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1,2 . \(\left[\frac{2,4x-0,23}{x}-0,05\right]=1,44\)
\(\left[\frac{2,4x-0,23}{x}-0,05\right]=1,2\)
\(\frac{2,4x-0,23}{x}=1,25\)
Làm tiếp nha
a) 3/x-7 = 27/135
3/x-7 = 3/15
x - 7 = 15
x = 15 + 7
x = 22
a) \(\frac{3}{x-7}=\frac{27}{135}\)
\(\Rightarrow\)\(\left(x-7\right).27=3.135\)
\(\Rightarrow\)\(\left(x-7\right).27=405\)
\(\Rightarrow\)\(x-7=15\)
\(\Rightarrow\)\(x=22\)
Vậy \(x=22\)
b ) \(71+65.4=\frac{x+140}{x}+260\)
\(71+260=\frac{x+140}{x}+260\)
\(331=\frac{x+140}{x}+260\)
\(331-260=\frac{x+140}{x}\)
\(71=\frac{x+140}{x}\)
\(71=\frac{x}{x}+\frac{140}{x}\)
\(71=1+\frac{140}{x}\)
\(70=\frac{140}{x}\)
\(x=140\div70\)
\(x=20\)
Vậy \(x=20\)
#TQY
1,2 x ( 
- 0,05 ) = 1,44
(
- 0,05) = 1,44 : 1,2
- 0,05 = 1,2
= 1,2 + 0,05
= 1,25
2,4 x X – 0,23 = 1,25 x X
2,4 x X –1,25 x X = 0,23
X x (2,4 -1,25 ) = 0,23
X = 0,23 : 1,15
X= 0,2
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+....+\frac{1}{97.100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{97}-\frac{1}{100}\right)=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\cdot\left(1-\frac{1}{100}\right)=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\cdot\frac{99}{100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{33}{100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow x=\frac{0,33\times100}{0,33}=100\)
\(\left(x^2.y\right)^5.\left(x^2.y^2\right)^7.\left(x.y^2\right)^6.x^3\)
\(=x^{10}.y^5.x^{14}.y^{14}.x^6.y^{12}.x^3\)
\(=x^{33}.y^{31}\)
a) \(\left(\dfrac{1}{2}x-3\right)\left(-\dfrac{1}{3}+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-3=0\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=0+3\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3:\dfrac{1}{2}\\x=0-\left(-\dfrac{1}{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{1}{3}\end{matrix}\right.\)
d) \(9x^2=1\)
\(\Leftrightarrow x^2=1:9\)
\(\Leftrightarrow x^2=\dfrac{1}{9}\)
\(\Leftrightarrow x^2=\left(\dfrac{1}{3}\right)^2\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
Ta có: \(1,2\left(\dfrac{2,4y-0,23}{y}-0,05\right)=1,44\)
\(\Rightarrow\dfrac{2,4y-0,23}{y}-0,05=1,2\)
\(\Rightarrow\dfrac{2,4y-0,23}{y}=1,25\)
\(\Rightarrow2,4y-0,23=1,25y\)
\(\Rightarrow2,4y-1,25y=0,23\)
\(\Rightarrow1,15y=0,23\)
\(\Rightarrow y=0,2=\dfrac{1}{5}\)
Vậy \(y=\dfrac{1}{5}.\)
\(1,2.\left(\dfrac{2,4y-0,23}{y}-0,05\right)=1,44\)
\(\Rightarrow\dfrac{2,4y-0,23}{y}-0,05=1,2\)
\(\Rightarrow\dfrac{2,4y-0,23}{y}=1,25\)
\(\Rightarrow2,4y-0,23=1,25y\)
\(\Rightarrow2,4y-1,25y=0,23\)
\(\Rightarrow1,15y=0,23\)
\(\Rightarrow y=\dfrac{0,23}{1,15}=0,2\)
Vậy y = 0,2