d: =>4x+6=15x-12

=>4x-15x=-12-6=-18

=>-11x=-18

hay x=18/11

e: =>\(45x+27=12+24x\)

=>21x=-15

hay x=-5/7

f: =>35x-5=96-6x

=>41x=101

hay x=101/41

g: =>3(x-3)=90-5(1-2x)

=>3x-9=90-5+10x

=>3x-9=10x+85

=>-7x=94

hay x=-94/7

làm rõ ra giúp với ạ, ghi v k hỉu j hết ;-;

Ta có: \(\frac{4}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}=\frac{4x-8+2x+4+5x-6}{4-x^2}=\frac{11x-2}{4-x^2}\)

\(\frac{4}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}\)

\(=\frac{4}{x+2}+\frac{2}{x-2}+\frac{6-5x}{x^2-4}\)

\(=\frac{4\left(x-2\right)+2\left(x+2\right)+6-5x}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{4x-8+2x+4+6-5x}{\left(x-2\right)\left(x+2\right)}\)

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

\(=\frac{1}{x-2}\)

\(b,\)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)

\(\Rightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)

\(\Rightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)

\(\Rightarrow\left(x+9\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\right)=\left(x+9\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)\)

\(\Rightarrow\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}=\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\left(KTM\right)\)

\(\text{Giải}\)

\(b,\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)

\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)

\(\Leftrightarrow x+2009=0\Leftrightarrow x=-2009\)

a) \(\frac{1}{x+2}+\frac{2}{x+3}=\frac{6}{x+4}\)

ĐKXĐ \(x\ne-2,-3,-4\)

=> \(\frac{1}{x+2}+\frac{2}{x+3}-\frac{6}{x+4}=0\)

=> \(\frac{3x+7}{\left(x+2\right)\left(x+3\right)}-\frac{6}{x+4}=0\)

=> \(\frac{\left(3x+7\right)\left(x+4\right)-6\left(x+2\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)\left(x+4\right)}=0\)

=> (3x + 7)(x + 4) - 6(x² + 5x + 6) = 0

=> 3x² + 19x + 28 - 6x² - 30x - 36 = 0

=> -3x² - 11x - 8 = 0

=> -3x² - 3x - 8x - 8 = 0

=> -3x(x + 1) - 8(x + 1) = 0

=> (x + 1)(-3x - 8) = 0

=> \(\orbr{\begin{cases}x=-1\\x=-\frac{8}{3}\end{cases}}\)

Vậy ...

b) Thiếu dữ liệu cuả đề 

c) \(\frac{6x+22}{x+2}-\frac{2x+7}{x+3}=\frac{x+4}{x^2+5x+6}\)

ĐKXĐ \(x\ne-2;-3\)

=> \(\frac{\left(6x+22\right)\left(x+3\right)-\left(x+2\right)\left(2x+7\right)}{\left(x+2\right)\left(x+3\right)}=\frac{x+4}{\left(x+2\right)\left(x+3\right)}\)

=> \(6x^2+40x+66-x\left(2x+7\right)-2\left(2x+7\right)=x+4\)

=> \(6x^2+40x+66-2x^2-7x-4x-14=x+4\)

=> 4x² + 29x + 52 = x + 4

=> 4x² + 29x + 52 - x - 4 = 0

=> 4x² + 28x + 48 = 0

=> 4(x² + 7x + 12) = 0

=> x² + 7x +12 = 0

=> x² + 3x + 4x + 12 = 0

=> x(x + 3) + 4(x + 3) = 0

=> (x + 3)(x + 4) = 0

=> \(\orbr{\begin{cases}x=-3\\x=-4\end{cases}}\) 

Mà \(x\ne-2,-3\)nên x = -3 loại

Vậy x = -4

1)\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3x}{\left(2x+6\right)x}-\frac{x-6}{2x^2+6x}\\ =\frac{3x}{2x^2+6x}-\frac{x-6}{2x^2+6x}=\frac{3x-\left(x-6\right)}{2x^2+6x}=\frac{2x+6}{x\left(2x+6\right)}=\frac{1}{x}\)

\(\frac{4}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}\)

\(=\frac{4}{x+2}+\frac{2}{x-2}-\frac{5x-6}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{4x-8+2x+4-5x+6}{\left(x-2\right)\left(x+2\right)}\)

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

\(=\frac{1}{x-2}\)

\(\frac{4}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}\)

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

\(=\frac{4x-8+2x+4-5x+6}{x^2-4}\)

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

điều kiện x khác {2,-2}

\(\frac{4}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}=\frac{4x-8+2x+4-5x+6}{x^2-4}\)

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

Đây là những bài cơ bản mà bạn!

bạn ấy muốn thách xem bạn nào đủ kiên nhẫn đánh hết chỗ này