. Ai đó giúp tôi đi mà ._.

bài khó quá bạn ạ

a) \(\left(x+8\right)\left(x+6\right)=104+x^2\Leftrightarrow x^2+6x+8x+48=104+x^2\)

\(\Leftrightarrow x^2+6x+8x-x^2=104-48\Leftrightarrow14x=56\Leftrightarrow x=\dfrac{56}{14}=4\)

vậy \(x=4\)

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

\(\Leftrightarrow x^2+2x+x+2-\left(x^2+4x-3x-12\right)=6\)

\(\Leftrightarrow x^2+2x+x+2-x^2-4x+3x+12=6\)

\(\Leftrightarrow2x+14=6\Leftrightarrow2x=6-14=-8\Leftrightarrow x=\dfrac{-8}{2}=-4\)

vậy \(x=-4\)

c) \(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=5\)

\(\Leftrightarrow4x^2-20x-\left(4x^2-3x-4x+3\right)=5\)

\(\Leftrightarrow4x^2-20x-4x^2+3x+4x-3=5\)

\(\Leftrightarrow-13x-3=5\Leftrightarrow-13x=5+3=8\Leftrightarrow x=\dfrac{8}{-13}=\dfrac{-8}{13}\)

vậy \(x=\dfrac{-8}{13}\)

d) \(\left(3x-4\right)\left(x-2\right)=3x\left(x-9\right)-3\)

\(\Leftrightarrow3x^2-6x-4x+8=3x^2-27x-3\)

\(\Leftrightarrow3x^2-6x-4x-3x^2+27x=-3-8\)

\(\Leftrightarrow17x=-11\Leftrightarrow x=\dfrac{-11}{17}\) vậy \(x=\dfrac{-11}{17}\)

e) câu này đề bị thiếu rồi nha bn

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

\(\Leftrightarrow5x^2-15x=5x^2-x-10x+2-5\)

\(\Leftrightarrow5x^2-15x-5x^2+x+10x=2-5\)

\(\Leftrightarrow-4x=-3\Leftrightarrow x=\dfrac{-3}{-4}=\dfrac{3}{4}\) vậy \(x=\dfrac{3}{4}\)

a) \(\left(x+8\right)\left(x+6\right)=104+x^2\)

\(\Leftrightarrow x^2+14x+48=104+x^2\)

\(\Leftrightarrow14x=56\)

\(\Rightarrow x=4\)

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

\(\Leftrightarrow x^2+3x+2-x^2-7x+12=6\)

\(\Leftrightarrow-4x=-8\)

\(\Rightarrow x=2\)

c) \(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=5\)

\(\Leftrightarrow4x^2-20x-4x^2+3x+4x-3=5\)

\(\Leftrightarrow-13x=8\)

\(\Rightarrow x=\dfrac{-8}{13}\)

d) \(\left(3x-4\right)\left(x-2\right)=3x\left(x-9\right)-3\)

\(\Leftrightarrow3x^2-10x+8=3x^2-27x-3\)

\(\Leftrightarrow17x=-11\)

\(\Rightarrow x=\dfrac{-11}{17}\)

e) \(\left(x-5\right)\left(x-4\right)-\left(x+1\right)\left(x-2\right)=7\)

\(\Leftrightarrow x^2-9x+20-x^2+x+2=7\)

\(\Leftrightarrow-8x=-15\)

\(\Rightarrow x=\dfrac{15}{8}\)

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

\(\Leftrightarrow5x^2-15x=5x^2-11x+2-5\)

\(\Leftrightarrow-4x=-3\)

\(\Rightarrow x=\dfrac{3}{4}\)

\(a,2x-5=-x+4\\ \Leftrightarrow3x=9\\ \Leftrightarrow x=3\\ b,\left(4x-10\right)\left(25+5x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}4x-10=0\\25+5x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-5\end{matrix}\right.\\ c,\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-x\\ \Leftrightarrow\dfrac{2x}{6}-\dfrac{3\left(2x+1\right)}{6}-\dfrac{x}{6}+\dfrac{6x}{6}=0\\ \Leftrightarrow2x-6x-3-x+6x=0\\ \Leftrightarrow x-3=0\\ \Leftrightarrow x=3\)

d, ĐKXĐ:\(x\ne-2,x\ne3\)

\(1+\dfrac{x}{3-x}=\dfrac{5x}{\left(x+2\right)\left(3-x\right)}+\dfrac{2}{x+2}\\ \Leftrightarrow\dfrac{\left(x+2\right)\left(3-x\right)}{\left(x+2\right)\left(3-x\right)}+\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(3-x\right)}-\dfrac{5x}{\left(x+2\right)\left(3-x\right)}-\dfrac{2\left(3-x\right)}{\left(x+2\right)\left(3-x\right)}=0\\ \Leftrightarrow\dfrac{-x^2+x+6}{\left(x+2\right)\left(3-x\right)}+\dfrac{x^2+2x}{\left(x+2\right)\left(3-x\right)}-\dfrac{5x}{\left(x+2\right)\left(3-x\right)}-\dfrac{6-2x}{\left(x+2\right)\left(3-x\right)}=0\)

\(\Leftrightarrow\dfrac{-x^2+x+6+x^2+2x-5x-6+2x}{\left(x+2\right)\left(3-x\right)}=0\\ \Rightarrow0=0\left(luôn.đúng\right)\)

a)x^2-(a+b)x+ab

= x^2 - ax - bx + ab

= (x^2 - ax) - (bx - ab)

= x(x-a) - b(x-a)

= (x-b)(x-a) 

b)7x^3-3xyz-21x^2+9z

= 

c)4x+4y-x^2(x+y)

= 4(x + y) - x^2(x+y)

= (4-x^2) (x+y)

= (2-x)(2+x)(x+y)

d) y^2+y-x^2+x

= (y^2 - x^2) + (x+y)

= (y-x)(y+x)+ (x+y)

= (y-x+1) (x+y)

e)4x^2-2x-y^2-y

= [(2x)^2 - y^2] - (2x +y)

= (2x-y)(2x+y) - (2x+y)

= (2x -y -1)(2x+y)

f)9x^2-25y^2-6x+10y

= 

ko biết làm

c \(\frac{\left(2x-4\right)\left(x-3\right)}{\left(x-2\right)\left(3x^2-27\right)}=\frac{2\left(x-2\right)\left(x-3\right)}{3\left(x-2\right)\left(x^2-9\right)}\)

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

d, \(\frac{x^2+5x+6}{x^2+4x+4}=\frac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\frac{x+3}{x+2}\)

Tương tự với a ; b 

roois vãi

-Đăng tách câu hỏi bạn nhé.

rối thế bn