8x=7,8x+25 tim x
|x+7|-3=2
[2x-5]:3=7
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\(a.x=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}\)
\(b.x=\frac{4}{7}:\frac{4}{5}=\frac{5}{7}\)
\(c.8x-7,8x=25\)
\(0,2x=25\Rightarrow x=125\)
1. \(\frac{4}{5}\cdot x=\frac{4}{7}\Rightarrow x=\frac{4}{7}:\frac{4}{5}=\frac{20}{28}=\frac{5}{7}\)
Vậy \(x=\frac{5}{7}\)
2.\(8x=7,8x+25\Rightarrow8x=\frac{39}{5}x+25=1000\)
\(1000:8=125\)và \(\left(1000-25\right):\frac{39}{5}=125\)
\(\Rightarrow x=125\)
1/ \(1+\frac{2}{x-1}+\frac{1}{x+3}=\frac{x^2+2x-7}{x^2+2x-3}\)
ĐKXĐ: \(\hept{\begin{cases}x-1\ne0\\x+3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-3\end{cases}}\)
<=> \(1+\frac{2\left(x+3\right)+x-1}{\left(x-1\right)\left(x+3\right)}=\frac{x^2+2x-3-5}{x^2+2x-3}\)
<=> \(1+\frac{2x+6+x-1}{x^2+2x-3}=1-\frac{5}{x^2+2x-3}\)
<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=1-1\)
<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=0\)
<=> \(\frac{3x+10}{x^2+2x-3}=0\)
<=> \(3x+10=0\)
<=> \(x=-\frac{10}{3}\)
a) <=> 15-5x-20=-12-3
<=> -5x=-12-3-15+20=-10
=>x=-10:(-5)=2
b)<=>7-x-25-7=-25
<=> -x=-25-7+25+7=0 =>x=0
c) /x+2/=0 => x+2=0 =>x=-2
d) sai đề
e)<=> /x-5/ = 7
<=> \(\orbr{\begin{cases}x-5=7\\x-5=-7\end{cases}}\)
<=> \(\orbr{\begin{cases}x=12\\x=-2\end{cases}}\)
g) <=> -x-20-8+2x=-15
<=> x=-15+20+8=13
Bài làm :
\(1\text{)}8x=7,8x+25\Leftrightarrow8x-7,8x=25\Leftrightarrow0,2x=25\Leftrightarrow x=25\div0,2=125\)
\(2\text{)}\left|x+7\right|-3=2\Leftrightarrow\left|x+7\right|=5\Leftrightarrow\orbr{\begin{cases}x+7=5\\x+7=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-12\end{cases}}\)
\(3\text{)}\left(2x-5\right)\div3=7\Leftrightarrow2x-5=21\Leftrightarrow2x=26\Leftrightarrow x=13\)
a,Ta có : 8x=7,8x+25
<=> 8x-7,8x=25
0,2x=25
x=25:0,2
x=25:1/5
x=25.5
=> x=125
b, Ta có : |x+7|-3=2
<=> |x+7|=2+3
|x+7|=5
TH1: x+7=5 TH2: x+7=-5
x=5-7 x=(-5)-7
x=-2 x=-12
c, Ta có: [2x-5]:3=7
<=> 2x-5=7.3
2x-5=21
2x=21+5
2x=26
x=26:2
=> x=13