cái bài 2 câu 1 câu 2 và câu 3 sửa cái vế phải lại thành 3/2-1-2x/4 và -15/5 và 2.(x-1)/5

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

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

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

Vậy ...

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

\(\Leftrightarrow x^2+2x+3x+6=0\)

\(\Leftrightarrow x^2+5x+6=0\)

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

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

Vậy ...

Còn cậu nữa chịu rồi !

câu 2 nhé :

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

câu này em phải sử dụng tam thức bậc 2 liệu em đã học chưa z :(????

1/ x² - 5x + 6 = 0 
⇔ x² - 2x - 3x + 6 = 0 
⇔ x(x - 2) - 3(x - 2) = 0 
⇔ (x - 2)(x - 3) = 0 
⇒S = {2 ; 3}.

1) \(x^2+5x+6=0\)

\(\Leftrightarrow x^2+2x+3x+6=0\)

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

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

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

2) \(2\left(x+3\right)-x^2-3x=0\)

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

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

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

3) \(x^2+4x+3=0\)

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

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

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

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

4) \(2x^2-3x-5=0\)

\(\Leftrightarrow2x^2+2x-5x-5=0\)

\(\Leftrightarrow2x\left(x+1\right)-5\left(x+1\right)=0\)

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+1=0\\2x-5=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-1\\x=\frac{5}{2}\end{array}\right.\)

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

<=> x-2=0 hoặc 2x+3 = 0

<=> x=2 hoặc x=-3/2

2. <=> x^2-4x+4-x^2+9 = 0

<=> 13-4x=0

<=> 4x=13

<=> x = 13/4

3.<=>4x^2-24x+36 - 4x^2+1 =  10

<=> 37-24x = 10

<=> 24x = 37 - 10 = 27

<=> x = 27 : 24 = 9/8

k mk nha

2. \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)

\(x^2-4x+4-x^2+9=0\)

\(-4x+13=0\)

\(-4x=-13\)

\(x=\frac{13}{4}\)

vậy \(x=\frac{13}{4}\)

1. 2x -8 =0

2. 4x + 274/21 = 0