(x+1/3)^2=1/25
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10 tháng 5 2022
1: \(\left(3x-\dfrac{1}{5}\right)^2=\left(-\dfrac{3}{25}\right)^2\)
=>3x-1/5=3/25 hoặc 3x-1/5=-3/25
=>3x=8/25 hoặc 3x=2/25
=>x=8/75 hoặc x=2/75
2: \(\left(2x-\dfrac{1}{3}\right)^2=\left(-\dfrac{2}{9}\right)^2\)
=>2x-1/3=2/9 hoặc 2x-1/3=-2/9
=>2x=5/9 hoặc 2x=1/9
=>x=5/18 hoặc x=1/18
8 tháng 7 2021
1) Ta có: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{-9}{8}-25\%\cdot\dfrac{-16}{5}\)
\(=\dfrac{4}{9}\cdot\dfrac{-9}{8}-\dfrac{1}{4}\cdot\dfrac{-16}{5}\)
\(=\dfrac{-1}{2}+\dfrac{4}{5}\)
\(=\dfrac{-5}{10}+\dfrac{8}{10}=\dfrac{3}{10}\)
2) Ta có: \(-1\dfrac{2}{5}\cdot75\%+\dfrac{-7}{5}\cdot25\%\)
\(=\dfrac{-7}{5}\cdot\dfrac{3}{4}+\dfrac{-7}{5}\cdot\dfrac{1}{4}\)
\(=\dfrac{-7}{5}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=-\dfrac{7}{5}\)
3) Ta có: \(-2\dfrac{3}{7}\cdot\left(-125\%\right)+\dfrac{-17}{7}\cdot25\%\)
\(=\dfrac{-17}{7}\cdot\dfrac{-5}{4}+\dfrac{-17}{7}\cdot\dfrac{1}{4}\)
\(=\dfrac{-17}{7}\cdot\left(\dfrac{-5}{4}+\dfrac{1}{4}\right)\)
\(=\dfrac{17}{7}\)
4) Ta có: \(\left(-2\right)^3\cdot\left(\dfrac{3}{4}\cdot0.25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
\(=\left(-8\right)\cdot\left(\dfrac{3}{4}\cdot\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=\left(-8\right)\cdot\dfrac{3}{16}:\dfrac{54-28}{24}\)
\(=\dfrac{-3}{2}\cdot\dfrac{24}{26}\)
\(=\dfrac{-72}{52}=\dfrac{-18}{13}\)
HC
1
16 tháng 6 2016
1) 31-3.(x+2) = 7
3.(x+2) = 31- 7
3.(x+2) = 24
x+2 = 24:3
x + 2 = 8
x = 8-2
x = 6
2) 175-25.(2x-1) = 175
25.(2x-1) = 175- 175 = 0
2x-1 = 0 : 25 = 0
2x = 0 + 1 = 1
x = 1 : 2 = 0,5
3) 25 - x/2 = 19
x/2 = 25-19
x/2 = 6
x : 2 = 6
x = 6 x 2 = 12
4) 2.(x-1)+3 = 15
2.(x-1) = 15 - 3 = 12
x - 1 = 12 : 2 = 6
x = 6+1 = 7
18 tháng 4 2015
tìm x biết:
(3x-1) [- 1/2x+5]=0
1/4+1/3:(2x-1)=-5
[2x+3/5]2 - 9/25=0
-5(x+1/5)-1/2(x-2/3)=3/2x - 5 /6
[x+1/2]x [2/3-2x]=0
17/2-|2x-3/4|=-7/4
2/3x-1/2x =5/12
(x+1/5)2+17/25=26/25
[x.44/7+3/7].11/5-3/7=-2
3[3x-1/2]+1/9=0
Toán lớp 6Tìm x
Trả lời Câu hỏi tương tự
Chưa có ai trả lời câu hỏi này,bạn hãy là người đâu tiên giúp nguyenvanhoang giải bài toán này !
11 tháng 12 2020
a) \(7x\left(x+1\right)-3\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(7x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\7x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{7}\end{matrix}\right.\)
b) 3(x + 8) - x2 - 8x = 0
=> 3(x + 8) - (x2 + 8x) = 0
=> 3(x + 8) - x(x + 8) = 0
=> (x + 8)(3 - x) = 0 => \(\left[{}\begin{matrix}x+8=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-8\\x=3\end{matrix}\right.\)
c) \(x^2-10x=-25\Rightarrow x^2-10x+25=0\Rightarrow\left(x-5\right)^2=0\Rightarrow x=5\)
d) Giống câu c
HO
19 tháng 8 2021
a)
b) 3(x + 8) - x2 - 8x = 0
=> 3(x + 8) - (x2 + 8x) = 0
=> 3(x + 8) - x(x + 8) = 0
=> (x + 8)(3 - x) = 0 =>
c)
15 tháng 4 2018
1,
\(\dfrac{6}{7}\) + \(\dfrac{-5}{8}\) : (-5) -\(\dfrac{3}{16}\) . \(\left(-2\right)^2\)
= \(\dfrac{6}{7}\) + \(\dfrac{-5}{8}\) . \(\dfrac{-1}{5}\) - \(\dfrac{3}{16}\) . 4
= \(\dfrac{6}{7}\) + \(\dfrac{1}{8}\) - \(\dfrac{3}{4}\)
= \(\dfrac{48}{56}\) + \(\dfrac{7}{56}\) - \(\dfrac{42}{56}\)
= \(\dfrac{13}{56}\)
3,
a,\(\dfrac{3}{5}\)x - \(\dfrac{7}{10}\)x = \(\dfrac{-1}{2}\)
x.(\(\dfrac{3}{5}\) - \(\dfrac{7}{10}\)) = \(\dfrac{-1}{2}\)
x.\(\dfrac{-1}{10}\)= \(\dfrac{-1}{2}\)
x =\(\dfrac{-1}{2}\):\(\dfrac{-1}{10}\)
x = 5
c, (2,5x - 3,6) : \(\dfrac{15}{7}\) = -1
( \(\dfrac{5}{2}\)x - \(\dfrac{18}{5}\)) : \(\dfrac{15}{7}\) = -1
\(\dfrac{5}{2}\)x - \(\dfrac{18}{5}\) = -1 . \(\dfrac{15}{7}\)
\(\dfrac{5}{2}\)x - \(\dfrac{18}{5}\) = \(\dfrac{-15}{7}\)
\(\dfrac{5}{2}\)x = \(\dfrac{-15}{7}\) + \(\dfrac{18}{5}\)
\(\dfrac{5}{2}\)x = \(\dfrac{51}{35}\)
x =\(\dfrac{51}{35}\) : \(\dfrac{5}{2}\)
x = \(\dfrac{102}{175}\)
Mình không biết có chỗ nào sai không nữa.
\(\left(x+\dfrac{1}{3}\right)^2=\dfrac{1}{25}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)^2=\left(\pm\dfrac{1}{5}\right)^2\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{1}{5}\\x+\dfrac{1}{3}=-\dfrac{1}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{15}\\x=-\dfrac{8}{15}\end{matrix}\right.\)
\(\left(x+\dfrac{1}{3}\right)^2=\dfrac{1}{25}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{1}{5}\\x+\dfrac{1}{3}=-\dfrac{1}{5}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{15}\\x=-\dfrac{8}{15}\end{matrix}\right.\)
Vậy phương trình có nghiệm \(S=\left\{-\dfrac{2}{15};-\dfrac{8}{15}\right\}\)