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Bài 11: Số vô tỉ. Khái niệm về căn bậc hai
1)
\(\sqrt{0,36}.\sqrt{\dfrac{25}{16}}+\left(-\dfrac{1}{2}\right)^2\)
\(=\dfrac{3}{5}.\dfrac{5}{4}+\dfrac{1}{4}\)
\(=\dfrac{3}{4}+\dfrac{1}{4}\)
\(=1\)
2)
\(\left(\dfrac{1}{4}-\dfrac{1}{\sqrt{64}}\right):\left(\dfrac{3}{2}\right)^2\)
\(=\left(\dfrac{1}{4}-\dfrac{1}{8}\right):\dfrac{9}{4}\)
\(=\left(\dfrac{2}{8}-\dfrac{1}{8}\right):\dfrac{9}{4}\)
\(=\dfrac{1}{8}:\dfrac{9}{4}\)
\(=\dfrac{1}{8}.\dfrac{4}{9}\)
\(=\dfrac{1}{18}\)
3)
\(\sqrt{\dfrac{4}{81}}.\dfrac{27}{8}-\sqrt{0,49}.\dfrac{1}{7}\)
\(=\dfrac{2}{9}.\dfrac{27}{8}-\dfrac{7}{10}.\dfrac{1}{7}\)
\(=\dfrac{3}{4}-\dfrac{7}{10}.\dfrac{1}{7}\)
\(=\dfrac{3}{4}-\dfrac{1}{10}\)
\(=\dfrac{30}{40}-\dfrac{4}{40}\)
\(=\dfrac{13}{20}\)
d)
\(\sqrt{\dfrac{16}{81}}.\dfrac{27}{8}-\sqrt{0,25}.\left(-2\right)^2\)
\(=\dfrac{4}{9}.\dfrac{27}{8}-\dfrac{1}{2}.4\)
\(=\dfrac{3}{2}-\dfrac{1}{2}.4\)
\(=\dfrac{3}{2}-2\)
\(=-\dfrac{1}{2}\)
\(#NguyễnNgọcKhánhLinh-VietNam\)
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\(3\sqrt{x+1-1=38}\)
Sửa đề: \(3\sqrt{x+1}-1=38\)
=>3 căn x+1=39
=>căn x+1=13
=>x+1=169
=>x=168
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\(\left(-\dfrac{1}{4}\right)^2X8+\sqrt{1\dfrac{9}{6}:}2\dfrac{1}{2}-\left|-\dfrac{3}{4}\right|\)
\(\left(-\dfrac{1}{4}\right)^2.8+\sqrt{1\dfrac{9}{6}}:2\dfrac{1}{2}-\left|-\dfrac{3}{4}\right|\)
\(=\dfrac{1}{16}.8+\sqrt{\dfrac{15}{6}}:\dfrac{5}{2}+\dfrac{3}{4}\)
\(=\dfrac{8}{16}+1,3:2,5+\dfrac{3}{4}\)
\(=\dfrac{1}{2}+0,52+0,75\)
\(=0,5+0,52+0,75\)
\(=1.77\)
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TÌM X BIẾT:
√66564x√16641=x-2^
=>x-2^9 - √(2^4)=√66564x√16641
=>x-2^9 - √(2^4)=258x129
=>x-2^9 - √(2^4)=33411
=>x-512-4=33411
=>x-(512+4)=33411
=>x-516=33411
=>x=33411+516
=>x=33927
Vậy x=33927
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\(\left(\dfrac{3}{2}\right)^2-\left[\dfrac{1}{2}:2-\dfrac{\sqrt{\left(-9\right)^2.}1}{3}\right]\)
\(=\dfrac{9}{4}-\left[\dfrac{1}{2}\times\dfrac{1}{2}-\dfrac{\sqrt{81}}{3}\right]\\ =\dfrac{9}{4}-\left(\dfrac{1}{4}-\dfrac{9}{3}\right)\\ =\dfrac{9}{4}-\left(\dfrac{1}{4}-3\right)\\ =\dfrac{9}{4}-\left(\dfrac{1-3\times4}{4}\right)\\ =\dfrac{9}{4}-\left(-\dfrac{11}{4}\right)\\ =\dfrac{9}{4}+\dfrac{11}{4}\\ =\dfrac{9+11}{4}\\ =\dfrac{20}{4}\\ =5\)
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\(=\dfrac{9}{4}-[\dfrac{1}{4}-\dfrac{9.1}{3}]\)
\(=\dfrac{9}{4}-[\dfrac{1}{4}-\dfrac{9}{3}]\)
\(=\dfrac{9}{4}--\dfrac{11}{4}\)
\(=5\)
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\(\sqrt{\dfrac{9}{16}-3.\sqrt{0,16}+\sqrt{\left(-2\right)^2}}\)
\(=\sqrt{\dfrac{9}{16}-3\cdot0.4+2}=\dfrac{\sqrt{545}}{20}\)
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a)x = 3 √ x; b) (x-1)^2=2.(x-1) tìm x
a: \(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-3\right)=0\)
=>x=0 hoặc x=9
b; =>(x-1)(x-3)=0
=>x=3 hoặc x=1
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1) \(\sqrt{x-1}=2\left(x\ge1\right)\)
2) \(\sqrt{2x+1=7}\left(x\ge\dfrac{-1}{2}\right)\)
3) \(2\sqrt{x-1}=0\left(\ge0\right)\)
1: =>x-1=4
=>x=5
2: =>2x+1=49
=>2x=48
=>x=24
3:=>x-1=0
=>x=1
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\(\dfrac{4}{5}.\sqrt{25}-\dfrac{7}{3}.\left(\sqrt{9-2^2}\right)\)
\(=\dfrac{4}{5}\cdot5-\dfrac{7}{3}\cdot\sqrt{5}=4-\dfrac{7\sqrt{5}}{3}\)
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\(\sqrt{\left[-\dfrac{2}{3}\right]^2}-\sqrt{0,09}+\sqrt{\dfrac{9}{25}}\)
=2/3-0,3+3/5
=10/15+9/15-3/10
=19/15-3/10
=38/30-9/30=29/30
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