1) (x+6)(3x-1)+x+6=0

⇔(x+6)(3x-1)+(x+6)=0

⇔(x+6)(3x-1+1)=0

⇔3x(x+6)=0

2) (x+4)(5x+9)-x-4=0

⇔(x+4)(5x+9)-(x+4)=0

⇔(x+4)(5x+9-1)=0

⇔(x+4)(5x+8)=0

3)(1-x)(5x+3)÷(3x-7)(x-1)

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

      c.   x^2-5x +6 = 0

<=> x^2 - 5x = -6

<=> - 4x = -6

<=> x= -6/-4

 Mình chỉ phân tích đa thức thành nhân tử thôi , phần còn lại bạn tự tính nha keo dài lắm

A)  2x²(x+3) - x(x+3) = 0  <=> x(x - 3)(2x-1)=0

B)  (2x+5)² - (x+2)²=0  <=>  (x+3)(3x+7)=0

C)  (x²-2x) - (3x-6)=0  <=> (x-2)(x-3)=0

D)  (2x-7)(2x-7-6x+18)=0   <=> (2x-7)(-4x+11)=0

E)  (x-2)(x+1) - (x-2)(x+2)=0   <=>  (x-2)*(-1)=0   <=> x-2=0

G)  (2x-3)(2x+2-5x)=0  <=> (2x-3)(-3x+2)=0

H)  (1-x)(5x+3+3x-7)=0     <=>  (1-x)(8x-4)=0

F)   (x+6)*3x=0

I)  (x-3)(4x-1-5x-2)=0  <=>  (x-3)(-x-3)=0

K)   (x+4)(5x+8)=0

H)  (x+3)(4x-9)=0

c. x^2-5x+6=0

<=> x^2-5x=-6

<=> -4x=-6

<=> x=-6/-4

vậy tập nghiệm của pt là s={-6/-4}

b: =>(x-4)(x-3)(x-1)>0

=>1<x<3 hoặc x>4

c: =>(2x-1)(x-1)(2x-3)<0

=>x<1/2 hoặc 1<x<3/2

20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)

Vậy...

\(2.\left(x+3\right)\left(x+5\right)+\left(x+3\right)\left(3x-4\right)=0\\ \Leftrightarrow x^2+5x+3x+15+3x^2-4x+9x-12=0\\ \Leftrightarrow x^2+3x^2+5x+3x-4x+9x+15-12=0\\\Leftrightarrow 4x^2+13x+3=0\\\Leftrightarrow 4\left(x^2+\frac{13}{4}x+\frac{3}{4}\right)=0\\\Leftrightarrow x^2+\frac{13}{4}x+\frac{3}{4}=0\\ \Leftrightarrow x^2+\frac{1}{4}x+3x+\frac{3}{4}=0\\\Leftrightarrow x\left(x+\frac{1}{4}\right)+3\left(x+\frac{1}{4}\right)=0\\\Leftrightarrow \left(x+3\right)\left(x+\frac{1}{4}\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x+3=0\\x+\frac{1}{4}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=-\frac{1}{4}\end{matrix}\right.\)

Vậy tập nghiệm của phương trình trên là: \(S=\left\{-3;-\frac{1}{4}\right\}\)

\(3.\left(x+6\right)\left(3x-1\right)+x+6=0\\ \Leftrightarrow3x^2-x+18x-6+x+6=0\\ \Leftrightarrow3x^2+18x=0\\ \Leftrightarrow3x\left(x+6\right)=0\\\Leftrightarrow \left[{}\begin{matrix}3x=0\\x+6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)

Vậy tập nghiệm của phương trình trên là \(S=\left\{0;-6\right\}\)

roois vãi

-Đăng tách câu hỏi bạn nhé.

a) Ta có: \(3x^2+2x-1=0\)

\(\Leftrightarrow3x^2+3x-x-1=0\)

\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-1;\dfrac{1}{3}\right\}\)

b) Ta có: \(x^2-5x+6=0\)

\(\Leftrightarrow x^2-2x-3x+6=0\)

\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

Vậy: S={2;3}

c) Ta có: \(x^2-3x+2=0\)

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

\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Vậy: S={1;2}

d) Ta có: \(2x^2-6x+1=0\)

\(\Leftrightarrow2\left(x^2-3x+\dfrac{1}{3}\right)=0\)

mà \(2\ne0\)

nên \(x^2-3x+\dfrac{1}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{23}{12}=0\)

\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\dfrac{23}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{\sqrt{69}}{6}\\x-\dfrac{3}{2}=\dfrac{-\sqrt{69}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9+\sqrt{69}}{6}\\x=\dfrac{9-\sqrt{69}}{6}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{9+\sqrt{69}}{6};\dfrac{9-\sqrt{69}}{6}\right\}\)

e) Ta có: \(4x^2-12x+5=0\)

\(\Leftrightarrow4x^2-10x-2x+5=0\)

\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{5}{2};\dfrac{1}{2}\right\}\)

cho vào máy tính là ra hết