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20 tháng 8 2021

1, \(2x^3-50x=0\Leftrightarrow2x\left(x^2-25\right)=0\Leftrightarrow x=0;x=\pm5\)

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

\(\Leftrightarrow5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left[5\left(x+1\right)-4\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+9\right)=0\Leftrightarrow x=-9;x=1\)

3, \(6x\left(x-2\right)=x-2\Leftrightarrow\left(6x-1\right)\left(x-2\right)=0\Leftrightarrow x=\frac{1}{6};x=2\)

4, \(7\left(x-2020\right)^2-x+2020=0\Leftrightarrow7\left(x-2020\right)^2-\left(x-2020\right)=0\)

\(\Leftrightarrow\left(x-2020\right)\left[7\left(x-2020\right)-1\right]=0\Leftrightarrow x=2020;x=\frac{14141}{7}\)

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

6, \(x^2-2x-3=0\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\Leftrightarrow x=-1;x=3\)

\(1,\)

\(2x^3-50x=0\)

\(\Leftrightarrow2x\left(x^2-25\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-25=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)

\(2,\)

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

\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)

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

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

\(\Leftrightarrow x\left(x-1\right)+9\left(x-1\right)=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+9=0\\x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-9\\x=1\end{cases}}\)

\(3,\)

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

\(\Leftrightarrow6x\left(x-2\right)-\left(x-2\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{6}\end{cases}}\)

\(4,\)

\(7\left(x-2020\right)^2-x+2020=0\)

\(\Leftrightarrow7\left(x-2020\right)^2-\left(x-2020\right)=0\)

\(\Leftrightarrow\left(x-2020\right)[7\left(x-2020\right)-1]=0\)

\(\Leftrightarrow\left(x-2020\right)[7x-14141]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2020\\7x=14141\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=\frac{14141}{7}\end{cases}}\)

\(5,\)

\(x^2-10x=-25\)

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

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

\(\Leftrightarrow x-5=0\)

\(\Leftrightarrow x=5\)

\(6,\)

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

12 tháng 8 2021

1/ x2-3x+2=0

⇒ (x2-2x)-(x-2)=0

⇒ x(x-2)-(x-2)=0

⇒ (x-1)(x-2)=0

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

2) x2-6x+5=0

⇒x2-6x+9-4=0

⇒(x2-6x+9)-22=0

⇒(x-3)2-22=0

⇒(x-3-2)(x-3+2)=0

⇒(x-5)(x-1)=0

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

3) 2x2+5x+3=0

⇒ (2x2+2x)+(3x+3)=0

⇒ 2x(x+1)+3(x+1)=0

⇒ (x+1)(2x+3)=0

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

4) x2-8x+15=0

⇒ (x2-8x+16)-1=0

⇒ (x-4)2-12=0

⇒ (x-4-1)(x-4+1)=0

⇒ (x-5)(x-3)=0

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

5) x2-x-12=0

⇒ (x2-4x)+(3x-12)=0

⇒ x(x-4)+3(x-4)=0

⇒ (x-4)(x+3)=0

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

1: Ta có: \(x^2-3x+2=0\)

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

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

2: Ta có: \(x^2-6x+5=0\)

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

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

3: Ta có: \(2x^2+5x+3=0\)

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

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

4: Ta có: \(x^2-8x+15=0\)

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

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

5: Ta có: \(x^2-x-12=0\)

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

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

14 tháng 6 2023

`1,(4x^3+3x^3):x^3+(15x^2+6x):(-3x)=0`

`<=> 4 + 3 + (-5x) + (-2)=0`

`<=> -5x+5=0`

`<=>-5x=-5`

`<=>x=1`

`2,(25x^2-10x):5x +3(x-2)=4`

`<=> 5x - 2 + 3x-6=4`

`<=> 8x -8=4`

`<=> 8x=12`

`<=>x=12/8`

`<=>x=3/2`

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

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

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

`<=>( x+1/2) (5x+3/2)=0`

`@ TH1`

`x+1/2=0`

`<=>x=0-1/2`

`<=>x=-1/2`

` @TH2`

`5x+3/2=0`

`<=> 5x=-3/2`

`<=>x=-3/2 : 5`

`<=>x=-15/2`

`4, x^2+8x+16=0`

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

`<=>x+4=0`

`<=>x=-4`

`5, 25-10x+x^2=0`

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

`<=>5-x=0`

`<=>x=5`

14 tháng 6 2023

\(x^2+8x+16=x^2+2.x.4+4^2=\left(x+4\right)^2\)

\(25-10x+x^2=5^2-2.5.x+x^2=\left(5-x\right)^2\)

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

14 tháng 7 2023

`@` `\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}.`

1: Ta có: \(2x\left(x+3\right)-6\left(x-3\right)=0\)

\(\Leftrightarrow2x^2+6x-6x+18=0\)

\(\Leftrightarrow2x^2+18=0\left(loại\right)\)

2: Ta có: \(2x^2\left(2x+3\right)+\left(2x+3\right)=0\)

\(\Leftrightarrow2x+3=0\)

hay \(x=-\dfrac{3}{2}\)

3: Ta có: \(\left(x-2\right)\left(x+1\right)-4x\left(x-2\right)=0\)

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

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

4: Ta có: \(2x\left(x-5\right)-3x+15=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)

5: Ta có: \(3x\left(x+4\right)-2x-8=0\)

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

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

6: Ta có: \(x^2\left(2x-6\right)+2x-6=0\)

\(\Leftrightarrow2x-6=0\)

hay x=3

14 tháng 10 2021

1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)

\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)

\(\Leftrightarrow x=2\)

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

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

\(\Leftrightarrow6x=6\)

hay x=1

22 tháng 7 2019

b) \(x^3+6x^2+9x=0\)

\(\Leftrightarrow x^3+3x^2+3x^2+9x=0\)

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

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

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

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

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

Vậy \(x\in\left\{-3;0\right\}\)

22 tháng 7 2019

a) \(2x\left(x-2\right)+x^2=4\)

\(\Leftrightarrow2x\left(x-2\right)+x^2-4=0\)

\(\Leftrightarrow2x\left(x-2\right)+\left(x-2\right)\left(x+2\right)=0\)

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

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

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

Vậy \(x\in\left\{\frac{-2}{3};2\right\}\)

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

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

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

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

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 

18 tháng 8 2021

1.

 \(x^2-5x+6=0\\ \Rightarrow x^2-2x-3x+6=0\\ \Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\\ \Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

2.

\(\left(x+4\right)^2-\left(3x-1\right)^2=0\\ \Rightarrow\left(x+4-3x+1\right)\left(x+4+3x-1\right)=0\\ \Rightarrow\left(-2x+5\right)\left(4x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}-2x+5=0\\4x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

3.

\(x^2-2x+24=0\\ \Rightarrow\left(x^2-2x+1\right)+23=0\\ \Rightarrow\left(x-1\right)^2+23=0\)

Vì (x-1)2≥0

23>0

\(\Rightarrow\left(x-1\right)^2+23>0\)

Vậy x vô nghiệm

4.

\(9x^2-4=0\\ \Rightarrow\left(3x-4\right)\left(3x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-4=0\\3x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)

5.

\(x^2+2x-8=0\\ \Rightarrow\left(x^2+2x+1\right)-9=0\\ \Rightarrow\left(x+1\right)^2-3^2=0\\ \Rightarrow\left(x-2\right)\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)