Question

1: 3/x+1 + 2/x+2 = 5x+4/x²+ 3x + 2

2: 2/3x + 1 - 15/6x²-x-1 = 3/2x - 1

3: 9/3x - 1 - 5-x/3x²-4x+1 = 4/x+ 1

4:5/x - 2 + 2/x+4 = 3x/x² + 2x - 8

5: 4/x+6 + 1/x - 3 = 9/x² + 3x - 18

6:x/x-3 - 2x² +9/2x² - 3x - 9= 1/2x + 3

Answer 1

\(\frac{3}{x+1}+\frac{2}{x+2}=\frac{5x+4}{x^2+3x+2}.\)ĐKXĐ: \(x\ne-1;-2\)

\(\Leftrightarrow\frac{3\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}+\frac{2\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}=\frac{5x+4}{\left(x+1\right)\left(x+2\right)}\)

\(\Leftrightarrow3x+6+2x+2=5x+4\)

\(\Leftrightarrow3x+2x-5x=-6-2+4\)

\(\Leftrightarrow0x=-4\)

=> PT vô nghiệm 

\(2;\frac{2}{3x-1}-\frac{15}{6x^2-x-1}=\frac{3}{2x-1}\)

\(\Leftrightarrow\frac{2\left(2x-1\right)}{\left(2x-1\right)\left(3x-1\right)}-\frac{15}{6x^2+3x-2x-1}=\frac{3\left(3x-1\right)}{\left(2x-1\right)\left(3x-1\right)}\)

\(\Leftrightarrow\frac{4x-2-15}{\left(2x-1\right)\left(3x-1\right)}=\frac{9x-3}{\left(2x-1\right)\left(3x-1\right)}\)

\(\Leftrightarrow4x-2-15=9x-3\)

\(\Leftrightarrow4x-9x=2+15-3\)

\(\Leftrightarrow-5x=14\)

.....

Answer 2

mấy cái này mẫu nào dài cậu phân tích ra : 

VD : câu  3 : \(3x^2-4x+1\)

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

\(=3x\left(x-1\right)-\left(x-1\right)\)

\(=\left(3x-1\right)\left(x-1\right)\)

r bắt đầu giải PHương trình :)) Mấy câu còn lại tương tự 

Answer 3

4; \(\frac{5}{x-2}+\frac{2}{x+4}=\frac{3x}{x^2+2x-8}.\)

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

\(\Leftrightarrow5x+20+2x-4=3x\)

\(\Leftrightarrow4x=-16\Leftrightarrow x=-2\left(TM\right)\)

KL ::

\(5;\frac{4}{x+6}+\frac{1}{x-3}=\frac{9}{x^2+3x-18}\)

\(\Leftrightarrow\frac{4\left(x-3\right)}{\left(x+6\right)\left(x-3\right)}+\frac{x+6}{\left(x-3\right)\left(x+6\right)}=\frac{9}{\left(x-3\right)\left(x+6\right)}\)

\(\Leftrightarrow4x+x=3+9-6\)

\(\Leftrightarrow5x=6\Leftrightarrow x=\frac{6}{5}\)

Answer 4

\(6;\frac{x}{x-3}-\frac{2x^2+9}{2x^2-3x-9}=\frac{1}{2x-3}.\)

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

\(\Leftrightarrow2x^2+3x-2x^2-9=x-3\)

\(\Leftrightarrow3x-x=9-3\)

\(\Leftrightarrow2x=6\Leftrightarrow x=3\left(Loại\right)\)

=> Vô nghiệm