e: ta có: \(4x^2+4x-6=2\)

\(\Leftrightarrow4x^2+4x-8=0\)

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

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

f: Ta có: \(2x^2+7x+3=0\)

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

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

a)   \(\left(2x-1\right)^2-25=0\)

⇔ \(\left(2x-1\right)^2-5^2=0\)

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

⇒  \(2x-1-5=0\) hoặc \(2x-1+5=0\)

⇔      \(x=3\)           hoặc  \(x=-2\)

Bài 1: Tìm x

a) (2x-1) ² - 25 = 0

<=> (2x-1)² =  25

<=>  2x-1 = 5  hay 2x-1 =-5

<=>  2x= 6      hay  2x=-4

<=>   x=3     hay    x= -2

Vậy S={3; -2}
b) 3x (x-1) + x - 1 = 0

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

<=> x-1=0  hay  3x+1=0

<=> x=1 hay 3x=-1

<=> x=1 hay x=\(\dfrac{-1}{3}\)

Vậy S={1;\(\dfrac{-1}{3}\)}

c) 2(x+3) - x ² - 3x = 0

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

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

<=> x+3=0 hay 2-x=0

<=> x=-3  hay  x=2

Vậy S={-3;2}
d) x(x - 2) + 3x - 6 = 0

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

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

<=> x-2=0 hay x+3=0

<=> x=2 hay x=-3

Vậy S={2;-3}
e) 4x ² - 4x +1 = 0

<=> (2x-1)²=0

<=> 2x-1=0

<=> 2x=1

<=> x=\(\dfrac{1}{2}\)

Vậy S={\(\dfrac{1}{2}\)}
f) x +5x²  = 0

<=> x(1+5x)=0

<=>x=0 hay 1+5x=0

<=> x=0 hay 5x=-1

<=> x=0 hay x= \(\dfrac{-1}{5}\)

Vậy S={0;\(\dfrac{-1}{5}\)}
g) x ²+ 2x -3 = 0

<=> x²-x+3x-3=0

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

<=>  (x-1)(x+3)=0

<=> x-1=0 hay x+3=0

<=> x=1  hay x=-3

Vậy S={1;-3}

a) 3x(4x-3)-2x(5-6x)=0

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow24x^2-19x=0\)

\(\Leftrightarrow x\left(24x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{19}{24}\)

b) 5(2x-3)+4x(x-2)+2x(3-2x)=0

\(\Leftrightarrow\)10x-15+4x²-8x+6x-4x²=0

\(\Leftrightarrow8x-15=0\)

\(\Leftrightarrow8x=15\)

\(\Leftrightarrow x=\dfrac{15}{8}\)

vậy x=\(\dfrac{15}{8}\)

(x-1)(2x^2-8)=0

\(\Leftrightarrow\left(x-1\right)\left(2x^2-8\right)=0\\ \left(2x^3-8x-2x^2+8\right)=0\)

\(\Leftrightarrow2x\left(x-1\right)-8\left(x-1\right)=0\)

\(\Leftrightarrow x=1;x=\dfrac{8}{2}\)

3x^2-8x+5=0

áp dụng công thức bậc 2 ta có:

\(x=\dfrac{-\left(-8\right)\pm\sqrt{\left(-8\right)^2-4.3.5}}{2.3}\)

\(\Rightarrow x=\dfrac{5}{3};x=1\)

(7x-1).2x-7x+1=0

\(\Leftrightarrow\left(7x-1\right)\left(2x-1\right)=0\)

\(\Leftrightarrow x=\dfrac{1}{7};x=\dfrac{1}{2}\)

Bài 1:

a) \(\Rightarrow3x^2+3x-2x^2-4x+x+1=0\)

\(\Rightarrow x^2=-1\left(VLý\right)\Rightarrow S=\varnothing\)

b) \(\Rightarrow\left(x-2020\right)\left(2x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2}\end{matrix}\right.\)

c) \(\Rightarrow\left(x-10\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)

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

e) \(\Rightarrow\left(x+6\right)\left(x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)

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

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

Bài 2:

a) \(\Rightarrow3x\left(x^2-4\right)=0\Rightarrow3x\left(x-2\right)\left(x+2\right)=0\)

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

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

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

a: \(x\left(x-3\right)+2x-6=0\)

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

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

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

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

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

mik cam on ban

2a) pt <=> (x + 6)^2 = 0

<=> x = -6

b) pt <=> (4x - 1)^2 = 0

<=> x = 1/4

c) pt<=> (x + 1)^3 = 0

<=> x = -1

Bài 1:

a: Ta có: \(A=\left(4x+3y\right)^2+\left(4x-3y\right)^2\)

\(=16x^2+24xy+9y^2+16x^2-24xy+9y^2\)

\(=32x^2+18y^2\)

b: Ta có: \(B=\left(x-2\right)^3-\left(x+2\right)^3\)

\(=x^3-6x^2+12x-8-x^3-6x^2-12x-8\)

\(=-12x^2-24\)

Bài 2: 

a: Ta có: \(x^2+12x+36=0\)

\(\Leftrightarrow x+6=0\)

hay x=-6

b: Ta có: \(16x^2-8x+1=0\)

\(\Leftrightarrow4x-1=0\)

hay \(x=\dfrac{1}{4}\)

Bài 1: 

a: Ta có: \(A=\left(4x+3y\right)^2+\left(4x-3y\right)^2\)

\(=16x^2+24xy+9y^2+16x^2-24xy+9y^2\)

\(=32x^2+18y^2\)

b: Ta có: \(B=\left(x-2\right)^3-\left(x+2\right)^3\)

\(=x^3-6x^2+12x-8-x^3-6x^2-12x-8\)

\(=-12x^2-24\)

c: Ta có: \(C=\left(x+2y\right)^2+2\left(x+2y\right)\left(x-2y\right)+\left(x-2y\right)^2\)

\(=\left(x+2y+x-2y\right)^2\)

\(=4x^2\)

a) \(3\left(x-1\right)^2\cdot3x\left(x-5\right)=0\)

\(\Rightarrow9x\left(x-1\right)^2\left(x-5\right)=0\)

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

b) \(\left(x+3\right)^2-5x-15=0\)

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

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

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

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

c) \(2x^5-4x^3+2x=0\)

\(\Rightarrow2x\left(x^4-2x^2+1\right)=0\)

\(\Rightarrow2x\left[\left(x^2\right)^2-2\cdot x^2\cdot1+1^2\right]=0\)

\(\Rightarrow2x\left(x^2-1\right)^2=0\)

\(\Rightarrow2x\left(x-1\right)^2\left(x+1\right)^2=0\)

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

\(\text{#}Toru\)