a: =>3,6-1,7x=2,3-1,4-4=0,9-4=-3,1

=>1,7x=6,7

hay x=67/17

b: \(\Leftrightarrow30\left(5x+4\right)-15\left(3x+5\right)=24\left(4x+9\right)-40\left(x-9\right)\)

=>150x+120-45x-75=96x+216-40x+360

=>105x+45=56x+576

=>49x=531

hay x=531/49

a: \(\Leftrightarrow7\left(7-3x\right)+12\left(5x+2\right)=84\left(x+13\right)\)

\(\Leftrightarrow49-21x+60x+24=84x+1092\)

\(\Leftrightarrow39x-84x=1092-73\)

=>-45x=1019

hay x=-1019/45

b: \(\Leftrightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)

=>21x+63-14=20x+36-49x+63

=>21x+49=-29x+99

=>50x=50

hay x=1

c: \(\Leftrightarrow7\left(2x+1\right)-3\left(5x+2\right)=21x+63\)

=>14x+7-15x-6-21x-63=0

=>-22x-64=0

hay x=-32/11

d: \(\Leftrightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-17\cdot105\)

=>70x-105-30x-45=84x+63-1785

=>40x-150-84x+1722=0

=>-44x+1572=0

hay x=393/11

a, msc 12.7=84 

Chuyển vế về =0 rồi làm

b,msc 28

c,làm tương tự

a, \(\Rightarrow49-21x+60x+24=84x+1092\)

\(\Leftrightarrow-45x=1019\Leftrightarrow x=-\dfrac{1019}{45}\)

b, \(\Rightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)

\(\Leftrightarrow21x+63-14=20x+36-49x+63\)

\(\Leftrightarrow50x=50\Leftrightarrow x=1\)

c, \(\Rightarrow14x+7-15x-6=21x+63\Leftrightarrow-22x=62\Leftrightarrow x=-\dfrac{31}{11}\)

d, \(\Rightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-105.17\)

\(\Leftrightarrow70x-105-30x-45=84x+63-1785\)

\(\Leftrightarrow-44x=-1572\Leftrightarrow x=\dfrac{393}{11}\)

b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)

\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)

\(=\dfrac{2y^2+8y+12}{y-1}\)

a) ĐKXĐ: x≠-5

Ta có: \(\dfrac{2x-5}{x+5}=4\)

\(\Leftrightarrow2x-5=4\left(x+5\right)\)

\(\Leftrightarrow2x-5=4x+20\)

\(\Leftrightarrow2x-5-4x-20=0\)

\(\Leftrightarrow-2x-25=0\)

\(\Leftrightarrow-2x=25\)

hay \(x=\dfrac{-25}{2}\)(nhận)

Vậy: \(S=\left\{-\dfrac{25}{2}\right\}\)

b) ĐKXĐ: x≠0

Ta có: \(\dfrac{x^2-4}{x}=\dfrac{2x+3}{2}\)

\(\Leftrightarrow2\left(x^2-4\right)=x\left(2x+3\right)\)

\(\Leftrightarrow2x^2-8=2x^2+3x\)

\(\Leftrightarrow2x^2-8-2x^2-3x=0\)

\(\Leftrightarrow-3x-8=0\)

\(\Leftrightarrow-3x=8\)

hay \(x=\dfrac{-8}{3}\)(nhận)

Vậy: \(S=\left\{-\dfrac{8}{3}\right\}\)

c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{2};-5\right\}\)

Ta có: \(\dfrac{2x+3}{2x-1}=\dfrac{x-3}{x+5}\)

\(\Leftrightarrow\left(2x+3\right)\left(x+5\right)=\left(2x-1\right)\left(x-3\right)\)

\(\Leftrightarrow2x^2+10x+3x+15=2x^2-6x-x+3\)

\(\Leftrightarrow2x^2+13x+15=2x^2-7x+3\)

\(\Leftrightarrow2x^2+13x+15-2x^2+7x-3=0\)

\(\Leftrightarrow20x+12=0\)

\(\Leftrightarrow20x=-12\)

hay \(x=-\dfrac{3}{5}\)(nhận)

Vậy: \(S=\left\{-\dfrac{3}{5}\right\}\)

d) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)

Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)

\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(x+7\right)\left(6x+1\right)\)

\(\Leftrightarrow6x^2-9x-4x+6=6x^2+x+42x+7\)

\(\Leftrightarrow6x^2-13x+6=6x^2+43x+7\)

\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)

\(\Leftrightarrow-56x-1=0\)

\(\Leftrightarrow-56x=1\)

hay \(x=-\dfrac{1}{56}\)(nhận)

Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)

\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)

a) Ta có: \(\left(x-1\right)\left(x-2\right)\left(x^2+x+1\right)\left(x^2+2x+4\right)-x^6+9x^3\)

\(=\left(x-1\right)\left(x^2+x+1\right)\left(x-2\right)\left(x^2+2x+4\right)-x^6+9x^3\)

\(=\left(x^3-1\right)\left(x^3-8\right)-x^6+9x^3\)

\(=x^6-9x^3+8-x^6+9x^3=8\)

b) Ta có: \(\left(\dfrac{1}{3}+2x\right)\left(\dfrac{1}{9}-\dfrac{2}{3}x+4x^2\right)-\left(2x-\dfrac{1}{3}\right)\left(4x^2+\dfrac{2}{3}x+\dfrac{1}{4}\right)\)

\(=\dfrac{1}{27}+8x^3-8x^3+\dfrac{1}{27}\)

\(=\dfrac{2}{27}\)

c) Ta có: \(\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(1-x\right)\)

\(=x^3-3x^2+3x-1-x^3+1-3x+3x^2\)

=0

d) Ta có: \(\left(x^2-y^2\right)\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)-x^6+y^6\)

\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)-x^6+y^6\)

\(=\left(x^3-y^3\right)\left(x^3+y^3\right)-x^6+y^6\)

\(=x^6-y^6-x^6+y^6=0\)

\(a,6x-4=5x\\ \Leftrightarrow x-4=0\\ \Leftrightarrow x=4\\ b,\dfrac{2x+3}{3}=\dfrac{5-4x}{2}\\ \Leftrightarrow2\left(2x+3\right)=3\left(5-4x\right)\\ \Leftrightarrow4x+6=15-12x\\ \Leftrightarrow16x-9=0\\ \Leftrightarrow x=\dfrac{9}{16}\\ c,\left(x+7\right)\left(x-10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-10=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-7\\x=10\end{matrix}\right.\)

d, ĐKXĐ:\(x\ne\pm3\)

\(\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{3x+5}{x^2-9}\\ \Leftrightarrow\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{3x+5}{\left(x+3\right)\left(x-3\right)}=0\\ \Leftrightarrow\dfrac{2x+6+3x-9-3x-5}{\left(x+3\right)\left(x-3\right)}=0\\ \Rightarrow2x-8=0\\ \Leftrightarrow x=4\left(tm\right)\)

a.6x-4=5x <=> x=4

b.\(\dfrac{2x+3}{3}=\dfrac{5-4x}{2}\)

\(\Leftrightarrow\dfrac{2\left(2x+3\right)}{6}=\dfrac{3\left(5-4x\right)}{6}\)

\(\Leftrightarrow2\left(2x+3\right)=3\left(5-4x\right)\)

\(\Leftrightarrow4x+6=15-12x\)

\(\Leftrightarrow16x=11\)

\(\Leftrightarrow x=\dfrac{11}{16}\)

c.(x+7)(x-10)=0

\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-10=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=10\end{matrix}\right.\)

d.\(ĐK:x\ne\pm3\)

\(\Rightarrow\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{3x+5}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow\dfrac{2\left(x+3\right)+3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x+5}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow2\left(x+3\right)+3\left(x-3\right)=3x+5\)

\(\Leftrightarrow2x+6+3x-9-3x-5=0\)

\(\Leftrightarrow2x-8=0\)

\(\Leftrightarrow2x=8\)

\(\Leftrightarrow x=4\left(tm\right)\)

a, \(6x-5x=4\Leftrightarrow x=4\)

b, \(4x+6=15-12x\Leftrightarrow16x=9\Leftrightarrow x=\dfrac{9}{16}\)

c, \(\left[{}\begin{matrix}x+7=0\\x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=10\end{matrix}\right.\)

d, đk : x khác -3 ; 3 

\(2x+6+3x-9=3x+5\Leftrightarrow2x=8\Leftrightarrow x=4\left(tmđk\right)\)