\(3\left(2x-3\right)\left(3x+2\right)-2\left(x+4\right)\left(4x-3\right)+9x\left(4-x\right)=0\)

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

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

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

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

xin lỗi toán lớp 8 thì mk chịu

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

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

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

vậy...

b) \(\left|5x+2\right|-4x=7\)

\(\left|5x+2\right|=7+4x\)

\(\Rightarrow\orbr{\begin{cases}5x+2=7+4x\\5x+2=-7-4x\end{cases}}\Rightarrow\orbr{\begin{cases}5x-4x=7-2\\5x+4x=-7-2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=5\\9x=-9\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)

vậy.... 

k nha 

a, \(\left(3-x\right)\left(5x+10\right)=0\)

\(\Rightarrow3-x=0\)   hoặc     \(5x+10=0\)

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

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

=> 4x^2-12x+9 - 4x^2 + 12x=0

=> 9=0 ( vô cmn lí )

=> vô nghiệm

Sai or đúng chưa rõ tự kiểm tra oke

a)

\(x^2-5x+4x-20=0.\)

\(x^2-x-20=0\)

\(\left(x^2-x+\frac{1}{4}\right)-20-\frac{1}{4}=0\)

\(\left(x-\frac{1}{2}\right)^2-\left(\frac{20.4+1}{4}\right)=0\)

\(\hept{\begin{cases}x-\frac{1}{2}-\left(\frac{20.4+1}{4}\right)=0\\x-\frac{1}{2}+\left(\frac{20.4+1}{4}\right)=0\end{cases}}\)

b)  \(x^2+6x-7x-42=0\)

\(x^2-x-42=0\)

\(x^2-x+\frac{1}{4}-42-\frac{1}{4}=0\)

\(\left(x-\frac{1}{2}\right)^2-\left(\frac{42.4+1}{4}\right)=0\)  " tương tự con A

\(x^3-16x=0\)

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

\(x=0,+4,-4\)

\(x^3-16x=0\)

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

\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-16=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=16\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}}\)

Vậy \(x=0\)hoặc \(x=\pm4\)

Tham khảo nhé~

a) 2x + 5 < 0 => 2x < - 5 => x < -2,5

b) -4 - 5x > 0 => -4 > 5x => -0,8 > x

c) -7x + 3 < 0 => -7x < -3 => x > 3/7

d) x - 7 > 0 => x > 7

e) -3 + 4x > 0 => 4x > 3 => x > 0,75

\(a,2x+5< 0\)                                                        \(b,-4-5x>0\)

\(\Rightarrow2x< -5\)                                                          \(\Rightarrow-4>5x\)

\(\Rightarrow x< -\frac{5}{2}\)                                                          \(\Rightarrow x< -\frac{4}{5}\)

\(c,-7x+3< 0\)                                                   \(d,x-7>0\)

\(\Rightarrow-7x< -3\)                                                      \(\Rightarrow x>7\)

\(\Rightarrow x>\frac{3}{7}\)

\(e,-3+4x>0\)

\(\Rightarrow4x>3\)

\(\Rightarrow x>\frac{3}{4}\)

đặt A=(x+1)+(X+2)+(x+3)+....+(x+99)

=> A= x+1+x+2+x+3+....+x+100

=x+x+x+x+...+x+(1+2+3+4+..+99)( có 99x)

=> 99x+4950=0

=> 99x=-4950

=> x=-50

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

x²+4x+4=0
(x+2)²=0
x+2=0
x=+-2
câu 1 giống câu 2
(x+1)²+2(x+1)=0
(x+1+2)(x+1)=0
Th1: x+3=0           Th2: x+1=0
            x=-3                      x=-1
vậy ...

a) x+3=12

x=12-3

x=9

b)(x-3):2=5¹⁴:5¹²

=>(x-3):2=5²

=>(x-3):2=25

=>x-3=25.2

=>x-3=50

=>x=50+3

=>x=53

c)4x+3x=30-20:10

=>x(4+3)=30-2

=>7x=28

=>x=28:7

=>x=4

d)2x-138=2³.3²

=>2x-138=8.9

=>2x-138=72

=>2x=72+138

=>2x=210

=>x=210:2

=>x=105

a) x + 3 = 12

x = 12 - 3

x = 9

b) ( x - 3 ) : 2 = 5¹⁴ : 5¹²

( x - 3 ) : 2 = 5^14-12

( x - 3 ) : 2 = 5²

( x - 3 ) : 2 = 25

x - 3 = 50

x = 53

c) 4x + 3x = 30 - 20 : 10

7x = 28

x = 4

d) 2x - 138 = 2³ x 3²

2x - 138 = 8 x 9

2x - 138 = 72

2x = 210

x = 105