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-3(x-7)=-2x+9
\(\Leftrightarrow\) -3x-(-3).7=-2x+9
\(\Leftrightarrow\)-3x-(-21)=-2x+9
\(\Leftrightarrow\) (-3x)-(-2x)=9+(-21)
\(\Leftrightarrow\) -1x=(-12)
\(\Leftrightarrow\)-x=(-12)
Số đối của x là (-12) \(\Rightarrow\) x=12
vậy x=12
A = -6 . ( -8 + 5 ) : 3 - 7 . ( -2 )
A = -6 . -3 : 3 - ( -14 )
A = 6 - ( -14 )
A = 20
\(\dfrac{1+x}{3-x}-\dfrac{1-2x}{3+x}-\dfrac{x\left(1-x\right)}{9-x^2}\)
\(=\dfrac{x^2+4x+3-\left(3-x\right)\left(1-2x\right)}{\left(3-x\right)\left(3+x\right)}-\dfrac{x\left(1-x\right)}{9-x^2}\)
\(=\dfrac{x^2+4x+3-3+7x-2x^2-x+x^2}{9-x^2}\)
\(=\dfrac{10x}{9-x^2}\)
ta có :
\(\dfrac{1+x}{3-x}-\dfrac{1-2x}{3+x}-\dfrac{x\cdot\left(1-x\right)}{9-x^2}\)
= \(\dfrac{\left(1+x\right)\cdot\left(3+x\right)}{\left(3-x\right)\cdot\left(3+x\right)}-\dfrac{\left(1-2x\right)\cdot\left(3-x\right)}{\left(3+x\right)\cdot\left(3-x\right)}-\dfrac{x\cdot\left(1-x\right)}{\left(3-x\right)\cdot\left(3+x\right)}\)
= \(\dfrac{3+x+3x+x^2-3+x+6x-2x^2-x+x^2}{\left(3+x\right)\cdot\left(3-x\right)}\)
= \(\dfrac{10x}{9-x^2}\)
bài lm có j sai sót mong bn thông cảm . 
Ta có: A= 3+3\(^2\)+3\(^3\)+3\(^4\)+3\(^5\)+3\(^6\)+3\(^7\)+3\(^8\)+3\(^9\)+3\(^{10}\)
\(\Rightarrow\)A= (3+3\(^2\)) +(3\(^3\)+3\(^4\))+(3\(^5\)+3\(^6\)) +(3\(^7\)+3\(^8\))+(3\(^9\)+3\(^{10}\))
\(\Rightarrow\) A= 12 + 3\(^2\)(3\(^1\)+3\(^2\))+3\(^4\)(3\(^1\)+3\(^2\)) +3\(^6\)(3\(^1\)+3\(^2\)) + 3\(^8\)(3\(^1\)+3\(^2\))
\(\Rightarrow\) A= 12 + 3\(^2\). 12+3\(^4\) . 12+3\(^6\) .12+ 3\(^8\) .12
\(\Rightarrow\)A= 12 . ( 3\(^2\)+3\(^4\) +3\(^6\)+ 3\(^8\))
Vì 12 \(⋮\)4 \(\Rightarrow\)12 . ( 3\(^2\)+3\(^4\) +3\(^6\)+ 3\(^8\)) \(⋮\)4 hay A \(⋮\)4
\(\frac{6}{x-3}=\frac{9}{2x-7}\)
\(\Leftrightarrow9\left(x-3\right)=6\left(2x-7\right)\)
\(9x-27=12x-42\)
\(9x-27-12x+42=0\)
\(9x-12x-27+42=0\)
\(-3x+15=0\)
\(-3x=0-15\)
\(-3x=-15\)
\(x=\left(-15\right)\div\left(-3\right)\)
\(x=5\)
\(\frac{6}{x-3}=\frac{9}{2x-7}\left(đkxđ:x\ne3;\frac{7}{2}\right)\)
\(< =>6\left(2x-7\right)=9\left(x-3\right)\)
\(< =>12x-42=9x-27\)
\(< =>12x-9x=-27+42\)
\(< =>3x=15< =>x=\frac{15}{3}=5\)