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-12(x-5)+7(3-x)=5
=> -12x+60+21-7x=5
=> -19x+81=5
=>-19x=5-81
=>-19x=-76
=>x=4
Vậy x=4
a, => -12+x-9 = 0
=> x-21=0
=> x=21
b, => -32-x+5 = 0
=> -x-27 = 0
=> -x=27
=> x=-27
c, => 11+15-x = 1
=> 26-x=1
=> x = 26-1 = 25
Tk mk nha
-12 + (x -9)=0 -32 - (x -5)=0 11 + (15- x )= 1
x -9 =0 +12 x -5 =0-32 15 - x =1 - 11
x -9 =12 x -5 = -32 15 -x =-10
x = 12+9 x = ( -32)+5 x = 15 +10
x =2 x = -27 x =25
a) \(\left(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.34}+...+\frac{11}{89.100}\right)-x=\frac{2}{3}\)
\(\left(1-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+\frac{1}{23}-\frac{1}{34}+...+\frac{1}{89}-\frac{1}{100}\right)-x=\frac{2}{3}\)
\(\left(1-\frac{1}{100}\right)-x=\frac{2}{3}\)
\(\frac{99}{100}-x=\frac{2}{3}\)
\(x=\frac{99}{100}-\frac{2}{3}\)
\(x=\frac{97}{300}\)
b) \(\frac{x+1}{9}+\frac{x+3}{7}+\frac{x+5}{5}+\frac{x+7}{3}=a\)
\(\Rightarrow\frac{x+1}{9}+1+\frac{x+3}{7}+1+\frac{x+5}{5}+1+\frac{x+7}{3}+1=a+4\)
\(\frac{x+10}{9}+\frac{x+10}{7}+\frac{x+10}{5}+\frac{x+10}{3}=a+4\)
\(\left(x+10\right).\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)=a+4\)
\(a,\left(\frac{11}{12}+\frac{11}{12\cdot23}+\frac{11}{23\cdot34}+...+\frac{11}{89\cdot100}\right)-x=\frac{2}{3}\)
\(\Rightarrow\left(1-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+\frac{1}{23}-\frac{1}{34}+...+\frac{1}{89}-\frac{1}{100}\right)-x=\frac{2}{3}\)
\(\Rightarrow1-\frac{1}{100}-x=\frac{2}{3}\)
\(\Rightarrow\frac{99}{100}-x=\frac{2}{3}\)
\(\Rightarrow x=\frac{99}{100}-\frac{2}{3}\)
\(\Rightarrow x=\frac{97}{300}\)
b, k hiểu đề :v
\(2x+3-3.2x+1=32\)
\(2x.\left(1-3\right)=32-1-3\)
\(2x.\left(-2\right)=28\)
\(2x=-14\)
\(x=-7\)
Ta có: \(2x+3-3\cdot2\left(x+1\right)=32\)
\(\Leftrightarrow2x+3-6x-6=32\)
\(\Leftrightarrow-4x=35\)
\(\Rightarrow x=-\frac{35}{4}\)
\(-32x+12x-5x=900\)
\(x\left(-32+12-5\right)=900\)
\(x\cdot\left(-25\right)=900\)
\(x=-36\)
\(\left(-32\right).x+12.x-5.x=900\)
\(\left(-32+12-5\right).x=900\)
\(-25x=900\)
\(x=900:\left(-25\right)\)
\(x=\left(-36\right)\)