a) Ta có: \(\frac{3x-11}{11}-\frac{x}{3}=\frac{3x-5}{7}-\frac{5x-3}{9}\)

\(\Leftrightarrow\frac{63\left(3x-11\right)}{693}-\frac{231x}{693}-\frac{99\left(3x-5\right)}{693}+\frac{77\left(5x-3\right)}{693}=0\)

\(\Leftrightarrow189x-693-231x-297x+495+385x-231=0\)

\(\Leftrightarrow46x-429=0\)

\(\Leftrightarrow46x=429\)

hay \(x=\frac{429}{46}\)

Vậy: \(x=\frac{429}{46}\)

b) Ta có: \(\frac{9x-0,7}{4}-\frac{5x-1,5}{7}=\frac{7x-1,1}{6}-\frac{5\left(0,4-2x\right)}{5}\)

\(\Leftrightarrow\frac{9x-0,7}{4}-\frac{5x-1,5}{7}-\frac{7x-1,1}{6}+\frac{5\left(0,4-2x\right)}{5}=0\)

\(\Leftrightarrow105\left(9x-0,7\right)-60\left(5x-1,5\right)-70\left(7x-1,1\right)+420\left(0,4-2x\right)=0\)

\(\Leftrightarrow945x-\frac{147}{2}-300x+90-490x+77+168-840x=0\)

\(\Leftrightarrow-685x+261.5=0\)

\(\Leftrightarrow-685x=-261.5\)

hay \(x=\frac{523}{1370}\)

Vậy: \(x=\frac{523}{1370}\)

c) Ta có: \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x-1\right)}{7}-5\)

\(\Leftrightarrow\frac{14\left(5x-3\right)}{84}-\frac{21\left(7x-1\right)}{84}-\frac{24\left(2x-1\right)}{84}+\frac{420}{84}=0\)

\(\Leftrightarrow70x-42-147x+21-48x+24+420=0\)

\(\Leftrightarrow-125x+423=0\)

\(\Leftrightarrow-125x=-423\)

hay \(x=\frac{423}{125}\)

Vậy: \(x=\frac{423}{125}\)

d) Ta có: \(14\frac{1}{2}-\frac{2\left(x+3\right)}{5}=\frac{3x}{2}-\frac{2\left(x-7\right)}{3}\)

\(\Leftrightarrow\frac{435}{30}-\frac{12\left(x+3\right)}{30}-\frac{45x}{30}+\frac{20\left(x-7\right)}{30}=0\)

\(\Leftrightarrow435-12x-36-45x+20x-140=0\)

\(\Leftrightarrow-37x+259=0\)

\(\Leftrightarrow-37x=-259\)

hay \(x=7\)

Vậy: x=7

bài 1

a) \(-\frac{1}{3}xy\).(3\(x^2yz^2\))

=\(\left(-\frac{1}{3}.3\right)\).\(\left(x.x^2\right)\).(y.y).\(z^2\)

=\(-x^3\).\(y^2z^2\)

b)-54\(y^2\).b.x

=(-54.b).\(y^2x\)

=-54b\(y^2x\)

c) -2.\(x^2y.\left(\frac{1}{2}\right)^2.x.\left(y^2.x\right)^3\)

=\(-2x^2y.\frac{1}{4}.x.y^6.x^3\)

=\(\left(-2.\frac{1}{4}\right).\left(x^2.x.x^3\right).\left(y.y^2\right)\)

=\(\frac{-1}{2}x^6y^3\)

Bài 3:

a) \(f\left(x\right)=-15x^2+5x^4-4x^2+8x^2-9x^3-x^4+15-7x^3\)

\(f\left(x\right)=\left(5x^4-x^4\right)-\left(9x^3+7x^3\right)-\left(15x^2+4x^2-8x^2\right)+15\)

\(f\left(x\right)=4x^4-16x^3-11x^2+15\)

b) 

\(f\left(x\right)=4x^4-16x^3-11x^2+15\)

\(f\left(1\right)=4\cdot1^4-16\cdot1^3-11\cdot1^2+15\)

\(f\left(1\right)=4\cdot1^4-16\cdot1^3-11\cdot1^2+15\)

\(f\left(1\right)=-8\)

\(f\left(x\right)=4x^4-16x^3-11x^2+15\)

\(f\left(-1\right)=4\cdot\left(-1\right)^4-16\cdot\left(-1\right)^3-11\cdot\left(-1\right)^2+15\)

\(f\left(-1\right)=24\)

Bài 1:

a) \(-\frac{1}{3}xy\cdot\left(3x^2yz^2\right)\)

\(=\left(-\frac{1}{3}\cdot3\right)\left(xx^2\right)\left(yy\right)z\)

\(=-x^3y^2z\)

b) \(-54y^2\cdot bx\)

\(=\left(-54b\right)xy^2\)

c) \(-2x^2y\cdot\left(\frac{1}{2}\right)^2\cdot x\cdot\left(y^2x\right)^3\)

\(=-2x^2y\cdot\frac{1}{4}\cdot x\cdot y^5x^3\)

\(=\left(-2\cdot\frac{1}{4}\right)\left(x^2xx^3\right)\left(yy^5\right)\)

\(=-\frac{1}{2}x^6y^6\)

1) \(A\left(x\right)=-5x^3+3x^4+\frac{5}{7}-8x^2-10x\)

\(A\left(x\right)=3x^4-5x^3-8x^2-10x+\frac{5}{7}\)

\(B\left(x\right)=-2x^4-\frac{2}{7}+7x^2+8x^3+6x\)

\(B\left(x\right)=-2x^4+8x^3+7x^2+6x-\frac{2}{7}\)

2)       \(A\left(x\right)=3x^4-5x^3-8x^2-10x+\frac{5}{7}\)

      +

          \(B\left(x\right)=-2x^4+8x^3+7x^2+6x-\frac{2}{7}\)

\(A\left(x\right)+B\left(x\right)=x^4+3x^3-x^2-4x+\frac{3}{7}\)

                \(A\left(x\right)=3x^4-5x^3-8x^2-10x+\frac{5}{7}\)

-

                \(B\left(x\right)=-2x^4+8x^3+7x^2+6x-\frac{2}{7}\)

\(A\left(x\right)-B\left(x\right)=5x^4-13x^3-15x^2-16x+1\)

\(\frac{25x-655}{95}-\frac{5\left(x-12\right)}{209}=\frac{89-3x-\frac{2\left(x-18\right)}{5}}{11}\)

\(< =>\frac{5x-131}{19}=\frac{1631-52x-\frac{38x-684}{5}}{209}\)

\(< =>\left(5x-131\right)209=\left(1631-52x-\frac{38x-684}{5}\right)19\)

\(< =>55x-1441=1631-52x-\frac{38x-684}{5}\)

\(< =>3072-107x=\frac{38x-684}{5}\)

\(< =>\left(3072-107x\right)5=38x-684\)

\(< =>15360-535x-38x-684=0\)

\(< =>14676=573x< =>x=\frac{14676}{573}=\frac{4892}{191}\)

nghệm xấu thế 

\(\frac{8\left(x+22\right)}{45}-\frac{7x+149+\frac{6\left(x+12\right)}{5}}{9}=\frac{x+35+\frac{2\left(x+50\right)}{9}}{5}\)

\(< =>\frac{8x+176}{45}-\frac{41x+817}{45}=\frac{11x+415}{45}\)

\(< =>993-33x-11x-415=0\)

\(< =>578=44x< =>x=\frac{289}{22}\)

Bài 1: 

b) Phương trình đã cho tương đương với phương trình:

\(\frac{8\left(x+22\right)-55\left(7x+149\right)-6\left(x+12\right)}{45}=\frac{9\left(x+35\right)+2\left(x+50\right)}{45}\)

\(\Leftrightarrow44x=-1056\)

\(\Leftrightarrow x=-24\)

Vậy x=-24 là nghiệm của phương trình

c) Phương trình đã cho tương đương với phương trình:

\(\frac{3x+6}{70}-\frac{x+4}{24}=\frac{32x+19}{60}+\frac{2}{3}\)

\(\Leftrightarrow12\left(3x+6\right)-35\left(x+4\right)=14\left(32x+19\right)+560\)

\(\Leftrightarrow-447x=894\)

\(\Leftrightarrow x=-2\)

Vậy x=-2 là nghiệm của phương trình

giup minh voi cac bạn

Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\) 

 \(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)

\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\) 

 \(\frac{x+2}{2-4x+2x^2}=\frac{x+2}{2\left(x-1\right)^2}=\frac{2x\left(x+2\right)}{2x\left(x-1\right)^2}\)

\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\) 

\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)

\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)

\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)