a)  \(x^3\)\(-\)\(\frac{1}{4}x\)\(=\)\(0\)

\(x\left(x^2-\frac{1}{4}\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x^2-\frac{1}{4}=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x^2=0,5^2\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=+-0,5\end{cases}}\)

Vậy .............................

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

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

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+2=0\\x-4=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}3x=-2\\x=4\end{cases}}\)\(\orbr{\begin{cases}x=\frac{-2}{3}\\x=4\end{cases}}\)

Vậy ................................

c)  \(x^2\)\(\left(x-3\right)\)\(+\)\(12\)\(-\)\(4x\)\(=\)\(0\)

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

\(\left(x^2-4\right)\left(x-3\right)\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2\\x-3=0\end{cases}-4=0}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2\\x=3\end{cases}=2^2}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=+-2\\x=3\end{cases}}\)

a)\(x^3-\frac{1}{4}x=0\)

\(\Leftrightarrow x\left(x^2-\frac{1}{4}\right)=0\)

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

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\3x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-\frac{2}{3}\end{cases}}}\)

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

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

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

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

\(\left[\begin{array}{nghiempt}x+1=0\\x+3=0\end{array}\right.\)

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

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

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

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

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

\(\left[\begin{array}{nghiempt}2x-1=0\\2x+3=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}2x=1\\2x=-3\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=\frac{1}{2}\\x=-\frac{3}{2}\end{array}\right.\)

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

\(x^2-4x+3x-12=0\)

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

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

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

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

\(x^2-25-\left(x-5\right)=0\)

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

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

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

\(\left[\begin{array}{nghiempt}x-5=0\\x+4=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=5\\x=-4\end{array}\right.\)

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

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

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

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

\(\left[\begin{array}{nghiempt}x-1=0\\x+1=0\end{array}\right.\) (vì \(x^2+1\ge1>0\))

\(\left[\begin{array}{nghiempt}x=1\\x=-1\end{array}\right.\) 

`2(x-3)-4x=0`

`<=> 2x-6-4x=0`

`<=> -2x-6=0`

`<=>-2x=6`

`<=>x=-3`

__

`x^2-2x+1=25`

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

a: 2(x-3)-4x=0

=>2x-6-4x=0

=>-2x-6=0

=>2x+6=0

=>2x=-6

=>x=-3

b: \(x^2-2x+1=25\)

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

=>\(\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

`Answer:`

Bài 1:

a) \(7+2x=22-3x\)

\(\Leftrightarrow2x+3x=22-7\)

\(\Leftrightarrow5x=15\)

\(\Leftrightarrow x=3\)

b) \(8x-3=5x+12\)

\(\Leftrightarrow8x-5x=12+3\)

\(\Leftrightarrow3x=15\)

\(\Leftrightarrow x=5\)

c) \(x-12+4x=25+2x-1\)

\(\Leftrightarrow x-12+4x-25-2x+1=0\)

\(\Leftrightarrow\left(x+4x-2x\right)+\left(1-12-25\right)=0\)

\(\Leftrightarrow3x-36=0\)

\(\Leftrightarrow x=12\)

d) \(x+2x+3x-19=3x+5\)

\(\Leftrightarrow6x-19=3x+5\)

\(\Leftrightarrow6x-3x=5+19\)

\(\Leftrightarrow3x=24\)

\(\Leftrightarrow x=8\)

Bài 2:

a) \(\left(2,3x-6,9\right)\left(0,1x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2,3x-6,9=0\\0,1x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-20\end{cases}}}\)

b) \(\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0\)

\(\Leftrightarrow2x+7=0\text{ hoặc }x-5=0\text{ hoặc }5x+1=0\)

\(\Leftrightarrow x=-\frac{7}{2}\text{ hoặc }x=5\text{ hoặc }x=-\frac{1}{5}\)

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

\(\Leftrightarrow\orbr{\begin{cases}4x+2=0\\x^2+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x^2=-1\text{(Loại)}\end{cases}}}\)

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

\(\Leftrightarrow x^2-4+\left(3x-2x^2-6+4x\right)=0\)

\(\Leftrightarrow x^2-4=\left(-2x^2+7x-6\right)=0\)

\(\Leftrightarrow x^2-4-2x^2+7x-6=0\)

\(\Leftrightarrow-x^2+7x-10=0\)

\(\Leftrightarrow x^2-5x-2x+10=0\)

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

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

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

Câu a check lại đề số xấu

a) \(...\left(x+3\right)^3=5^3\Rightarrow x+3=5\Rightarrow x=2\)

b) \(...\Rightarrow-4< x< 9\)

c) \(...\Rightarrow2x+12>0\Rightarrow2x>-12\Rightarrow x>-6\)

nhưng đè là thế mà!!!

a: =>25-4x=1

=>4x=24

hay x=6

b: =>2x-4=0

hay x=2

c: =>x-35=115

hay x=150

d: =>x-2=12

hay x=14

e: =>x-36=216

hay x=252

a)  (x - 3)² - 5.(x - 2) + 5 = 0.

<=> x^2 - 6x + 9 - 5x + 10 + 5 = 0

<=> x^2 - 11x + 24 = 0

<=> (x-3)(x-8)=0

<=> x = 3 hoặc x = 8

b) (2x - 1)² - 3.(x - 2).(x + 2) - 25 = 0.

<=> 4x^2 - 4x + 1 - 3x^2 + 12 - 25 = 0

<=> x² - 4x - 12 = 0

<=> (x+2)(x-6) = 0

<=> x = -2 hoặc x = 6

d)  x² - 4x + 5 = 0.

<=> (x - 2)² = -1 (vô lý)

Vậy phương trình vô nghiệm

a) \(\left(x+3\right)^2-\left(x-2\right)^3=\left(x+5\right)\left(x^2-5x+25\right)-108\)

\(\Leftrightarrow x^2+6x+9-x^2+4x-4=x^3-5x^2+25x+5x^2-25x+125-108\)

\(\Leftrightarrow x^3-10x+12=0\Leftrightarrow\left(x-2\right)\left(x^2+2x+6\right)=0\)

\(\Leftrightarrow x=2\)( do \(x^2+2x+6=\left(x+1\right)^2+4\ge4>0\))

Câu 1:

a) 1 + x = 0

c) 1 - 2t = 0

d) 3y = 0

câu 1 bạn tự làm nha

câu 2:

a) 4x-20=0

<=> 4x =20

<=>x =\(\dfrac{20}{4}\)

<=> x=5

b) 2x+x+12=0

<=>3x+12=0

<=>3x =-12

<=>x =\(\dfrac{-12}{3}\)

<=> x=-4

c) x-5=3-x

<=>x+x=3+5

<=>2x=8

<=>x =\(\dfrac{8}{2}\)

<=>x=4

d) 7-3x=9-x

<=>-3x+x =9-7

<=>-2x=2

<=>x=\(\dfrac{2}{-2}\)

<=>x=-1