ĐKXĐ: ...

\(\Leftrightarrow\frac{2x}{3x^2-4x+1}-\frac{7x}{3x^2+2x+1}=6\)

\(\Leftrightarrow\frac{2}{3x-4+\frac{1}{x}}-\frac{7}{3x+2+\frac{1}{x}}=6\)

Đặt \(3x-4+\frac{1}{x}=a\)

\(\frac{2}{a}-\frac{7}{a+6}=6\)

\(\Leftrightarrow2\left(a+6\right)-7a=6a\left(a+6\right)\)

\(\Leftrightarrow6a^2+41a-12=0\)

Nghiệm xấu, bạn coi lại đề

GPT

\(\frac{3}{3x^2-4x+1}+\frac{13}{3x^2+2x+1}=\frac{6}{x}\)

\(x=0\) không phải nghiệm, pt tương đương:

\(\frac{12}{x+4+\frac{2}{x}}-\frac{3}{x+2+\frac{2}{x}}=1\)

Đặt \(x+2+\frac{2}{x}=a\)

\(\frac{12}{a+2}-\frac{3}{a}=1\Leftrightarrow12a-3\left(a+2\right)=a\left(a+2\right)\)

\(\Leftrightarrow a^2-7a+6=0\Rightarrow\left[{}\begin{matrix}a=1\\a=6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+2+\frac{2}{x}=1\\x+2+\frac{2}{x}=6\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2+x+2=0\\x^2-4x+2=0\end{matrix}\right.\)

a, \(5\left(m+3x\right)\left(x+1\right)-4\left(1+2x\right)=80\)

Phương trình nhận \(x=2\)làm nghiệm nên :

\(5\left(m+3.2\right)\left(2+1\right)-4\left(1+2.2\right)=80\)

\(\Leftrightarrow15m+90-20=80\)

\(\Leftrightarrow15m=80+20-90\)

\(\Leftrightarrow15m=10\Leftrightarrow m=1,5\)

....

b, \(3\left(2x+m\right)\left(3x+2\right)-2\left(3x+1\right)^2=43\)

Phương trình nhận \(x=1\)làm nghiệm nên :

\(3\left(2.1+m\right)\left(3.1+2\right)-2\left(3.1+1\right)^2=43\)

\(\Leftrightarrow30+15m-32=43\)

\(\Leftrightarrow15m=43+32-30\)

\(\Leftrightarrow15m=45\Leftrightarrow m=3\)

....

\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)

\(\Leftrightarrow\frac{315-x}{101}+1+\frac{313-x}{103}+1+\frac{311-x}{105}+1+\frac{309-x}{107}+1=0\)

\(\Leftrightarrow\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)

\(\Leftrightarrow\left(416-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)

\(\Leftrightarrow416-x=0\)

\(\Leftrightarrow x=416\)

a) 5(m + 3x)(x + 1) - 4(1 + 2x) = 80

Phương trình có nghiệm x = 2:

5(m + 3.2)(2 + 1) - 4(1 + 2.2) = 80

<=> 5(m + 6).3 - 4.5 = 80

<=> 15(m + 6) - 4.5 = 80

<=> 15(m + 6) - 20 = 80

<=> 15(m + 6) = 80 + 20

<=> 15(m + 6) = 100

<=> m + 6 = 100 : 15

<=> m + 6 = 20/3

<=> m = 20/3 - 6

<=> m = 2/3

b) 3(2x + m)(3x + 2) - 2(3x + 1)² = 43

Phương trình có nghiệm x = 1:

3(2.1 + m)(3.1 + 2) - 2(3.1 + 1)² = 43

<=> 3(2 + m).5 - 2.16 = 43

<=> 15(2 + m) - 32 = 43

<=> 15(2 + m) = 43 + 32

<=> 15(2 + m) = 75

<=> 2 + m = 75 : 15

<=> 2 + m = 5

<=> m = 5 - 2

<=> m = 3

Phương trình trên 

\(< =>\frac{315-x}{101}+1+\frac{313-x}{103}+1+\frac{311-x}{105}+1+\frac{319-x}{107}+1=0\)

\(< =>\frac{315+101-x}{101}+\frac{313+103-x}{103}+\frac{311+105-x}{105}+\frac{309+107-x}{107}=0\)

\(< =>\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)

\(< =>\left(416-x\right).\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)

Do \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\ne0\)

\(< =>416-x=0\)

\(< =>x=416\)

\(\frac{3x-2}{x+7}=\frac{6x+1}{2x-3}\) (Đkxđ: \(x\ne-7;x\ne\frac{3}{2}\))

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

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

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

\(\Leftrightarrow-56x=1\)

\(\Leftrightarrow x=-\frac{1}{56}\) (t/m đkxđ)

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

ĐKXĐ: x khác -7 và 3/2

Từ đề bài <=> (3x-2)(2x-3) = (6x+1)(x+7)

<=> 6x^2-4x-9x+6 = 6x^2+x+42x+7

<=> -13x+6 = 43x+7

<=> 6-7 = 43x+13x

<=> 56x = -1

<=> x = -1/56 (TM)

Vậy ...

ĐKXĐ:x khác -7;x khác 1,5

=>(3x-2)(2x-3)=(6x+1)(x+7)

=>6x²-4x-9x+6=6x²+x+42x+7

=>6x²-13x+6=6x²+43x+7

=>6x²-6x²-13x-43x+6-7=0

=>-56x-1=0

=>-56x=1

=>x=\(\frac{-1}{56}\)