10x-138=262
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(10x + 120) - 138 = 262
10x+120=262+138
10x+120=400
10x=400-120
10x=280
x=280:10
x=28
\(2\left(x-5\right)+10=20\)
\(\Leftrightarrow2\left(x-5\right)=10\)
\(\Leftrightarrow x-5=5\)
\(\Leftrightarrow x=10\)
\(\left(3x+1\right)-17=5\)
\(\Leftrightarrow3x+1=22\)
\(\Leftrightarrow3x=21\)
\(\Leftrightarrow x=7\)
\(\left(10x+120\right)-138=262\)
\(\Leftrightarrow10x+120=400\)
\(\Leftrightarrow10x=280\)
\(\Leftrightarrow x=28\)
\(4\left(x+1\right)-12=8\)
\(\Leftrightarrow4\left(x+1\right)=20\)
\(\Leftrightarrow x+1=5\)
\(\Leftrightarrow x=4\)
\(2x+2\frac{1}{3}=\frac{1}{2}:\frac{3}{2}\)
\(\Leftrightarrow2x+\frac{7}{3}=\frac{1}{3}\)
\(\Leftrightarrow2x=-2\)
\(\Leftrightarrow x=-1\)
\(2\left(x-5\right)+10=20\)
\(2x-10+10=20\)
\(2x=20\)
\(x=10\)
\(\left(3x+1\right)-17=5\)
\(3x+1-17=5\)
\(3x-16=5\)
\(3x=21\)
\(x=\frac{21}{3}=7\)
\(\left(10x+120\right)-138=262\)
\(10x+120-138=262\)
\(10x-18=262\)
\(10x=280\)
\(x=28\)
\(4\left(x+1\right)-12=8\)
\(4x+4-12=8\)
\(4x-8=8\)
\(4x=16\)
\(x=4\)
2.(x-5)+10 = 20
2.(x-5) = 20 - 10
2.(x-5) = 10
x-5 = 10 : 2
x-5 = 5
x = 5 + 5
x = 10
(3x+1)-17 = 5
3x+1 = 5 + 17
3x+1 = 22
3x = 22 - 1
3x = 21
x = 21 : 3
x = 7
(10x+120)-138 = 262
10x+120 = 262 + 138
10x+120 = 400
10x = 400-120
10x = 280
x = 280 : 10
x = 28
4.(x+1)-12 = 8
4.(x+1) = 8 + 12
4.(x+1) = 20
x+1 = 20 : 4
x+1 = 5
x = 5 - 1
x = 4
\(2x+2\frac{1}{3}=\frac{1}{2}:\frac{3}{2}\)
\(2x+\frac{7}{3}=\frac{1}{3}\)
\(2x=\frac{1}{3}-\frac{7}{3}\)
\(2x=-2\)
x = (-2) : 2
x = -1
Ta có : \(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{10}=\frac{y}{15}\)
\(\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{y}{15}=\frac{z}{21}\)
Do đó : \(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)
Ta có : \(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x+y+z}{10+15+21}=\frac{138}{46}=3\)
Nên : \(\frac{x}{10}=3\Rightarrow x=30\)
\(\frac{y}{15}=3\Rightarrow y=45\)
\(\frac{z}{21}=3\Rightarrow z=63\)
Vậy x = 30 ; y = 45 ; z = 63
10x-138=262
10x =262+138
10x =400
x = 400:10
x = 40