a) 2x² - 3x - 2 = 0.

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

<=> 2x + 1 = 0 hoặc x - 2 = 0

<=> x = -1/2 hoặc x = 2

b) 3x² - 7x - 10 = 0.

<=> (x + 1)(3x - 10) = 0

<=> x = -1 hoặc x = 10/3

c) 2x² - 5x + 3 = 0.

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

<=> x = 1 hoặc x = 3/2

B = {3/2;1}

\(2x^2+5x+3=0\)

\(\Leftrightarrow2x^2+2x+3x+3=0\)

\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=-1\end{matrix}\right.\)

Vậy \(S=\left\{-1;-\dfrac{3}{2}\right\}\)

\(\left(x-\sqrt{2}\right)-3\left(x^2-2\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{1}{6}-\dfrac{\sqrt{73-3\sqrt{32}}}{6}\\x_2=\dfrac{\sqrt{73-3\sqrt{32}}}{6}+\dfrac{1}{6}\end{matrix}\right.\)

kh hiểu bn ơi

`4x=2+xx+1x<=>4x=2+3x<=>4x-3x=2<=>1x=2<=>x=2`

bạn tách từng bài ra bn

tl câu hỏi trên cho mik ik

giải phương trình làm dell gì, lớp 8 mà

\(a,\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)

\(\left(x-1\right)\left(5x+3-3x+8\right)=0\)

\(\left(x-1\right)\left(2x+11\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\2x+11=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\2x=-11\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=-\frac{11}{2}\end{cases}}}\)

\(b,3x\left(25x+15\right)-35\left(5x+3\right)=0\)

\(15x\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\left(5x+3\right).5\left(3x-7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x+3=0\\5\left(3x-7\right)=0\end{cases}\Rightarrow\orbr{\begin{cases}5x=-3\\3x-7=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\x=\frac{7}{3}\end{cases}}}\)

\(c,\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)

\(\left(3x-2\right)\left(2-5x\right)+\left(3x-2\right)\left(x+11\right)=0\)

\(\left(3x-2\right)\left(2-5x+x+11\right)=0\)

\(\left(3x-2\right)\left(13-4x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-2=0\\13-4x=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=2\\4x=13\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{13}{4}\end{cases}}}\)

còn đâu tự lm lười :_# 

Chọn C

Ta có f(x) + g(x) = (2x2 - 5x - 3) + (-2x2 - 2x + 1) = -7x - 2

Cho -7x - 2 = 0 ⇒ x = -2/7

\(a,2x^2+x=0\)

\(x\left(2x+1\right)=0\)

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

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

\(b,-0,4x^2+1,2x=0\)

\(x\left[\left(0,4x\right)-\left(1,2\right)\right]=0\)

\(\left[{}\begin{matrix}x=0\\0,4x-1,2=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0\\0,4x=1,2\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0\\x=\frac{3}{10}\end{matrix}\right.\)

\(c,7x^2-5x=0\)

\(x\left(7x-5\right)=0\)

\(\left[{}\begin{matrix}x=0\\7x-5=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0\\x=\frac{5}{7}\end{matrix}\right.\)

\(e,-2x^2-11x=0\)

\(x\left(2x+11\right)=0\)

\(\left[{}\begin{matrix}x=0\\2x+11=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0\\x=\frac{11}{2}\end{matrix}\right.\)

1.

a) \(2x^4-4x^3+2x^2\)

\(=2x^2\left(x^2-2x+1\right)\)

\(=2x^2\left(x-1\right)^2\)

b) \(2x^2-2xy+5x-5y\)

\(=\left(2x^2-2xy\right)+\left(5x-5y\right)\)

\(=2x\left(x-y\right)+5\left(x-y\right)\)

\(=\left(x-y\right)\cdot\left(2x+5\right)\)

2 . 

a,

\(4x\left(x-3\right)-x+3=0\)

⇒\(4x\left(x-3\right)-\left(x-3\right)=0\)

⇒\(\left(x-3\right)\left(4x-1\right)=0\)

⇒\(\left[{}\begin{matrix}x-3=0\\4x-1=0\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=3\\4x=1\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{4}\end{matrix}\right.\)

vậy \(x\in\left\{3;\dfrac{1}{4}\right\}\)

b, 

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

⇒\(\left(2x-3-x-1\right)\left(2x-3+x+1\right)\) = 0

⇒\(\left(x-4\right)\left(3x-2\right)=0\)

⇔\(\left[{}\begin{matrix}x-4=0\\3x-2=0\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=4\\3x=2\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=4\\x=\dfrac{2}{3}\end{matrix}\right.\)

vậy \(x\in\left\{4;\dfrac{2}{3}\right\}\)

ban tích cho mk vs nha

Theo định lí Vi-et ta có:

x1.x2 = c/a = 3/2 ⇒ x2 = 3/2