\(\left(4-3x\right)\left(10x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)

\(\left(7-2x\right)\left(4+8x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)

rồi thực hiện đến hết ... 

Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>

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

\(2x^2-7x+3=4x^2+4x-3\)

\(2x^2-7x+3-4x^2-4x+3=0\)

\(-2x^2-11x+6=0\)

\(2x^2+11x-6=0\)

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

\(2x\left(x+6\right)-\left(x+6\right)=0\)

\(\left(x+6\right)\left(2x-1\right)=0\)

\(x+6=0\Leftrightarrow x=-6\)

\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)

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

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

\(x=0\)

\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)

Tự lm tiếp nha 

chiu lop 3 ma

a,

đoạn 9x-6-> 2x-6=0

=> x=3

b,6x^2+13x+5=6x^2-20x+6

33x=1

=>x=1/33

a) (x+1)(x+9)=(x+3)(x+5) 

<=>x^2+10x+9=x^2+8x+15

<=>x^2+10x+9-x^2-8x-15=0

<=>9x-6=0 phải là 2x - 6

<=>9x=6

<=>x=6/9=2/3 => S= 2/3

d) (3x+5)(2x+1)=(6x-2)(x-3)

<=>6x^2+13x+5=6x^2-16x+6 phải là 6x^2 - 20x + 6

<=>6x^2+13x+5-6x^2+16x-6=0

<=>29x-1=0

<=>29x=1

<=>x=1/29

a) \(\left(x+1\right)\left(x+9\right)=\left(x+3\right)\left(x+5\right)\)

\(\Leftrightarrow x^2+10x+9=x^2+8x+15\)

\(\Leftrightarrow x^2+10x+9-x^2-8x-15=0\)

\(\Leftrightarrow2x-6=0\)

\(\Leftrightarrow x=3\)

Vây tập nghiệm của phương trình là \(S=\left\{3\right\}\)

d) \(\left(3x+5\right)\left(2x+1\right)=\left(6x-2\right)\left(x-3\right)\)

\(\Leftrightarrow6x^2+13x+5=6x^2-20x+6\)

\(\Leftrightarrow6x^2+13x+5-6x^2+20x-6=0\)

\(\Leftrightarrow33x-1=0\)

\(\Leftrightarrow x=\frac{1}{33}\)

Vây tập nghiệm của phương trình là \(S=\left\{\frac{1}{33}\right\}\)

\(x\left(3x-5\right)-9x+15=0\)

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

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

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}3x+2=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-2}{3}\\x=5\end{cases}}\)

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

1/ \(1+\frac{2}{x-1}+\frac{1}{x+3}=\frac{x^2+2x-7}{x^2+2x-3}\)

ĐKXĐ: \(\hept{\begin{cases}x-1\ne0\\x+3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-3\end{cases}}\)

<=> \(1+\frac{2\left(x+3\right)+x-1}{\left(x-1\right)\left(x+3\right)}=\frac{x^2+2x-3-5}{x^2+2x-3}\)

<=> \(1+\frac{2x+6+x-1}{x^2+2x-3}=1-\frac{5}{x^2+2x-3}\)

<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=1-1\)

<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=0\)

<=> \(\frac{3x+10}{x^2+2x-3}=0\)

<=> \(3x+10=0\)

<=> \(x=-\frac{10}{3}\)

\(a,x=3x^2\Rightarrow x-3x^2=0\Rightarrow x\left(1-3x\right)=0\Rightarrow\orbr{\begin{cases}x=0\\1-3x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{3}\end{cases}}\)

\(b,\left(2x-6\right)\left(x+4\right)+2\left(2x-6\right)=0\)

\(\Rightarrow\left(2x-6\right)\left(x+4+2\right)=0\)

\(\Rightarrow\left(2x-6\right)\left(x+6\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-6=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)

\(c,\left(2x-5\right)\left(x+9\right)+6x-15=0\)

\(\Rightarrow\left(2x-5\right)\left(x+9\right)+3\left(2x-5\right)=0\)

\(\Rightarrow\left(2x-5\right)\left(x+9+3\right)=0\)

\(\Rightarrow\left(2x-5\right)\left(x+12\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-5=0\\x+12=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-12\end{cases}}\)

a)x^3+5x-6=0

x^3-x+6x-6

x(x^2-1)+6(x-1)=0

(x-1)(x(x+1)+6)

(x-1)(x^2+x+6)=0

=>x-1 hoac x^2+x+6

TH1

x-1=0

x=1

TH2

x^2+x+6=0

(x^2+2.1/2 x +1/4)-1/4+6=0

(x+1/2)^2=-23/4

vi (x+1/2)^2 >/0

=>pt vo nghiem

=>nghiem cua pt la 1

a) \(5\left(x-7\right)=0\)

\(\Rightarrow x-7=0\)

\(\Rightarrow x=7\)

b) \(25\left(x-4\right)=0\)

\(\Rightarrow x-4=0\)

\(\Rightarrow x=4\)

c) \(\left(34-2x\right)\left(2x-6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)

d) \(\left(2019-x\right)\left(3x-12\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=\dfrac{12}{3}=4\end{matrix}\right.\)

e) \(57\left(9x-27\right)=0\)

\(\Rightarrow9x-27=0\)

\(\Rightarrow9\left(x-3\right)=0\)

\(\Rightarrow x-3=0\)

\(\Rightarrow x=3\)

a) 5.(x-7)=0⇔x-7=0⇔x=7

b) 25(x-4)=0⇔x-4=0⇔x=4

c) (34-2x).(2x-6)=0

⇔ 34-2x=0 hoặc 2x-6=0

⇔2x=34 hoặc 2x=6

⇔ x=17 hoặc x=3

d) (2019-x).(3x-12)=0

⇔ 2019-x=0 hoặc 3x-12=0

⇔ x=2019 hoặc x=4

e) 57.(9x-27)=0

⇔ 9x-27=0

⇔ x=3

f) 25+(15-x)=30

⇔ 15-x=5

⇔ x=10

g) 43-(24-x)=20

⇔ 24-x=23

⇔ x=1

h) 2.(x-5)-17=25

⇔ 2(x-5)=42

⇔x-5=21

⇔ x=26

i) 3(x+7)-15=27

⇔ 3(x+7)=42

⇔ x+7=14

⇔ x=7

j) 15+4(x-2)=95

⇔ 4(x-2)=80

⇔ x-2=20

⇔ x=22

k) 20-(x+14)=5

⇔ x+14=15

⇔ x=1

l) 14+3(5-x)=27

⇔ 3(5-x)=13

⇔ 5-x=13/3

⇔ x=5-13/3

⇔ x=2/3