt{0,5}+5\sqrt{0,5}=9\sqrt{0,5}\)
c, \(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
\(=2\sqrt{5}-3\sqrt{5}+3\sqrt{18}+2\sqrt{18}\)
\(=-\sqrt{5}+5\sqrt{18}\)
d, \(0,1.\sqrt{200}+2\sqrt{0,08}+0,4\sqrt{50}\)
\(=\sqrt{0,01.200}+0,2.\sqrt{2}+0,4.5\sqrt{2}\)
\(=\sqrt{2}+0,2\sqrt{2}+2\sqrt{2}=3,2\sqrt{2}\)
Chúc bạn học tốt!!!
a) \(\dfrac{1}{2}\sqrt{20}+5=\dfrac{1}{2}\cdot2\sqrt{5}+5=5+\sqrt{5}\)
b) \(\sqrt{16}+\sqrt{64}=4+8=12\)
c) \(\sqrt{20}-\sqrt{45}+3\sqrt{18}=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}=9\sqrt{2}-\sqrt{5}\)
d) \(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{2}=2-\sqrt{2}+\sqrt{2}=2\)
a: \(=10\sqrt{2}-4\sqrt{2}+6\sqrt{2}=12\sqrt{2}\)
b: \(=5\sqrt{7}-4\sqrt{7}+3\sqrt{7}=4\sqrt{7}\)
c: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}=\dfrac{1}{6}\sqrt{6}\)
d: \(=8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}=5\sqrt{5}\)
e: \(=\sqrt{5}+\dfrac{2}{5}\sqrt{5}+\sqrt{5}=2.4\sqrt{5}\)
f: \(=\dfrac{1}{5}\sqrt{5}+\dfrac{3}{2}\sqrt{2}+\dfrac{5}{2}\sqrt{2}=\dfrac{1}{5}\sqrt{5}+4\sqrt{2}\)
\(a,=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}\)
\(=\sqrt{2}\left(3-12+8-5\right)=-6\sqrt{2}\)
\(b,=\left|\sqrt{2}-\sqrt{3}\right|+3\sqrt{2}=\sqrt{3}-\sqrt{2}+3\sqrt{2}=\sqrt{3}+2\sqrt{2}\)
\(c,=\sqrt{5}+\sqrt{5}+\dfrac{5}{\sqrt{5}}-1=3\sqrt{5}-1\)
\(d,=\sqrt{3-2.2\sqrt{3}+4}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+1+\sqrt{3}=2\)
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) \(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
= \(2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}\)
= \(-\sqrt{5}+15\sqrt{2}\)
b) \(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)
= \(\left(2\sqrt{7}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
= \(2.7-2\sqrt{21}+7+2\sqrt{21}=14+7=21\)
c) \(\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}\)
= \(6+2\sqrt{6}.\sqrt{5}+5-2\sqrt{30}\)
= \(11+2\sqrt{30}-2\sqrt{30}=11\)
d) \(\left(\dfrac{1}{2}-\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{200}\right):\dfrac{1}{8}\)
= \(\left(\dfrac{1}{2}-\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+8\sqrt{2}\right).8\)
= \(4-4\sqrt{2}-12\sqrt{2}+64\sqrt{2}=4+48\sqrt{2}\)
Bài này dễ ẹc ( đâu có khó đâu :)) )
a) \(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
\(=\sqrt{2^2.5}-\sqrt{3^2.5}+3\sqrt{3^2.2}+\sqrt{6^2.2}\)
\(=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}\)
\(=\left(2-3\right)\sqrt{5}+\left(9+6\right)\sqrt{2}\)
\(=15\sqrt{2}-\sqrt{5}\)
b) \(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)
\(=\sqrt{2^2.7}.\sqrt{7}-2\sqrt{3}.\sqrt{7}+\sqrt{7}.\sqrt{7}+\sqrt{2^2.21}\)
\(=2.7-2\sqrt{21}+7+2\sqrt{21}\)
\(=14+7+\left(2-2\right)\sqrt{21}=21\)
c) \(\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}\)
\(=6+2\sqrt{30}+5-\sqrt{2^2.30}\)
\(=6+5+2\sqrt{30}-2\sqrt{30}=11\)
d) \(\left(\dfrac{1}{2}\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{200}\right):\dfrac{1}{8}\)
\(=\left(\dfrac{1}{2}\sqrt{\dfrac{2}{2^2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{10^2.2}\right):\dfrac{1}{8}\)
\(=\left(\dfrac{1}{4}\sqrt{2}-\dfrac{3}{2}\sqrt{2}+8\sqrt{2}\right).8\)
\(=2\sqrt{2}-12\sqrt{2}+64\sqrt{2}=54\sqrt{2}\)
Hok tốt
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: \(B=\sqrt{125}-10\sqrt{\dfrac{1}{20}}+\dfrac{5-\sqrt{5}}{\sqrt{5}}\)
\(=5\sqrt{5}-\sqrt{5}+\sqrt{5}-1\)
\(=5\sqrt{5}-1\)
Bài 1:
a/
$\sqrt{(\sqrt{7}-4)^2}+\sqrt{8-2\sqrt{7}}$
$=|\sqrt{7}-4|+\sqrt{7+1-2\sqrt{7}}=|\sqrt{7}-4|+\sqrt{(\sqrt{7}-1)^2}$
$=4-\sqrt{7}+|\sqrt{7}-1|=4-\sqrt{7}+\sqrt{7}-1=3$
b/
\(\sqrt{(\sqrt{5}-2)^2}+\sqrt{6+2\sqrt{5}}\\ =|\sqrt{5}-2|+\sqrt{5+1+2\sqrt{5}}\\ =\sqrt{5}-2+\sqrt{(\sqrt{5}+1)^2}\\ =\sqrt{5}-2+|\sqrt{5}+1|=\sqrt{5}-2+\sqrt{5}+1=2\sqrt{5}-1\)
Bài 2:
a. $=\sqrt{5}+\sqrt{5}+\sqrt{5}=3\sqrt{5}$
b. $=\frac{\sqrt{2}}{2}+\frac{3\sqrt{2}}{2}+\frac{5\sqrt{2}}{2}$
$=\frac{\sqrt{2}+3\sqrt{2}+5\sqrt{2}}{2}=\frac{9\sqrt{2}}{2}$
c.
$=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}$
$=-\sqrt{5}+15\sqrt{2}$
d.
$=0,1.10\sqrt{2}+2.\frac{\sqrt{2}}{5}+0,4.5\sqrt{2}$
$=\sqrt{2}+0,4\sqrt{2}+2\sqrt{2}$
$=\sqrt{2}(1+0,4+2)=3,4\sqrt{2}$
a, \(3\sqrt{5}\)
b, \(\dfrac{9\sqrt{2}}{2}\)
c, \(15\sqrt{2}-\sqrt{5}\)
d, \(\dfrac{17\sqrt{2}}{5}\)