Bài làm

a) \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x-4}\)

\(\Leftrightarrow\frac{3x+2}{3x-2}-\frac{6}{3x+2}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Leftrightarrow\frac{(3x+2)\left(3x+2\right)}{(3x-2)\left(3x+2\right)}-\frac{6\left(3x-2\right)}{(3x+2)\left(3x-2\right)}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Rightarrow\left(3x+2\right)^2-\left(18x-12\right)=9x^2\)

\(\Leftrightarrow9x^2+12x+4-18x+12x-9x^2=0\)

\(\Leftrightarrow6x+4=0\)

\(\Leftrightarrow x=-\frac{4}{6}\)

\(\Leftrightarrow x=-\frac{2}{3}\)

Vậy x = -2/3 là nghiệm.

@Tao Ngu :))@ 9x-4 không tách thành (3x+4)(3x-4) được đâu bạn. Chỗ đó phải là: 9x²-4

Bài thiếu đkxđ của x \(\hept{\begin{cases}3x-2\ne0\\2+3x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}3x\ne2\\3x\ne-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne\frac{2}{3}\\x\ne\frac{-2}{3}\end{cases}\Leftrightarrow}x\ne\pm\frac{2}{3}}\)

Giải:

a) $\frac{}{}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}$⇔⇔ 9x² + 12x + 4 - 18x + 12 = 9x² ⇔ 9x² + 12x + 4 - 18x + 12 - 9x² = 0

⇔ 16 + 6x = 0 ⇔ 2(8 + 3x) = 0 ⇔ 8 + 3x = 0 ⇔ x = \(\frac{-8}{3}\)

Vậy nghiệm của phương trình là x = \(\frac{-8}{3}\) .

b) \(\frac{3}{5x-1}+\frac{3}{3-5x}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\text{⇔ }\frac{-3}{1-5x}+\frac{-3}{5x-3}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\)

⇔ \(\frac{9-15x}{\left(1-5x\right)\left(5x-3\right)}+\frac{15x-3}{\left(1-5x\right)\left(5x-3\right)}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\) ⇔ 9 - 15x + 15x - 3 = 4

⇔ 8 = 4 ( vô lí)

Vậy phương trình trên vô nghiệm.

Mình chỉ làm 2 câu a, b thôi nhé! Các bài tập này cách làm giống nhau, bạn tự hoàn thành những bài còn lại nhé!

ĐKXĐ đâu?

\(1.\frac{7x-3}{x-1}=\frac{2}{3}\)   ( \(x\ne1\))

\(\Leftrightarrow\frac{3\left(7x-1\right)}{3\left(x-1\right)}=\frac{2\left(x-1\right)}{3\left(x-1\right)}\)

\(\Rightarrow3\left(7x-3\right)=2\left(x-1\right)\)

\(\Leftrightarrow21x-9=2x-2\)

\(\Leftrightarrow19x=7\)

\(\Leftrightarrow x=\frac{7}{19}\)

\(2.\frac{5x-1}{3x+2}=\frac{5x-7}{3x-1}\)

\(\Leftrightarrow\frac{\left(5x-1\right)\left(3x-1\right)}{\left(3x+2\right)\left(3x-1\right)}=\frac{\left(5x-7\right)\left(3x+2\right)}{\left(3x-1\right)\left(3x+2\right)}\)

\(\Rightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\)

\(\Leftrightarrow15x^2-5x-3x+1=15x^2+10x-21x-14\)

\(\Leftrightarrow15x^2-8x+1=15x^2-11x-14\)

\(\Leftrightarrow\left(15x^2-15x^2\right)+\left(-8x+11x\right)=-14-1\)

\(\Leftrightarrow3x=-15\)

\(\Leftrightarrow x=-5\)

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

\(\Leftrightarrow\frac{\left(1-x\right)\left(3x-1\right)}{\left(x+1\right)\left(3x-1\right)}+\frac{3\left(x+1\right)\left(3x-1\right)}{\left(x+1\right)\left(3x-1\right)}=\frac{\left(2x+3\right)\left(x+1\right)}{\left(3x-1\right)\left(0+1\right)}\)

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

\(\Leftrightarrow3x-1-3x^2+x+3\left(3x^2-x+3x-1\right)=2x^2+2x+3x+3\)

\(\Leftrightarrow3x-1-3x^2+x+9x^2-3x+9x-3=2x^2+2x+3x+3\)

\(\Leftrightarrow6x^2+10x-4=2x^2+5x+3\)

\(\Leftrightarrow\left(6x^2-2x^2\right)+\left(10x-5x\right)=7\)

\(\Leftrightarrow4x^2+5x-7=0\)

\(\Leftrightarrow\left(2x\right)^2+4x.\frac{5}{4}+\frac{16}{25}+\frac{191}{25}=0\)

\(\Leftrightarrow\left(2x+\frac{5}{4}\right)^2-\frac{191}{25}=0\)

\(\left(2x+\frac{5}{4}\right)^2>0\)

\(\Rightarrow\left(2x+\frac{5}{4}\right)^2+\frac{191}{25}>0\)

=> PT vô nghiệm 

\(4.\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)

\(\Leftrightarrow\frac{\left(1-6x\right)\left(x+2\right)}{x^2-4}+\frac{\left(9x+4\right)\left(x-2\right)}{x^2-4}=\frac{2\left(3x-2\right)+1}{x^2-4}\)

\(\Rightarrow\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3\left(3x-2\right)+1\)

\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8=3x^2-2x+1\)

\(\Leftrightarrow3x^2-25x-6=3x^2-2x+1\)

\(\Leftrightarrow\left(3x^2-3x^2\right)+\left(-25x+2x\right)+\left(-6-1\right)=0\)

\(\Leftrightarrow-23x-7=0\)

\(\Leftrightarrow-23x=7\)

\(\Leftrightarrow x=\frac{-7}{23}\)

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

\(\Leftrightarrow\frac{\left(3x+2\right)^2}{9x^2-4}-\frac{6\left(3x-2\right)}{9x^2-4}=\frac{9x^2}{9x^2-4}\)

\(\Rightarrow\left(3x+2\right)^2-6\left(3x-2\right)=9x^2\)

\(\Leftrightarrow9x^2+12x+4-18x+12=9x^2\)

\(\Leftrightarrow\left(9x^2-9x^2\right)+\left(12x-18x\right)+\left(4+12\right)=0\)

\(\Leftrightarrow-6x+16=0\)

\(\Leftrightarrow-6x=-16\)

\(\Leftrightarrow x=\frac{16}{6}\)

\(6.1+\frac{1}{x+2}=\frac{12}{8-x^3}\)

\(\Leftrightarrow\frac{\left(x+2\right)\left(8-x^3\right)}{\left(x+2\right)\left(8-x^3\right)}+\frac{1\left(8-x^3\right)}{\left(x+2\right)\left(8-x^3\right)}=\frac{12\left(x+2\right)}{\left(x+2\right)\left(8-x^3\right)}\)

\(\Rightarrow\left(x+2\right)\left(8-x^3\right)+1\left(8-x^3\right)=12\left(x+2\right)\)

\(\Leftrightarrow8x+x^4+16+2x^3+8-x^3=12x+24\)

\(\Leftrightarrow x^4+\left(2x^3-x^3\right)+\left(8x-12x\right)+\left(16-24\right)=0\)

\(\Leftrightarrow x^4+x^3-4x-8=0\)

\(\Leftrightarrow\left(x^4-4x\right)+\left(x^3-8\right)=0\)

Đến đấy mk tắc r xl bạn nhé 

Cho x,y,z là các sô dương.Chứng minh rằng x/2x+y+z+y/2y+z+x+z/2z+x+y<=3/4

\(\frac{3x-1}{2}-\frac{2-6x}{5}=\frac{1}{2}+\left(3x-1\right)\)

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

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

\(\Leftrightarrow\frac{-1}{10}\left(3x-1\right)=\frac{1}{2}\)

\(\Leftrightarrow3x-1=-5\)

\(\Leftrightarrow3x=-4\Leftrightarrow x=\frac{-4}{3}\)

Vậy nghiệm duy nhất của phương trình là\(x=\frac{-4}{3}\)

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

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

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

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

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

\(\Leftrightarrow\left(x+1\right)\left(x+1-\frac{1}{3}-\frac{6x+1}{6}\right)=0\)

\(\Leftrightarrow\frac{1}{2}\left(x+1\right)=0\)

\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

Vậy nghiệm duy nhất của phương trình là\(x=-1\)