(x-1/5)2=4/25
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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: \(\left(-2\right)^3\cdot\left(\dfrac{3}{4}\cdot0.25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
\(=\left(-8\right)\cdot\left(\dfrac{3}{4}\cdot\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=\left(-8\right)\cdot\dfrac{3}{16}:\dfrac{54-28}{24}\)
\(=\dfrac{-3}{2}\cdot\dfrac{24}{26}\)
\(=\dfrac{-72}{52}=\dfrac{-18}{13}\)
a: \(=\dfrac{5}{4}\cdot\dfrac{2}{5}\cdot\dfrac{17}{21}\cdot\dfrac{7}{34}=\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot\dfrac{1}{3}=\dfrac{1}{12}\)
b: =>0,5^x(0,5+1)=1,5
=>0,5^x=1
=>x=0
c: =>x*0,05-0,25*x=-1,2
=>-0,3*x=-1,2
=>x=4
d: =>x-1=1 hoặc x-1=-1
=>x=0 hoặc x=2
e: =5+4=9
1. 444 x 5 = 222 x 2 x 5 = 222 x 10 = 2220
a.Đúng b.Sai
2. 282 x 5 = 280 + 2 x 5 = 280 x 10 = 2800
a.Đúng b. Sai
3. 4 x 8 x 7 x 25 = (8 x 7) x (25 x4) = 56 x 100 = 5600
a.Đúng b.Sai
4. 25 x 8 x 9 = (25 x 4) x (4 x 9) = 100 x 36 = 3600
a.Đúng b.Sai
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 !
Tìm X
a) \(2x+\dfrac{3}{24}=3x-\dfrac{1}{32}\)
\(\Leftrightarrow\left(2x+\dfrac{3}{24}\right)-\left(3x-\dfrac{1}{32}\right)=0\)
\(\Leftrightarrow2x+\dfrac{3}{24}-3x+\dfrac{1}{32}=0\)
\(\Leftrightarrow\left(\dfrac{3}{24}+\dfrac{1}{32}\right)+\left(2x-3x\right)=0\)
\(\Leftrightarrow\dfrac{5}{32}-x=0\)
\(\Leftrightarrow x=\dfrac{5}{32}\)
g. \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{3}\end{cases}}\)
Vậy \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)
f. \(\frac{2}{3}x-\frac{1}{2}x=\frac{5}{12}\)
\(\Leftrightarrow x\left(\frac{2}{3}-\frac{1}{2}\right)=\frac{5}{12}\)
\(\Leftrightarrow x\left(\frac{4}{6}-\frac{3}{6}\right)=\frac{5}{12}\)
\(\Leftrightarrow\frac{1}{6}x=\frac{5}{12}\)
\(\Leftrightarrow x=\frac{5}{12}\div\frac{1}{6}\)
\(\Leftrightarrow x=\frac{30}{12}=\frac{5}{2}\)
<=> x=-1/5
<=> x=3/5
ok nhé
\(\left(x-\frac{1}{5}\right)^2=\frac{4}{25}\)
\(\left(x-\frac{1}{5}\right)^2=\left(\pm\frac{2}{5}\right)^2\)
\(\Rightarrow x-\frac{1}{5}=\frac{2}{5}\text{ hoặc }x-\frac{1}{5}=\left(-\frac{2}{5}\right)\)
\(x=\frac{3}{5}\) \(x=-\frac{1}{5}\)
Vậy x = ....