\(x^2 + 10x + 25 + 4x^2 - 12x + 9 - 5(x^2 - 4) = 0\\ ⇔ -2x + 54 = 0\\ ⇔ x = 27\)

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

\(\Leftrightarrow x^2+10x+25+4x^2-12x+9-5\left(x^2-4\right)=0\)

\(\Leftrightarrow5x^2-2x+34-5x^2+20=0\)

\(\Leftrightarrow-2x+54=0\)

hay x=27

Vậy: S={27}

Ta có : x(x + 3) = 0

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

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

1.

a) x . ( x + 3 ) = 0

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

b) ( x - 2 ) . ( 5 - x ) = 0

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

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

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x^2+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\\text{x không tồn tại}\end{cases}}}\)

2.

-12 . ( x - 5 ) + 7 . ( 3 - x ) = 5

-12x - ( -60 ) + 21 - 7x = 5

-12x + 60 + 21 - 7x = 5

( -12x - 7x ) + ( 60 + 21 ) = 5

-19x + 81 = 5

-19x = 5 - 81

-19x = -76

x = ( -76 ) : (-19 )

x = 4

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

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

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

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

b)\(\left(3x-5\right)\left(5x-7\right)+\left(5x+1\right)\left(2-3x\right)=4\)

\(\Leftrightarrow15x^2-46x+35-15x^2+7x+2-4=0\)

\(\Leftrightarrow33-39x=0\Leftrightarrow33=39x\Leftrightarrow x=\frac{33}{39}\)

a) \(x\left(x-3\right)-2x+6=0\)

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

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

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

b) \((3x-5)(5x-7)+(5x+1)(2-3x)=4\)

\(15x^2-46x+35+10x-15x^2+2-3x-4=0\)

\(33-39x=0\)

\(3\left(11-13x\right)=0\)

\(11-13x=0\)

\(13x=11\)

\(x=\frac{11}{13}\)

lớp11

Bn viet bai nay co moi tay ko???

neu co thi cho mk

con neu ko cug cho mk nha

va ket bn vs mk nha

bn mắc rảnh ạ mk còn ko muốn đọc đừng nói là giải...

a: =>10+3x-3=6x+10

=>3x-3=6x

=>-3x=3

=>x=-1

b: =>x+1=0 hoặc x-2=0

=>x=-1 hoặc x=2

a) \(10+3\left(x-1\right)=10+6x\)

\(\Rightarrow10+3x-3=10+6x\)

\(\Rightarrow3x-6x=10-10+3\)

\(\Rightarrow-3x=3\)

\(\Rightarrow x=-\dfrac{3}{3}\)

\(\Rightarrow x=-1\)

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

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

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

Ta có: \(x^2=x^5\)

\(\Rightarrow x\in\left\{0;1\right\}\)

Vậy ...........

\(Ta \)  \(có : x^2 =x^5\)

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

\(\Leftrightarrow\)\(x^2 . (1 - x^3 )=0\)

\(\Rightarrow\)\(x^2 = 0 \)  \(hoặc \)  \(1 - x^3 = 0\)

\(\Rightarrow\)\(x = 0 \)  \(hoặc\)  \(x^3=1\)

\(\Rightarrow\)\(x = 0\)  \(hoặc\)  \(x = 1\)

\(Vậy : x = 0\)  \(hoặc \)  \(x = 1\)

x=1 hoặc x=0