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

a) x+15 = 20-4x

=> x=1

b) 17-x=7-6x

=> x=-2

c) -12+x=5x-20

=> x=2

d) 4x-5=15-x

=> x=4

e) 9x-7=20-6x

=>x= \(\frac{9}{5}\)

g)2.(x-10)=3.(x-20)=x-4

=> x thuộc ∅

h) (x^2+2).(x-3) <0

=> x=3,...

`@` `\text {Ans}`

`\downarrow`

`c)`

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

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

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

`=>`\(\left[{}\begin{matrix}x=34\div2\\x=6\div2\end{matrix}\right.\)

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

Vậy, `x \in {17; 3}`

`d)`

`( 2019 - x ) . ( 3x - 12 ) =0` `?`

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

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

`=>`\(\left[{}\begin{matrix}x=2019\\x=12\div3\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=2019\\x=4\end{matrix}\right.\)

Vậy, `x \in {2019; 4}`

`e) `

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

`=>`\(9x-27=0\div57\)

`=> 9x - 27 = 0`

`=> 9x = 27`

`=> x = 27 \div 9`

`=> x = 3`

Vậy, `x = 3`

`f)`

`25 + ( 15 - x ) = 30`

`=> 15 - x = 30 - 25`

`=> 15 - x = 5`

`=> x = 15 -5 `

`=> x = 10`

Vậy, `x = 10`

`g) `

`43 - ( 24 - x ) = 20`

`=> 24 - x = 43 - 20`

`=> 24 - x = 23`

`=> x = 24 - 23`

`=> x = 1`

Vậy, `x = 1`

`h) `

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

`=> 2 ( x - 5) = 25+17`

`=> 2 ( x - 5) = 42`

`=> x - 5 = 42 \div 2`

`=> x - 5 = 21`

`=> x = 21 + 5`

`=> x = 26`

Vậy, `x = 26`

`i)`

`3 . ( x + 7 ) - 15 = 27`

`=> 3(x + 7) = 27 + 15`

`=> 3(x + 7) = 42`

`=> x +7 = 42 \div 3`

`=> x + 7 = 14`

`=> x = 14 - 7`

`=> x = 7`

Vậy, `x = 7`

`j)`

`15 + 4 . ( x - 2 ) = 95`

`=> 4(x - 2) = 95 - 15`

`=> 4(x - 2) = 80`

`=> x - 2 = 80 \div 4`

`=> x - 2 = 20`

`=> x = 20 + 2`

`=> x = 22`

Vậy, `x = 22`

`k)`

`20 - ( x + 14 ) = 5`

`=> x + 14 = 20 - 5`

`=> x + 14 = 15`

`=> x = 15 - 14`

`=> x = 1`

Vậy, `x = 1`

`l) `

`14 + 3 . ( 5 - x ) = 27`

`=> 3(5 - x) = 27 - 14`

`=> 3(5 - x) = 13`

`=> 5 - x = 13 \div 3`

`=> 5 - x = 13/3`

`=> x = 5- 13/3`

`=> x = 2/3`

Vậy, `x = 2/3.`

`@` `\text {Kaizuu lv uuu}`

nhanh mik tick cho nha

a) x : 1/4 = 7/5
    x         = 7/5 x 1/4
    x         = 7/20
b) 9/2 - x = 3/5
            x = 9/2 - 3/5
            x = 39/10
c) 12/5 : x = 7/2
              x = 12/5 : 7/2
              x = 24/35

a) \(x\div\dfrac{1}{4}=\dfrac{7}{5}\)

    \(x=\dfrac{7}{5}\times\dfrac{1}{4}\)

    \(x=\dfrac{7}{20}\)

b) \(\dfrac{9}{2}-x=\dfrac{3}{5}\)

            \(x=\dfrac{9}{2}-\dfrac{3}{5}\)

            \(x=\dfrac{39}{10}\)

c) \(\dfrac{12}{5}\div x=\dfrac{7}{2}\)

             \(x=\dfrac{12}{5}\div\dfrac{7}{2}\)

             \(x=\dfrac{12}{5}\times\dfrac{2}{7}\)

             \(x=\dfrac{24}{35}\)

a)\(x=\dfrac{1}{4}\)x\(\dfrac{7}{5}=\dfrac{7}{20}\)

b)\(x=\dfrac{9}{2}-\dfrac{3}{5}=\dfrac{39}{10}\)

c)\(x=\dfrac{12}{5}:\dfrac{7}{2}=\dfrac{24}{35}\)

Bài 3 là hỗn số hả em?

Bài 1: 

a) Ta có: \(\dfrac{2}{5}\cdot x+\dfrac{1}{3}=\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{2}{5}\cdot x=\dfrac{1}{5}-\dfrac{1}{3}=\dfrac{-2}{15}\)

\(\Leftrightarrow x=\dfrac{-2}{15}:\dfrac{2}{5}=\dfrac{-2}{15}\cdot\dfrac{5}{2}\)

hay \(x=-\dfrac{1}{3}\)

Vậy: \(x=-\dfrac{1}{3}\)

b) Ta có: \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}=\dfrac{3}{10}\)

\(\Leftrightarrow x=\dfrac{5}{3}:\dfrac{3}{10}=\dfrac{5}{3}\cdot\dfrac{10}{3}\)

hay \(x=\dfrac{50}{9}\)

Vậy: \(x=\dfrac{50}{9}\)

c) Ta có: \(\dfrac{4}{9}-\dfrac{5}{3}\cdot x=-2\)

\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{4}{9}+2=\dfrac{22}{9}\)

\(\Leftrightarrow x=\dfrac{22}{9}:\dfrac{5}{3}=\dfrac{22}{9}\cdot\dfrac{3}{5}\)

hay \(x=\dfrac{22}{15}\)

Vậy: \(x=\dfrac{22}{15}\)

d) Ta có: \(\dfrac{5}{7}:x-3=\dfrac{-2}{7}\)

\(\Leftrightarrow\dfrac{5}{7}:x=\dfrac{-2}{7}+3=\dfrac{19}{21}\)

\(\Leftrightarrow x=\dfrac{5}{7}:\dfrac{19}{21}=\dfrac{5}{7}\cdot\dfrac{21}{19}\)

hay \(x=\dfrac{15}{19}\)

Vậy:\(x=\dfrac{15}{19}\)

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

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

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

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

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

Trả lời:

7, 5( x + y )² + 15( x + y )

= 5( x + y )( x + y + 3 )

9, 7x( y - 4 )² - ( 4 - y )³ 

= 7x ( 4 - y )² - ( 4 - y )

= ( 4 - y )² ( 7x - 4 + y )

11, ( x + 1 )( y - 2 ) - ( 2 - y )²

= ( x + 1 )( y - 2 ) - ( y - 2 )²

= ( y - 2 )( x + 1 - y + 2 )

= ( y - 2 )( x - y + 3 )

8, 9x ( x - y ) - 10 ( y - x )² 

= 9x ( x - y ) - 10 ( x - y )²

= ( x - y )[ ( 9x - 10 ( x - y ) ]

= ( x - y )( 9x - 10x + 10y )

= ( x - y )( 10y - x )

10, ( a - b )² - ( a + b )( b - a ) 

= ( b - a )² - ( a + b )( b - a )

= ( b - a )( b - a - a - b )

= - 2a( b - a )

= 2a ( a - b )

12, 2x ( x - 3 ) + y ( x - 3 ) + ( 3 - x )

= 2x ( x - 3 ) + y ( x - 3 ) - ( x - 3 )

= ( x - 3 )( 2x + y - 1 )

a) Ta có: \(x^2-9x+20=0\)

\(\Leftrightarrow x^2-5x-4x+20=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)

Vậy: x∈{4;5}

b) Ta có: \(x^3-4x^2+5x=0\)

\(\Leftrightarrow x\left(x^2-4x+5\right)=0\)(1)

Ta có: \(x^2-4x+5\)

\(=x^2-4x+4+1=\left(x-2\right)^2+1\)

Ta có: \(\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-2\right)^2+1\ge1>0\forall x\)

hay \(x^2-4x+5>0\forall x\)(2)

Từ (1) và (2) suy ra x=0

Vậy: x=0

c) Sửa đề: \(x^2-2x-15=0\)

Ta có: \(x^2-2x-15=0\)

\(\Leftrightarrow x^2+3x-5x-15=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)

Vậy: x∈{-3;5}

d) Ta có: \(\left(x^2-1\right)^2=4x+1\)

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

\(\Leftrightarrow x^4-2x^2-4x=0\)

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

\(\Leftrightarrow x\left(x^3+2x^2+2x-2x^2-4x-4\right)=0\)

\(\Leftrightarrow x\cdot\left[x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\right]=0\)

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

Ta có: \(x^2+2x+2\)

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

Ta có: \(\left(x+1\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+1\right)^2+1\ge1>0\forall x\)

hay \(x^2+2x+2>0\forall x\)(4)

Từ (3) và (4) suy ra

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

Vậy: x∈{0;2}