bài 1:
\(a.\frac{4x - 8 + (4 - 2x)}{x^2 + 1}=0\) (đkxđ: x thuộc R)
\(\frac{4x - 8 + 4 - 2x}{x^2 + 1}=0\Leftrightarrow\frac{2x - 4}{x^2 + 1}=0\)
\(\Rightarrow2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)
\(b.\frac{x^2 + 2x + 1}{x + 1}=0\left(x\neq-1\right)\)
\(\Leftrightarrow\frac{(x + 1)^2}{x + 1}=0\Rightarrow x+1=0\Rightarrow x=-1\left(L\right)\)
vậy phương trình vô nghiệm
c. \(\frac{2x - 5}{x + 5}=3\left(x\neq-5\right)\)
\(\Rightarrow2x-5=3(x+5)\Leftrightarrow2x-5=3x+15\)
\(\Leftrightarrow2x-3x=15+5\Leftrightarrow-x=20\)
\(\Rightarrow x=-20\left(TM\right)\)
d. \(\frac{4}{x - 2}-2=0\left(x\neq2\right)\)
\(\Leftrightarrow\frac{4}{x - 2}=2\Rightarrow4=2(x-2)\)
\(\Leftrightarrow4=2x-4\Leftrightarrow2x=8\Rightarrow x=4\left(TM\right)\)
bài 2:
a. \(\frac{x^2 + 6x - 16}{x - 2}=x+8\left(x\neq2\right)\)
\(\Leftrightarrow \frac{(x - 2)(x + 8)}{x - 2} = x + 8\)
\(\Leftrightarrow x + 8 = x + 8\)
vậy phương trình đúng với mọi x khác 2
b. \(3x-\frac{1}{x - 2}=\frac{x - 1}{2 - x}\left(x\neq2\right)\)
\(\Leftrightarrow3x-\frac{1}{x - 2}=\frac{1 - x}{x - 2}\Leftrightarrow3x=\frac{1 - x + 1}{x - 2}\)
\(\Leftrightarrow3x=\frac{2 - x}{x - 2}\Leftrightarrow3x=-1\Rightarrow x=-\frac13\left(TM\right)\)
c. \(\frac{x^2 - 15x + 1}{x + 17}=x-2\left(x\neq-17\right)\)
\(\Rightarrow x^2-15x+1=(x-2)(x+17)\Leftrightarrow x^2-15x+1=x^2+15x-34\)
\(\Leftrightarrow-15x-15x=-34-1\Leftrightarrow-30x=-35\Rightarrow x=\frac76\left(TM\right)\)
d. \(\frac{x - 1}{x - 2}-3+x=\frac{1}{x - 2}\left(x\neq2\right)\)
\(\Leftrightarrow \frac{x - 1}{x - 2} - \frac{1}{x - 2} + x - 3 = 0\)
\(\Leftrightarrow \frac{x - 2}{x - 2} + x - 3 = 0\)
\(\Leftrightarrow1+x-3=0\Leftrightarrow x-2=0\Rightarrow x=2\left(L\right)\)
vậy phương trình vô nghiệm
bài 3:
\(a.\frac{x^3 - (x - 1)^3}{(4x + 3)(x - 5)}=\frac{7x - 1}{4x + 3}-\frac{x}{x - 5}\left(x\neq-\frac{3}{4};x\neq5\right)\)
\(\Leftrightarrow \frac{x^3 - (x^3 - 3x^2 + 3x - 1)}{(4x + 3)(x - 5)} = \frac{(7x - 1)(x - 5) - x(4x + 3)}{(4x + 3)(x - 5)}\)
\(\Leftrightarrow \frac{3x^2 - 3x + 1}{(4x + 3)(x - 5)} = \frac{7x^2 - 36x + 5 - 4x^2 - 3x}{(4x + 3)(x - 5)}\)
\(\Leftrightarrow \frac{3x^2 - 3x + 1}{(4x + 3)(x - 5)} = \frac{3x^2 - 39x + 5}{(4x + 3)(x - 5)}\)
\(\Rightarrow3x^2-3x+1=3x^2-39x+5\Leftrightarrow36x=4\)
\(\Rightarrow x=\frac19\left(TM\right)\)
\(b.1+\frac{2x - 5}{x - 2}-\frac{3x - 5}{x - 1}=0\left(x\neq2;x\neq1\right)\)
\(\Rightarrow (x - 2)(x - 1) + (2x - 5)(x - 1) - (3x - 5)(x - 2) = 0\)
\(\Leftrightarrow (x^2 - 3x + 2) + (2x^2 - 7x + 5) - (3x^2 - 11x + 10) = 0\)
\(⇔3x
2
−10x+7−3x
2
+11x−10=0\)
\(\Leftrightarrow x-3=0\Rightarrow x=3\left(TM\right)\)
\(c.\frac{x + 2}{x - 2}-\frac{2}{x^2 - 2x}=\frac{1}{x}\left(x\neq0;x\neq2\right)\)
\(\Leftrightarrow \frac{x + 2}{x - 2} - \frac{2}{x(x - 2)} = \frac{1}{x}\)
\(\Rightarrow x(x + 2) - 2 = x - 2\)
\(\Leftrightarrow x^2 + 2x - 2 = x - 2\)
\(\Leftrightarrow x^2+x=0\Leftrightarrow x(x+1)=0\)
\(\Leftrightarrow\left[\begin{array}{l}x=0\left(L\right)\\ x=-1\left(TM\right)\end{array}\right.\)
\(d.\frac{x + 2}{x - 3}+\frac{x - 2}{x + 3}-\frac{2(x^2 + 6)}{x^2 - 9}=0\left(x\neq\pm3\right)\)
\(\Leftrightarrow \frac{x + 2}{x - 3} + \frac{x - 2}{x + 3} - \frac{2(x^2 + 6)}{(x - 3)(x + 3)} = 0\)
\(\Rightarrow (x + 2)(x + 3) + (x - 2)(x - 3) - 2(x^2 + 6) = 0\)
\(\Leftrightarrow (x^2 + 5x + 6) + (x^2 - 5x + 6) - 2x^2 - 12 = 0\)
\(\Leftrightarrow 2x^2 + 12 - 2x^2 - 12 = 0\)
⇒ 0x = 0
vậy phương trình luôn đúng với mọi x khác +-3
bài 4:
\(a.x+\frac{2x - 1}{x - 2}=3x+\frac{3}{x - 2}\) (x khác 2)
\(\Leftrightarrow \frac{2x - 1}{x - 2} - \frac{3}{x - 2} = 3x - x\)
\(\Leftrightarrow \frac{2x - 4}{x - 2} = 2x\)
\(\Leftrightarrow\frac{2(x - 2)}{x - 2}=2x\Leftrightarrow2=2x\Rightarrow x=1\left(TM\right)\)
\(b.\frac{5x + 1}{5}-\frac{2x - 1}{2x + 2}=2+\frac{x^2 + 4x + 1}{x + 1}\) (x khác -1)
\(\Leftrightarrow \frac{5x + 1}{5} - \frac{2x - 1}{2(x + 1)} = 2 + \frac{x^2 + 4x + 1}{x + 1}\)
\(\Rightarrow 2(x + 1)(5x + 1) - 5(2x - 1) = 20(x + 1) + 10(x^2 + 4x + 1)\)
\(\Leftrightarrow 2(5x^2 + 6x + 1) - 10x + 5 = 20x + 20 + 10x^2 + 40x + 10\)
\(\Leftrightarrow 10x^2 + 12x + 2 - 10x + 5 = 10x^2 + 60x + 30\)
\(\Leftrightarrow 2x + 7 = 60x + 30\)
\(\Leftrightarrow-58x=23\Rightarrow x=-\frac{23}{58}\left(TM\right)\)
\(c.\frac{1}{x + 2}+\frac{1}{x^2 - 2x}=\frac{8}{x^3 - 4x}\left(x\neq0;x\neq\pm2\right)\)
\(\Leftrightarrow \frac{1}{x + 2} + \frac{1}{x(x - 2)} = \frac{8}{x(x - 2)(x + 2)}\)
\(\Rightarrow x(x - 2) + (x + 2) = 8\)
\(\Leftrightarrow x^2 - 2x + x + 2 = 8\)
\(\Leftrightarrow x^2-x-6=0\Leftrightarrow(x-3)(x+2)=0\)
\(\Leftrightarrow\left[\begin{array}{l}x=3\left(TM\right)\\ x=-2\left(L_{}\right)\end{array}\right.\)
\(d.\frac{x + 5}{x^2 - 5x}+\frac{5 - x}{2x^2 + 10x}=\frac{x - 5}{2x^2 - 50}\left(x\neq\pm5;x\neq0\right)\)
\(\Leftrightarrow \frac{x + 5}{x(x - 5)} + \frac{5 - x}{2x(x + 5)} = \frac{x - 5}{2(x - 5)(x + 5)}\)
\(\Rightarrow 2(x + 5)^2 + (5 - x)(x - 5) = x(x - 5)\)
\(\Leftrightarrow 2(x^2 + 10x + 25) - (x - 5)^2 = x^2 - 5x\)
\(\Leftrightarrow 2x^2 + 20x + 50 - (x^2 - 10x + 25) = x^2 - 5x\)
\(\Leftrightarrow x^2 + 30x + 25 = x^2 - 5x\)
\(\Leftrightarrow35x=-25\Rightarrow x=-\frac57\left(TM\right)\)
