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

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

phần cuối mk chụp ko đc hết . chỗ cuối là bằng \(\frac{-5}{-3}\)=\(\frac{5}{3}\)

\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)

\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)

\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)

\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)

1) \(x^2-9x=0\Rightarrow x\left(x-9\right)=0\Rightarrow x=0;9\)

2) \(x\left(x-4\right)-x^2=7\Rightarrow-4x=7\Rightarrow x=-\dfrac{7}{4}\)

3) \(3x+2\left(x-5\right)=5\Rightarrow5x-10=5\Rightarrow5x=15\Rightarrow x=3\)

4) \(25x^2-1=0\Rightarrow x^2=\dfrac{1}{25}\Rightarrow x=\pm\dfrac{1}{5}\)

5) \(3x\left(x-2\right)-5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(3x-5\right)=0\Rightarrow x=2;\dfrac{5}{3}\)

6) \(3x\left(x-7\right)+4\left(x-7\right)\Rightarrow\left(3x+4\right)\left(x-7\right)=0\Rightarrow x=-\dfrac{4}{3};7\)

7) \(4x^2-9=0\Rightarrow x^2=\dfrac{9}{4}\Rightarrow x=\pm\dfrac{3}{2}\)

8) \(10x\left(x-4\right)+2x-8=0\Rightarrow2\left(x-4\right)\left(5x+1\right)=0\Rightarrow x=4;-\dfrac{1}{5}\)

9) \(x\left(2x-5\right)-2x^2=0\Rightarrow x\left(2x-5-2x=0\right)\Rightarrow x=0\)

10) \(2x^2-4x=0\Rightarrow2x\left(x-2\right)=0\Rightarrow x=0;2\)

11) \(2x\left(3-4x\right)+3\left(4x-3\right)=0\Rightarrow2x\left(4x-3\right)-3\left(4x-3\right)=0\Rightarrow\left(4x-3\right)\left(2x-3\right)=0\Rightarrow x=\dfrac{3}{4};\dfrac{3}{2}\)

12) \(2x\left(x-5\right)-2x^2=3\Rightarrow-10x=3\Rightarrow x=-\dfrac{3}{10}\)

Đường thẳng song song d nên nhận (3;-4) là 1 vtpt

Phương trình:

\(3\left(x-2\right)-4\left(y-1\right)=0\Leftrightarrow3x-4y-2=0\)

a) \(2\left(x+3\right)-x^2-3x=0\)

\(\Leftrightarrow2\left(x+3\right)-\left(x^2+3x\right)=0\)

\(\Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}2-x=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{-3;2\right\}\)

c) \(3x\left(x-5\right)-x^2+25=0\)

\(\Leftrightarrow3x\left(x-5\right)-\left(x^2-25\right)=0\)

\(\Leftrightarrow3x\left(x-5\right)-\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(3x-x-5\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\2x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\2x=5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{5}{2}\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{5;\frac{5}{2}\right\}\) 

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

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

\(\Leftrightarrow\left(3x-5\right)\left(1+2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-5=0\\1+2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=5\\2x=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\) \(x^2+3x-10=0\)

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

a, \(3x-5-2x\left(5-3x\right)=0\)

\(\Rightarrow3x-5-10x+6x^2=0\)

\(\Rightarrow6x^2-7x-5=0\)

\(\Rightarrow6x^2+3x-10x-5=0\)

\(\Rightarrow3x\left(2x+1\right)-5\left(2x+1\right)=0\)

\(\Rightarrow\left(2x+1\right)\left(3x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\3x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\)

b, \(x^2+3x-10=0\)

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

Chúc bạn hoc tốt!!!

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

Vậy...

1) 16 - 8x = 0 ⇔ 8(2 - x) = 0⇔ 2 - x = 0 ⇔ x = 2

Vậy phương trình có nghiệm là x = 2

a: (2x-3)(3x+6)>0

=>(2x-3)(x+2)>0

=>x<-2 hoặc x>3/2

b: (3x+4)(2x-6)<0

=>(3x+4)(x-3)<0

=>-4/3<x<3

c: (3x+5)(2x+4)>4

\(\Leftrightarrow6x^2+12x+10x+20-4>0\)

\(\Leftrightarrow6x^2+22x+16>0\)

=>\(6x^2+6x+16x+16>0\)

=>(x+1)(3x+8)>0

=>x>-1 hoặc x<-8/3

f: (4x-8)(2x+5)<0

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

=>-5/2<x<2

h: (3x-7)(x+1)<=0

=>x+1>=0 và 3x-7<=0

=>-1<=x<=7/3

a) \(\frac{2}{3}x+\frac{1}{4}=0\)

\(\Leftrightarrow\frac{2}{3}x=-\frac{1}{4}\)

\(\Leftrightarrow x=\frac{-3}{8}\)

Vậy pt có nghiệm x=-3/8

b) \(4x-\frac{5}{6}=0\)

\(\Leftrightarrow4x=\frac{6}{5}\)

\(\Leftrightarrow x=\frac{3}{10}\)

vậy ...

c) \(4x+\frac{1}{2}=0\Leftrightarrow4x=-\frac{1}{2}\Leftrightarrow x=\frac{-1}{8}\)

vậy...

d) \(\frac{5}{9}-x=0\Leftrightarrow x=\frac{5}{9}\)

vậy...

e) \(\frac{4}{3}x+\frac{1}{4}=\frac{2}{3}x-\frac{5}{4}\)

\(\Leftrightarrow\frac{4}{3}x-\frac{2}{3}x=-\frac{5}{4}-\frac{1}{4}\)

<=> x=...

vậy ...

\(x\left(3x-5\right)=0\)

\(\Rightarrow\hept{\begin{cases}x=0\\3x-5=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=\frac{5}{3}\end{cases}}}\)

Vậy \(x\in\left\{0;\frac{5}{3}\right\}\)

a) \(x\left(3x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\3x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{5}{3}\end{cases}}}\)

b) \(3x^2-27=0\)

\(\Leftrightarrow3x^2=27\)

\(\Leftrightarrow x^2=9\)

\(\Leftrightarrow x=\pm3\)

c) \(\left(x-5\right)^2=x-5\)

\(\Leftrightarrow x^2-10x+25-x+5=0\)

\(\Leftrightarrow x^2-11x+30=0\)

\(\Leftrightarrow\left(x-6\right)\left(x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=6\\x=5\end{cases}}}\)

d) \(2\left(x+7\right)-x^2-7x=0\)

\(\Leftrightarrow2x+14-x^2-7x=0\)

\(\Leftrightarrow-x^2-5x+14=0\)

\(\Leftrightarrow\left(x-7\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=2\end{cases}}}\)

e)\(7x\left(x-3\right)+2.3x=0\)

\(\Leftrightarrow7x^2-21x+6x=0\)

\(\Leftrightarrow7x^2-15x=0\)

\(\Leftrightarrow x\left(7x-15\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\7x-15=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{15}{7}\end{cases}}}\)

#H

\(3x^2-27=0\)

\(3x^2=27\)

\(x^2=9\)

\(x\in\left\{3;-3\right\}\)