Bài 2: Tìm x

a)ĐKXĐ: \(x\ne0\)

Ta có: \(\left(4x^4+3x^3\right):\left(-x^3\right)+\left(15x^2+6x\right):3x=0\)

\(\Leftrightarrow\frac{-x^3\left(4x+3\right)}{x^3}+\frac{3x\left(5x+2\right)}{3x}=0\)

\(\Leftrightarrow-4x-3+5x+2=0\)

\(\Leftrightarrow x-1=0\)

hay x=1(nhận)

Vậy: x=1

b) ĐKXĐ: \(x\notin\left\{0;\frac{1}{3}\right\}\)

Ta có: \(\left(x^2-12x\right):2x-\left(3x-1\right)^2:\left(3x-1\right)=0\)

\(\Leftrightarrow\frac{x\left(x-12\right)}{2x}-\frac{\left(3x-1\right)^2}{\left(3x-1\right)}=0\)

\(\Leftrightarrow\frac{x-12}{x}-3x+1=0\)

\(\Leftrightarrow\frac{x-12}{x}=3x-1\)

\(\Leftrightarrow x-12=x\left(3x-1\right)\)

\(\Leftrightarrow3x^2-x+x-12=0\)

\(\Leftrightarrow3x^2-12=0\)

\(\Leftrightarrow3x^2=12\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-2\left(nhận\right)\end{matrix}\right.\)

Vậy: \(x\in\left\{2;-2\right\}\)

`1)<=> -4x-3 + 5x+ 2 =0`

`<=> 5x-4x = -2+3`

`<=> x =1`

`2)<=> -5x +2-3x+6 =4`

`<=> -5x-3x = 4-6-2`

`<=> -8x=-4`

`<=> x=1/2`

`3) <=> -7x^2 +2 +7x^2 +14x =8`

`<=> 14x +2 =8`

`<=> 14x = 6`

`<=> x=3/7`

giup minh voi dang can

\(8x^3+12x^2+6x-26=0\)

<=>  \(4x^3+6x^2+3x-13=0\)

<=>  \(4x^3-4x^2+10x^2-10x+13x-13=0\)

<=>  \(4x^2\left(x-1\right)+10x\left(x-1\right)+13\left(x-1\right)=0\)

<=>  \(\left(x-1\right)\left(4x^2+10x+13\right)=0\)

<=> \(x-1=0\)

<=>  \(x=1\)

Vậy...

Cam on ban nhieu nhe <3

        4x² - 12x -7 = 0

<=>  4x² -14x +2x -7 = 0

<=>  2x(2x-7) + (2x-7) = 0

<=>  (2x-7)(2x+1) = 0

<=>  2x-7 = 0 hoặc 2x+1 = 0

<=>  2x = 7     hoặc 2x = -1

<=>  x= \(\frac{7}{2}\)hoặc  x= \(\frac{-1}{2}\)

Vậy tập nghiệm của phương trình là S={\(\frac{7}{2}\);\(\frac{-1}{2}\)}

Đây nhé! Tích giúp mình nha

Ta có: \(\dfrac{4x^4+3x^3}{-x^3}+\dfrac{15x^2+6x}{3x}=0\)

\(\Leftrightarrow-4x-3+5x+2=0\)

\(\Leftrightarrow x-1=0\)

hay x=1

\(15x^2+30=0\\ \Rightarrow x^2+2=0\left(vô.lí\right)\\ \Rightarrow x\in\varnothing\)

\(\left(2x-1\right)^2.4=1\\ \Rightarrow\left(2x-1\right)^2=\dfrac{1}{4}\\ \Rightarrow\left[{}\begin{matrix}2x-1=\dfrac{1}{2}\\2x-1=-\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)

4x²-12x=0

=>4x(x-3)=0

=>x=0 hoặc x-3=0

=>x=0 hoặc x=3

a: Ta có: \(4x\left(x-7\right)-4x^2=56\)

\(\Leftrightarrow4x^2-7x-4x^2=56\)

hay x=-8

b: Ta có: \(12x\left(3x-2\right)-\left(4-6x\right)=0\)

\(\Leftrightarrow36x^2-24x-4+6x=0\)

\(\Leftrightarrow36x^2-18x-4=0\)

\(\text{Δ}=\left(-18\right)^2-4\cdot36\cdot\left(-4\right)=900\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{18-30}{72}=\dfrac{-1}{6}\\x_2=\dfrac{18+30}{72}=\dfrac{2}{3}\end{matrix}\right.\)

c: Ta có: \(4\left(x-5\right)-\left(x-5\right)^2=0\)

\(\Leftrightarrow\left(x-5\right)\left(4-x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=9\end{matrix}\right.\)