Question

Giải phương trình :a,\(\dfrac{2x 1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)b,\(\dfrac{3\left(2x 1\right)}{4}-5-\dfrac{3x 2}{10}=\dfrac{2\left(3x-1\right)}{5}\)\(c,\dfrac{x 1}{2009} \dfrac{x 3}{2007}=\dfrac{x 5}{2005} \dfrac{x 7}{2003}\)\(d,\dfrac{392-x...

Answer 1

a) Ta có: \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)

\(\Leftrightarrow\dfrac{2\left(2x+1\right)}{12}-\dfrac{3\left(x-2\right)}{12}=\dfrac{4\left(3-2x\right)}{12}-\dfrac{12x}{12}\)

\(\Leftrightarrow4x+2-3x+6=12-8x-12x\)

\(\Leftrightarrow x+8-12+20x=0\)

\(\Leftrightarrow21x-4=0\)

\(\Leftrightarrow21x=4\)

\(\Leftrightarrow x=\dfrac{4}{21}\)

Vậy: \(S=\left\{\dfrac{4}{21}\right\}\)

Answer 2

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Answer 3

a) 

PT \(\Leftrightarrow \frac{4x+2}{12}-\frac{3x-6}{12}=\frac{12-8x}{12}-\frac{12x}{12}\)

\(\Leftrightarrow 4x+2-3x+6=12-8x-12x\)

\(\Leftrightarrow 21x=4\Leftrightarrow x=\frac{4}{21}\)

b) 

PT \(\Leftrightarrow \frac{30x+15}{20}-\frac{100}{20}-\frac{6x+4}{20}=\frac{24x-12}{20}\)

\(\Leftrightarrow 30x+15-100-6x-4=24x-12\Leftrightarrow -89=-12\) (vô lý)

Vậy pt vô nghiệm.

Answer 4

c) 

PT \(\Leftrightarrow \frac{x+1}{2009}+1+\frac{x+3}{2007}+1=\frac{x+5}{2005}+1+\frac{x+7}{2003}+1\)

\(\Leftrightarrow \frac{x+2010}{2009}+\frac{x+2010}{2007}=\frac{x+2010}{2005}+\frac{2010}{2003}\)

\(\Leftrightarrow (x+2010)\left(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{2003}\right)=0\)

Dễ thấy \(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{2003}< 0\) nên $x+2010=0\Leftrightarrow x=-2010$

d) 

\(\Leftrightarrow \frac{392-x}{32}+1+\frac{390-x}{34}+1+\frac{388-x}{36}+1+\frac{386-x}{38}+1+\frac{384-x}{40}+1=0\)

\(\Leftrightarrow \frac{424-x}{32}+\frac{424-x}{34}+\frac{424-x}{36}+\frac{424-x}{38}+\frac{424-x}{40}=0\)

\(\Leftrightarrow (424-x)\left(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{38}+\frac{1}{40}\right)=0\)

Dễ thấy \(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{38}+\frac{1}{40}>0\) nên $424-x=0\Leftrightarrow x=424$

Answer 5

e) ĐK: $x\neq 0; 2$

PT $\Leftrightarrow \frac{x-3}{x-2}+\frac{x+2}{x}=2$

$\Leftrightarrow 1-\frac{1}{x-2}+1+\frac{2}{x}=2$

$\Leftrightarrow \frac{2}{x}-\frac{1}{x-2}=0$

$\Leftrightarrow \frac{x-4}{x(x-2)}=0\Rightarrow x=4$ (thỏa mãn)

f) 

$(x-2)(\frac{2}{3}x-6)=0$

$\Rightarrow x-2=0$ hoặc $\frac{2}{3}x-6=0$

$\Rightarrow x=2$ hoặc $x=9$

Answer 6

g.

PT $\Leftrightarrow (x+1)(x+4)(x+2)(x+3)=1680$

$\Leftrightarrow (x^2+5x+4)(x^2+5x+6)=1680$

$\Leftrightarrow a(a+2)=1680$ (đặt $a=x^2+5x+4$)

$\Leftrightarrow (a+1)^2=1681$

$\Rightarrow a+1=\pm 41$

$\Rightarrow a-40=0$ hoặc $a+42=0$

$\Leftrightarrow x^2+5x-36=0(1)$ hoặc $x^2+5x+46=0(2)$

Hiển nhiên PT (2) vô nghiệm

PT $(1)\Leftrightarrow (x-4)(x+9)=0\Rightarrow x=4$ hoặc $x=-9$

 

Answer 7

h, $(x-1)(x-2)(x-3)(x-4)=24$

PT $\Leftrightarrow (x-1)(x-4)(x-2)(x-3)=24$

$\Leftrightarrow (x^2-5x+4)(x^2-5x+6)=24$

$\Leftrightarrow a(a+2)=24$ (đặt $x^2-5x+4=a$)

$\Leftrightarrow (a+1)^2-25=0$

$\Leftrightarrow (a-4)(a+6)=0$

$\Rightarrow a-4=0$ hoặc $a+6=0$

$\Leftrightarrow x^2-5x=0(1)$ hoặc $x^2-5x+10=0(2)$

Dễ thấy PT $(2)$ vô nghiệm

Đối với PT $(1)\Leftrightarrow x(x-5)=0\Rightarrow x=0$ hoặc $x=5$