Nhác quá mấy bài này hỏi làm j

ngu thế à bạn

$(3x-1)\left(-\dfrac14-5x\right)\left(-\dfrac27x+3\right)=0$

$3x-1=0$ hoặc $-\dfrac14-5x=0$ hoặc $-\dfrac27x+3=0$

$x=\dfrac13$ hoặc $x=-\dfrac1{20}$ hoặc $x=\dfrac{21}{2}$

Vậy $x\in\left{\dfrac13,-\dfrac1{20},\dfrac{21}{2}\right}$.

$\left|\dfrac34x-5\right|-\dfrac23=\left|-\dfrac14\right|$

$\left|\dfrac34x-5\right|-\dfrac23=\dfrac14$

$\left|\dfrac34x-5\right|=\dfrac14+\dfrac23$

$\left|\dfrac34x-5\right|=\dfrac{11}{12}$

$\dfrac34x-5=\dfrac{11}{12}$ hoặc $\dfrac34x-5=-\dfrac{11}{12}$

$\dfrac34x=\dfrac{71}{12}$ hoặc $\dfrac34x=\dfrac{49}{12}$

$x=\dfrac{71}{9}$ hoặc $x=\dfrac{49}{9}$

Vậy $x\in\left{\dfrac{49}{9},\dfrac{71}{9}\right}$.

Bài 1 :

\(\frac{x-1}{x-5}=\frac{6}{7}\Leftrightarrow7x-7=6x-30\)

\(\Leftrightarrow x=-23\)

\(\frac{x-2}{x-1}=\frac{x+4}{x+7}\)ĐK : \(x\ne1;-7\)

\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=\left(x+4\right)\left(x-1\right)\)

\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)

\(\Leftrightarrow2x-10=0\Leftrightarrow x=5\)

Dễ thế mà không làm được

$\left|\frac54x-\frac72\right|=\frac58+\frac35$

$\left|\frac54x-\frac72\right|=\frac{49}{40}$

$\frac54x-\frac72=\pm\frac{49}{40}$

TH1:

$\frac54x-\frac72=\frac{49}{40}$

$\frac54x=\frac{189}{40}$

$x=\frac{189}{50}$

TH2:

$\frac54x-\frac72=-\frac{49}{40}$

$\frac54x=\frac{91}{40}$

$x=\frac{91}{50}$

Vậy $\boxed{x=\frac{91}{50};\frac{189}{50}}$.

a: \(\Leftrightarrow12x^2-10x-12x^2-28x=7\)

=>-38x=7

hay x=-7/38

b: \(\Leftrightarrow-10x^2-5x+9x^2+6x+x^2-\dfrac{1}{2}x=0\)

=>1/2x=0

hay x=0

c: \(\Leftrightarrow18x^2-15x-18x^2-14x=15\)

=>-29x=15

hay x=-15/29

d: \(\Leftrightarrow x^2+2x-x-3=5\)

\(\Leftrightarrow x^2+x-8=0\)

\(\text{Δ}=1^2-4\cdot1\cdot\left(-8\right)=33>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{33}}{2}\\x_2=\dfrac{-1+\sqrt{33}}{2}\end{matrix}\right.\)

e: \(\Leftrightarrow-15x^2+10x-10x^2-5x-5x=4\)

\(\Leftrightarrow-25x^2=4\)

\(\Leftrightarrow x^2=-\dfrac{4}{25}\left(loại\right)\)