Mình ms lớp 7 neh, mình giải theo ý hiểu của mình thôi nha :

Có \(\frac{5x}{\left(x-2\right)\left(x+3\right)}=\frac{a}{x-2}+\frac{b}{x+3}=\frac{a\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}+\frac{b\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)

\(\Leftrightarrow\frac{5x}{\left(x-2\right)\left(x+3\right)}=\frac{ax+3a}{\left(x-2\right)\left(x+3\right)}+\frac{bx-2b}{\left(x-2\right)\left(x+3\right)}=\frac{ax+3a+bx-2b}{\left(x-2\right)\left(x+3\right)}\)

\(\Leftrightarrow5x=ax+3a+bx-2b\Leftrightarrow5x-\left(ax+bx\right)=3a-2b\)

\(\Leftrightarrow5x-ax-bx=3a-2b\Leftrightarrow x\left(5-a-b\right)=3a-2b\)

Em lậy a Minh ạ,...

Cái deck j thê snày ? 

\(\frac{a\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}+\frac{b\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)

Thôi thôi mắt cận thì đừng cs đụng tay vào cái đấy của ng ta là x + 3 đấy !

\(\Leftrightarrow\left(x+1\right)\left(x^4+x^3+2x^2+x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)[x^2\left(x^2+x+1\right)+x^2+x+1]=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)\left(x^2+x+1\right)=0\)

\(=>x=-1\)

Học tốt

\(\left(2x-5\right)^3-\left(3x-4\right)^3+\left(x+1\right)^3\)

\(=\left(2x-5-3x+4\right)\left[\left(2x-5\right)^2+\left(2x-5\right)\left(3x-4\right)+\left(3x-4\right)^2\right]+\left(x+1\right)^3\)

\(=-\left(x+1\right)\left(4x^2-20x+25+6x^2-23x+20+9x^2-24x+16\right)+\left(x+1\right)^3\)

\(=-\left(x+1\right)\left(19x^2-67x+61\right)+\left(x+1\right)\left(x^2+2x+1\right)=\)

\(-\left(x+1\right)\left(18x^2-69x+60\right)=0\Rightarrow\left(x+1\right)\left(6x^2-23x+20\right)=0.....\)

giúp mik vs

ĐKXĐ:\(x\ne1\)

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

\(\Leftrightarrow\frac{x^2+x+1+2\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}\)

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

\(\Leftrightarrow x^2+3x-1=3x^2\)\(\Leftrightarrow2x^2-3x+1=0\)

\(\Leftrightarrow2x^2-2x-x+1=0\)\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\left(KTMĐK\right)\\x=\frac{1}{2}\left(TMĐK\right)\end{cases}}}\)

Vậy nghiệm của pt là \(x=\frac{1}{2}\)

\(a,2x\left(x+5\right)=x+5\)

\(2x^2+10x=x+5\)

\(2x^2+10x-x-5=0\)

\(2x^2+9x-5=0\)

\(2x^2+x-10x-5=0\)

\(x\left(2x+1\right)-5\left(2x+1\right)=0\)

\(\left(x-5\right)\left(2x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\2x=-1\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-\frac{1}{2}\end{cases}}}\)

\(pt\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x^2+3x-1}{x^3-1}=\frac{3x^2}{x^3-1}\)

\(\Rightarrow x^2+3x-1=3x^2\Leftrightarrow3x-1=2x^2\Leftrightarrow2x^2-3x+1=0\Leftrightarrow x^2-\frac{3}{2}x+\frac{1}{2}=0\)

đến đây là pt bậc 2

\(\left(2x+1\right)^2-3\left(x-1\right)^2-\left(x+1\right)\left(x-1\right)\)

\(=\left(2.\left(-\frac{1}{2}\right)+1\right)^2-3\left(-\frac{1}{2}-1\right)^2-\left(-\frac{1}{2}+1\right)\left(-\frac{1}{2}-1\right)\)

\(=-3\left(-\frac{9}{4}\right)-\frac{1}{2}.\left(-\frac{3}{2}\right)\)

\(=\frac{27}{4}+\frac{3}{4}=\frac{31}{4}\)

còn đâu tự lm nha ! 

\(x\left(x-y\right)+y\left(y-x\right)=x\left(x-y\right)-y\left(x-y\right)\)

\(=\left(x-y\right)\left(x-y\right)=\left(x-y\right)^2=\left(124-24\right)^2=100^2=10000\)

\(\left(x-3\right)^2-\left(x+1\right)^3+12x\left(x-1\right)=\frac{49}{4}-\frac{1}{8}+\frac{\left(-6\right).\left(-3\right)}{2}\)

\(=\frac{97}{8}+9=\frac{169}{8}\)

X(X-Y)+Y(Y-X)=X² -XY +Y² -XY=(X-Y)² =(124-24)² =100² =10000

(x-3)² -(x+1)³ +12x(x-1)=x² -6x+9-x³ -3x² -3x-1+12x² -12x=-x³ +10x² -9x+8