a: \(8x\left(x-2017\right)-2x+4034=0\)

\(\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)

c) =>x mũ 2 -9<0 hoăc x mũ 2 +1 <0

Mà x mũ 2+1>0 => x mũ 2+1 ko thể <0 

=>x mũ 2 -9< 0

=>x mũ 2 <9

=>x<3 hoặc x> -3 

Mk làm nhầm bạn sửa dấu > và < thành dấu bằng là đc nhé sorry

a) -3(x-4)+5(x-1)=-7

=>-3x+12+5x-5=-7

=>2x+7=-7

=>2x=-14=>x=-7

b) -4./x-8/+12=0

=>/x-8/=3

=>x-8=3 hoặc -3

(tự tính) 

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

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

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

a) (x+2)(x+3)-(x-2)(x+5)=0

  \(x^2+3x+2x+6-x^2-5x+2x+10=0\) 

\(2x+16=0\) 

\(2x=-16\) 

\(x=-8\) 

Vậy......

b) (8-5x)(x+2)+4(x-2)(x+1)+2(x-2)(x+2)=0

  \(8x+16-5x^2-10x+4x^2+4x-8x-8+2x^2+4x-4x-8=0\) 

  \(-6x+x^2=0\) 

 \(x\left(-6+x\right)=0\) 

=> x=0   hoặc  -6+x=0  <=>x=6

Vậy \(x\in\left\{0;6\right\}\)

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

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

\(\Leftrightarrow x^2+2x+3x+6-x^2+5x+2x+10=0\)

\(\Leftrightarrow12x+16=0\)

\(\Leftrightarrow12x=-16\)

\(\Leftrightarrow x=\frac{-4}{3}\)

Vậy...

Bài tui sai rồi 2x ghi nhầm 12x

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

\(x^2+5x+6-x^2-3x+10=0\)

\(2x+16=0\)

\(2x=-16\)

\(x=-8\)

\(b,\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)

\(8x+16-5x^2-10x+4x^2-4x-8+2x^2-8=0\)

\(x^2-6x=0\)

\(x\left(x-6\right)=0\)

\(\orbr{\begin{cases}x=0\\x-6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}}\)

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

\(\Rightarrow x^2+5x+6-x^2-3x+10=0\)

\(\Rightarrow2x=-16\Leftrightarrow x=-8\)

\(b,\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)

\(\Rightarrow8x+16-5x^2-10x+4\left(x^2-x+2\right)+2\left(x^2-4\right)=0\)

\(\Rightarrow8x+16x-5x^2-10x+4x^2-4x+8+2x^2-8=0\)

\(\Rightarrow x^2+10x=0\Rightarrow x\left(x+10\right)=0\Rightarrow x\in\left\{0;-10\right\}\)

a) Ta có: \(x^4-16x^2=0\)

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

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

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

b) Ta có: \(x^8+36x^4=0\)

\(\Leftrightarrow x^4\left(x^4+36\right)=0\)

\(\Leftrightarrow x^4=0\)

hay x=0

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

\(\Leftrightarrow\left(x-5\right)\cdot\left[\left(x-5\right)^2-1\right]=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-4\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)

d) Ta có: \(5\left(x-2\right)-x^2+4=0\)

\(\Leftrightarrow5\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)

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

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

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

a. x.(x - 4) - (x² - 8) = 0

<=> x² - 4x - x² + 8 = 0

<=> 8 - 4x = 0

<=> 4x = 8

<=> x = 2

Vậy x = 2.

a ) \(x\left(x-4\right)-\left(x^2-8\right)=0\)

\(\Leftrightarrow x^2-4x-\left(x^2-8\right)=0\)

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

\(\Leftrightarrow-4x+8=0\)

\(\Rightarrow-4x=-8\)

\(\Leftrightarrow x=-2\)

\(x\left(x-4\right)-\left(x^2-8\right)=0\)

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

\(\Rightarrow\left(x^2-x^2\right)-4x=0-8\)

\(\Rightarrow-4x=-8\)

\(\Rightarrow x=\left(-8\right):\left(-4\right)\)

\(\Rightarrow x=2\)

Vậy x = 2