a) \(\left(x-1\right)^2=0\)

=> \(x-1=0\)

=> \(x=1\)

b) \(x\left(2x+1\right)=0\)

=> \(\orbr{\begin{cases}x=0\\2x+1=0\end{cases}}\)

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

c) \(\left(-12\right)^2.x=56+10.13x\)

=> \(144x=56+130x\)

=> \(144x-130x=56\)

=> \(14x=56\)

=> \(x=56:14\)

=> \(x=4\)

d) \(x-\left(17-x\right)=x-7\)

=> \(x-17+x=x-7\)

=> \(2x-17=x-7\)

=> \(2x-x=-7+17\)

=> \(x=10\)

a) (x-1)² = 0

=> x - 1 = 0

=> x = 1

vậy_

b) x(2x+1)=0

=> x = 0 hoặc 2x + 1 = 0

=> x = 0 hoặc 2x = -1

=> x = 0 hoặc x = -1/2

vậy_

c) (-12)^2.x=56+10.13.x

=> 144x = 56 + 130x

=> 144x - 130x = 56

=> 14x = 56

=> x = 4

vậy_

d) x-(17-x)=x-7

=> x - 17 + x = x - 7

=> x + x - x = -7 + 17

=>x = 10

vậy_

e) (x+4) ⋮ (x+1)

=> x + 1 + 3 ⋮ x + 1

=> 3 ⋮ x + 1

=> x + 1 thuộc Ư(3)

=> x + 1 thuộc {-1; 1; -3; 3}

=> x thuộc {-2; 0; -4; 2}

vậy_

g) (4x+3)⋮(x-2)

=> 4x  - 4 + 7 ⋮ x - 2

=> 2(x - 2) + 7 ⋮ x - 2

=> 7 ⋮ x - 2

=> x - 2 thuộc Ư(7)

=> x - 2 thuộc {-1; 1; -7; 7}

=> x thuộc {1; 3; -5; 9}

vậy_

#)Giải :

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

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

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

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

c)\(\left(-12\right)^2.x=56+10.13x\)

\(\Rightarrow144x=56+130x\)

\(\Rightarrow14x=56\)

\(\Rightarrow x=4\)

d)\(x-\left(17-x\right)=x-7\)

\(\Rightarrow x-17+x=x-7\)

\(\Rightarrow x+x-17=x-7\)

\(\Rightarrow x+x-x=-7+17\)

\(\Rightarrow x=10\)

\(1,\Leftrightarrow x^2+10x+25=x^2-4x-21\\ \Leftrightarrow14x=-46\\ \Leftrightarrow x=-\dfrac{23}{7}\\ 2,\Leftrightarrow x^3+8=15+x^3+2x\\ \Leftrightarrow2x=-7\Leftrightarrow x=-\dfrac{7}{2}\\ 3,\Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x=-3\\ 4,\Leftrightarrow x^3-9x^2+27x-27=0\\ \Leftrightarrow\left(x-3\right)^3=0\\ \Leftrightarrow x-3=0\Leftrightarrow x=3\\ 5,\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\\ \Leftrightarrow-12x=24\Leftrightarrow x=-2\\ 6,\Leftrightarrow x^2-3x+5x-15=0\\ \Leftrightarrow\left(x-3\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

5: =>4x^2-1/9=0

=>(2x-1/3)(2x+1/3)=0

=>x=1/6 hoặc x=-1/6

6: =>x-1=2

=>x=3

7:=>(2x-1)^3=-27

=>2x-1=-3

=>2x=-2

=>x=-1

8: =>1/8(x-1)^3=-125

=>(x-1)^3=-1000

=>x-1=-10

=>x=-9

3: =>(5x-5)^2-4=0

=>(5x-7)(5x-3)=0

=>x=3/5 hoặc x=7/5

4: =>(5x-1)^2=0

=>5x-1=0

=>x=1/5

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

=>x=1/3 hoặc x=1/2

2: =>x^2(2x-3)-4(2x-3)=0

=>(2x-3)(x^2-4)=0

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

=>x=3/2;x=2;x=-2

`@` `\text {Answer}`

`\downarrow`

`1,`

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

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

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

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

`<=>`\(\left[{}\begin{matrix}2x-1=0\\3x-1=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}2x=1\\3x=1\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy,  `S = {1/2; 1/3}`

`2,`

\(x^2\left(2x-3\right)+12-8x=0\)

`<=> x^2(2x - 3) - 8x + 12 =0`

`<=> x^2(2x - 3) - (8x - 12) = 0`

`<=> x^2(2x - 3) - 4(2x - 3) = 0`

`<=> (x^2 - 4)(2x - 3) = 0`

`<=>`\(\left[{}\begin{matrix}x^2-4=0\\2x-3=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x^2=4\\2x=3\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x^2=\left(\pm2\right)^2\\x=\dfrac{3}{2}\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=\pm2\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy, `S = {+-2; 3/2}`

`3,`

\(25\left(x-1\right)^2-4=0\)

`<=> 25(x-1)(x-1) - 4 = 0`

`<=> 25(x^2 - 2x + 1) - 4 = 0`

`<=> 25x^2 - 50x + 25 - 4 = 0`

`<=> 25x^2 - 15x - 35x + 21 = 0`

`<=> (25x^2 - 15x) - (35x - 21) = 0`

`<=> 5x(5x - 3) - 7(5x - 3) = 0`

`<=> (5x - 7)(5x - 3) = 0`

`<=>`\(\left[{}\begin{matrix}5x-7=0\\5x-3=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}5x=7\\5x=3\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=\dfrac{7}{5}\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy, `S = {7/5; 3/5}`

`4,`

\(25x^2-10x+1=0\)

`<=> 25x^2 - 5x - 5x + 1 = 0`

`<=> (25x^2 - 5x) - (5x - 1) = 0`

`<=> 5x(5x - 1) - (5x - 1) = 0`

`<=> (5x - 1)(5x-1)=0`

`<=> (5x-1)^2 = 0`

`<=> 5x - 1 = 0`

`<=> 5x = 1`

`<=> x = 1/5`

Vậy,` S = {1/5}.`

`@` `\text {Ans}`

`\downarrow`

`5,`

`-4x^2 + 1/9 = 0`

`<=> -4x^2 = 0 - 1/9`

`<=> -4x^2 = -1/9`

`<=> 4x^2 = 1/9`

`<=> x^2 = 1/9 \div 4`

`<=> x^2 = 1/36`

`<=> x^2 = (+-1/6)^2`

`<=> x = +-1/36`

Vậy, `S = {1/36; -1/36}`

`6,`

`(x-1)^3 = 8`

`<=> (x-1)^3 = 2^3`

`<=> x-1=2`

`<=> x = 2 + 1`

`<=> x = 3`

Vậy, `S = {3}`

`7,`

`(2x-1)^3 + 27 = 0`

`<=> (2x - 1)^3 = -27`

`<=> (2x-1)^3 = (-3)^3`

`<=> 2x - 1 = -3`

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

`<=> 2x = -2`

`<=> x = -1`

Vậy,` S = {-1}`

`8,`

`125 + 1/8(x-1)^3 = 0`

`<=> 1/8(x-1)^3 = - 125`

`<=> (x-1)^3 = -125 \div 1/8`

`<=> (x-1)^3 = -1000`

`<=> (x-1)^3 = (-10)^3`

`<=> x - 1 = - 10`

`<=> x = -10+1`

`<=> x = -9`

Vậy, `S = {-9}.`

1: =>(x+3)(x-5)=0

=>x=5 hoặc x=-3

2: =>(x-1)(5x-1)=0

=>x=1/5 hoặc x=1

5: =>(x-4)*x=0

=>x=0 hoặc x=4

10: =>(x+5)(x-3)=0

=>x=3 hoặc x=-5

9: =>(x-2)(x-4)=0

=>x=2 hoặc x=4

7: =>(x-6)(2x-1)=0

=>x=1/2 hoặc x=6

8: =>(2x-1)(3x-12)=0

=>x=4 hoặc x=1/2

a)  \(7x^2-16x=2x^3-56\)

\(\Leftrightarrow\)\(2x^3-7x^2+16x-56=0\)

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

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

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

\(\Leftrightarrow\)\(x=3,5\)

Vậy...

b)  \(x^7+x^3+2x^5+2x=0\)

\(\Leftrightarrow\)\(x.\left(x^6+x^2+2x^4+2\right)=0\)

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

\(\Leftrightarrow\)\(x=0\)

Vậy...

c)  \(\left(2x+1\right)x-5\left(x+\frac{1}{2}\right)=0\)

\(\Leftrightarrow\)\(2x\left(x+\frac{1}{2}\right)-5\left(x+\frac{1}{2}\right)=0\)

\(\Leftrightarrow\)\(\left(2x-5\right)\left(x+\frac{1}{2}\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}2x-5=0\\x+\frac{1}{2}=0\end{cases}}\)

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

Vậy...

a. x- (17 - x) = x-7

x-17+x = x+7

2x-17=x+7

2x-x= 7+17

x= 24

b. 9-25=(7-x)-(25+7)

-16=7-x-25-7

-16=-x-25

x= -25 + 16

x = -9

c. (-12)^2 .x = 56 + 10.13.x

144x = 56 + 130x

144x - 130x = 56

14x=56

x= 56 : 14

x= 4

d. 4/24 : (3x-2) = -3

4/24 = -3.(3x-2)

4/24 = -9x + 6

9x = 6 - 4/24

9x = 35/6

x= 35/6 :9

x= 35/54

e. 5/-45 chia hay nhân với (-3-2x) nhỉ

chia bạn nhé

1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)

=>-13x=0

=>x=0

2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

=>3x=13

=>x=13/3

3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)

=>-2x^2=0

=>x=0

4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

=>-8x=6-14=-8

=>x=1

`1)2x(x-5)-(3x+2x^2)=0`

`<=>2x^2-10x-3x-2x^2=0`

`<=>-13x=0`

`<=>x=0`

___________________________________________________

`2)x(5-2x)+2x(x-1)=13`

`<=>5x-2x^2+2x^2-2x=13`

`<=>3x=13<=>x=13/3`

___________________________________________________

`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`

`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`

`<=>x=0`

___________________________________________________

`4)5x(x-1)-(x+2)(5x-7)=0`

`<=>5x^2-5x-5x^2+7x-10x+14=0`

`<=>-8x=-14`

`<=>x=7/4`

___________________________________________________

`5)6x^2-(2x-3)(3x+2)=1`

`<=>6x^2-6x^2-4x+9x+6=1`

`<=>5x=-5<=>x=-1`

___________________________________________________

`6)2x(1-x)+5=9-2x^2`

`<=>2x-2x^2+5=9-2x^2`

`<=>2x=4<=>x=2`

Δ=(-2)^2-4(m-3)

=4-4m+12=16-4m

Để phương trình có hai nghiệm dương phân biệt thì 16-4m>0 và m-3>0

=>m>3 và m<4

x1^2+x2^2=(x1+x2)^2-2x1x2

=2^2-2(m-3)=4-2m+6=10-2m

=>x1^2=10-2m-x2^2

x1^2+12=2x2-x1x2

=>10-2m-x2^2+12=2x2-m+3

=>\(-x_2^2+22-2m-2x_2+m-3=0\)

=>\(-x_2^2-2x_2-m+19=0\)

=>\(x_2^2+2x_2+m-19=0\)(1)

Để (1) có nghiệmthì 2^2-4(m-19)>0

=>4-4m+76>0

=>80-4m>0

=>m<20

=>3<m<4