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.

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

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

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

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

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

2.

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

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

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

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

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

3.

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\2x+1=0\end{matrix}\right.\)

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

4.

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x+2=0\end{matrix}\right.\)

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

5.

\(\left(x+5\right)^2-16x^2=0\)

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

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

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

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

6.

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

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

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

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

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

a) x = 1; x = - 1 3                 b) x = 2.

c) x = 3; x = -2.                 d) x = -3; x = 0; x = 2.

a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

                 \(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

                \(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

           \(=>2x+\frac{3}{5}=\frac{3}{5}\)

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

                                    \(2x=0\)

                                      \(x=0:2\)

                                      \(x=0\)

b) \(\left(3x-1\right).\left(-\frac{1}{2x}+5\right)=0\)

=> \(\left(3x-1\right)=0\)hoặc \(\left(-\frac{1}{2x}+5\right)=0\)hoặc \(\left(3x-1\right)\)và\(\left(-\frac{1}{2x}+5\right)\)cùng bằng 0.

\(\orbr{\begin{cases}3x-1=0\\-\frac{1}{2x}+5=0\end{cases}}=>\orbr{\begin{cases}3x=1\\-\frac{1}{2x}=-5\end{cases}}=>\orbr{\begin{cases}x\in\varnothing\\2x=\frac{1}{5}\end{cases}}=>x=\frac{1}{5}:2=>x=\frac{1}{10}\)

a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

\(\left(2x+\frac{3}{5}\right)^2=0+\frac{9}{25}\)

\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

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

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

b) \(\left(3x-1\right)\left(-\frac{1}{2}x+5\right)=0\)

\(\left(3x-1\right)\left(-\frac{x}{2}+5\right)=0\)

\(\left(3x-1\right)\left(5-\frac{x}{2}\right)=0\)

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

\(\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)

1) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\)

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

\(\Leftrightarrow-4x+10=0\)

\(\Leftrightarrow-4x=-10\)

\(\Leftrightarrow x=\frac{5}{2}.\)

Vậy \(S=\left\{\frac{5}{2}\right\}\)

2)\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

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

\(\Leftrightarrow\left(x+3\right).\left(x^2-3x+9+x-9\right)=0\)

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

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

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

\(\Leftrightarrow x=-3\)hoặc \(x=0\)hoặc \(x=2\)

Vậy \(S=\left\{-3;0;2\right\}\)

Một. Khai triển vế trái của phương trình:
(x-3)(x+3) = x(x+3) - 3(x+3) = x^2 + 3x - 3x - 9 = x^2 - 9

Khai triển vế phải của phương trình:
(x-5)^2 = (x-5)(x-5) = x(x-5) - 5(x-5) = x^2 - 5x - 5x + 25 = x^2 - 10x + 25

Đặt hai cạnh bằng nhau:
x^2 - 9 = x^2 - 10x + 25

Trừ x^2 từ cả hai phía:
-9 = -10x + 25

Trừ 25 từ cả hai vế:
-34 = -10 lần

Chia cả hai vế cho -10:
x = 3,4

b. Khai triển vế trái của phương trình:
(2x+1)^2 - 4x(x-1) = (2x+1)(2x+1) - 4x^2 + 4x = 4x^2 + 2x + 2x + 1 - 4x^2 + 4x = 8x + 1

Đặt vế trái bằng 17:
8x + 1 = 17

Trừ 1 cho cả hai vế:
8x = 16

Chia cả hai vế cho 8:
x = 2

c. Khai triển vế trái của phương trình:
(3x-2)(3x+2) - 9(x-1)x = (9x^2 - 4) - 9x^2 + 9x - 9x = -4 + 9x

Đặt vế trái bằng 0:
-4 + 9x = 0

Thêm 4 vào cả hai bên:
9x = 4

Chia cả hai vế cho 9:
x = 4/9

d. Khai triển vế trái của phương trình:
(3-x)^3 - (x+3)^3 = (27 - 9x + x^2) - (x^3 + 9x^2 + 27) = 27 - 9x + x^2 - x^3 - 9x^2 - 27 = -x^3 - 8x^2 - 9x

Đặt vế trái bằng 36x^2 - 54x:
-x^3 - 8x^2 - 9x = 36x^2 - 54x

Cộng x^3 + 8x^2 + 9x vào cả hai vế:
0 = 37x^2 - 63x

Chia cả hai vế cho x:
0 = 37x - 63

Thêm 63 vào cả hai bên:
63 = 37 lần

Chia cả hai vế cho 37:
x = 63/37

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

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

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

\(\Leftrightarrow y=3\)

\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

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

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

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

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

\(\Leftrightarrow x\in\left\{0;-3;2\right\}\)

Bài làm

Vì ( y - 2 ) . ( y - 3 ) + ( y - 2 ) - 1 = 0

=> ( y - 2 ) = 0 hoặc ( y - 3 ) + ( y - 2 ) - 1 = 0

=> y = 2 hoặc y = 3 

Vậy y = 2 hoặc y = 3

~ Mấy câu còn lại làm tương tự. Làm theo mẫu câu a . b = 0 , => a = 0 hoặc b = 0. ~
# Chúc bạn học tốt # 

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

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

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

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

\(\left(x-7\right)\left(x+3\right)=\left(x+3\right)\left(2x-9\right)\)

\(\Leftrightarrow\left(x-7\right)\left(x+3\right)-\left(x+3\right)\left(2x-9\right)=0\)

\(\Leftrightarrow\left(x-7-2x+9\right)\left(x+3\right)=0\)

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

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