\(5,4x^2-36=0\\ \Leftrightarrow\left(2x\right)^2-6^2=0\\ \Leftrightarrow\left(2x-6\right)\left(2x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\2x+6=0\end{matrix}\right.\)

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

Vậy \(S=\left\{3;-3\right\}\)

\(7,\left(3x+1\right)^2-16=0\\ \Leftrightarrow\left(3x+1\right)^2-4^2=0\\ \Leftrightarrow\left(3x+1-4\right)\left(3x+1+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-3=0\\3x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\)

Vậy \(S=\left\{1;-\dfrac{5}{3}\right\}\)

\(8,\left(2x-3\right)^2-49=0\\ \Leftrightarrow\left(2x-3\right)^2-7^2=0\\ \Leftrightarrow\left(2x-3-7\right)\left(2x-3+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-10=0\\2x+4=0\end{matrix}\right.\)

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

Vậy \(S=\left\{-2;5\right\}\)

c) x( 2x - 3 ) - 2( 3 - 2x) =0

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

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

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

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

d) 25x² - 36 =0

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

\(\Leftrightarrow\left(5x-6\right)\left(5x+6\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}5x-6=0\\5x+6=0\end{array}\right.\)

\(\Leftrightarrow x=\pm\frac{6}{5}\)

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

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

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

b) \(25x^2-36=0\)

\(\Leftrightarrow\left(5x-6\right)\left(5x+6\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}5x-6=0\\5x+6=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{6}{5}\\x=-\frac{6}{5}\end{array}\right.\)

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

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

\(\Leftrightarrow\left(y-3\right)^2=0\)

\(\Leftrightarrow y=3\)

\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

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

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

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

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

\(\Leftrightarrow x\in\left\{0;-3;2\right\}\)

Bài làm

Vì ( y - 2 ) . ( y - 3 ) + ( y - 2 ) - 1 = 0

=> ( y - 2 ) = 0 hoặc ( y - 3 ) + ( y - 2 ) - 1 = 0

=> y = 2 hoặc y = 3 

Vậy y = 2 hoặc y = 3

~ Mấy câu còn lại làm tương tự. Làm theo mẫu câu a . b = 0 , => a = 0 hoặc b = 0. ~
# Chúc bạn học tốt # 

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

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

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

\(\Leftrightarrow x\in\left\{2;-3\right\}\)

\(\left(x-7\right)\left(x+3\right)=\left(x+3\right)\left(2x-9\right)\)

\(\Leftrightarrow\left(x-7\right)\left(x+3\right)-\left(x+3\right)\left(2x-9\right)=0\)

\(\Leftrightarrow\left(x-7-2x+9\right)\left(x+3\right)=0\)

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

\(\Leftrightarrow x\in\left\{2;-3\right\}\)

\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

\(2x+\left(1+2+3+...+100\right)=15150\)

\(2x+\left[\left(1+100\right)+\left(2+99\right)+...+\left(50+51\right)\right]=15150\)

\(2x+\left[101+101+...+101\right]=15150\)CÓ 50 SỐ 101

\(2x+\left[101\times50\right]=15150\)

\(2x=15150:5050\)

\(2x=3\)

\(x=3:2\)

\(x=1.5\)

a, 2x + (1+2+3+4+...+100) = 15150 

=> 2x + \(\frac{\left(1+100\right).\left[\left(100-1\right)+1\right]}{2}\)= 15150 

=> 2x + \(\frac{101.100}{2}\)= 15150 

=> 2x + 5050 = 15150 

=> 2x             = 15150 - 5050 

=> 2x             = 10100

=> x              =  10100 : 2 

=> x              = 5050 

Vậy x = 5050 

b, .(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)+(x+7)+(x+8)=36 

=> (x + x + x + x +x + x +x +x ) + (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) = 36 

=> 8x + 36 = 36 

=> 8x         = 0 

=>  x          = 0 

Vậy x = 0 

c, 0+0+4+6+8+...+2x=110 

Sửa đề :0 + 2 + 4 + 6 + 8 + ... + 2x = 110 = 2 + 4 + 6 + 8 + ... + 2x = 110 

SSH  : \(\frac{\left(2\text{x}-2\right)}{2}+1=x-1+1=x\)

Tổng : \(\frac{\left(2\text{x}+2\right).x}{2}=110\Leftrightarrow\frac{2.\left(x+1\right).x}{2}=110\)

                                                    \(\Leftrightarrow\left(x+1\right)x=110\)

                                                     \(\Leftrightarrow\left(10+1\right).10=110\)

 => x = 10 

Vậy x = 10 

#)Giải :

a) 2x + ( 1 + 2 + 3 + 4 + ... + 100 ) = 15150

=> 2x + [ ( 100 + 1 ) x 100 : 2 ] = 15150

=> 2x + 5050 = 15150

=> 2x = 10100

=> x = 5050

b) ( x + 1 ) + ( x + 2 ) + ... + ( x + 8 ) + ( x + 9 ) = 36

=> ( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 9 ) = 36

=> ( x + x + x + ... + x ) + 45 = 36

=> ( x + x + x + ... + x ) = -9

=> x = -1

c) Đề kiểu j v ? xem lại đi bạn ???

\(\left(x-5\right)\left(x-7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=7\end{cases}}}\)

tíc mình nha

dễ mà bạn

giúp mình giải bài này với ^_​​

3^{39 và 11 mũ 21}

​