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

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

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

\(\left(x-1\right)\left(x-4\right)\left(x-5\right)\left(x-8\right)+36=0\)

\(\Rightarrow\left(x-1\right)\left(x-8\right)\left(x-4\right)\left(x-5\right)+36=0\)

\(\Leftrightarrow\left(x^2-8x-x+8\right)\left(x^2-4x-5x+20\right)+36=0\)

\(\Rightarrow\left(x^2-9x+8\right)\left(x^2-9x+20\right)+36=0\)

\(\Rightarrow\left(x^2-9x+14-6\right)\left(x^2-9x+14+6\right)+36=0\)

\(\Rightarrow\left(x^2-9x+14\right)^2-36+36=0\)

\(\Rightarrow x^2-9x+14=0\Leftrightarrow\left(x-2\right)\left(x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=7\end{matrix}\right.\)

Rảnh rỗi thật sự .-.

Bài 1.

a) \(15-5\left|x+4\right|=-12-3\)

\(\Leftrightarrow5\left|x+4\right|=30\)

\(\Leftrightarrow\left|x+4\right|=6\)

\(\Leftrightarrow\orbr{\begin{cases}x+4=6\\x+4=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-10\end{cases}}\)

b) \(\left(4x-8\right)\left(7-x\right)=0\Leftrightarrow\orbr{\begin{cases}4x-8=0\\7-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=7\end{cases}}\)

c) \(\left(x^2-36\right)\left(x^2+5\right)=0\Rightarrow\left(x-6\right)\left(x+6\right)=0\)

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

d) \(-3\left(x+7\right)-11=2\left(x+5\right)\)

\(\Leftrightarrow-3x-32=2x+10\)

\(\Leftrightarrow5x=-42\Rightarrow x=-\frac{42}{5}\)

\(a,\Leftrightarrow2x^2+10x-2x^2=12\Leftrightarrow x=\dfrac{12}{10}=\dfrac{6}{5}\\ b,\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(9-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\\ d,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ e,\Leftrightarrow4x^2-4x+1-4x^2+196=0\\ \Leftrightarrow-4x=-197\Leftrightarrow x=\dfrac{197}{4}\)

\(f,\Leftrightarrow x^2+8x+16-x^2+1=16\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\\ g,Sửa:\left(3x+1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x+1-x-1\right)\left(3x+1+x+1\right)=0\\ \Leftrightarrow2x\left(4x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\ h,\Leftrightarrow x^2+8x-x-8=0\\ \Leftrightarrow\left(x+8\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\\ i,\Leftrightarrow2x^2-13x+15=0\\ \Leftrightarrow2x^2+2x-15x-15=0\\ \Leftrightarrow\left(x+1\right)\left(2x-15\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{15}{2}\end{matrix}\right.\)

1) x - 36 + 12 = - x+ 10

=> x + x  = 10 + 24

=> 2x = 34

=> x = 34/2 = 17

2) (x + 15) - (11 - x) = (-2)²

=> x + 15 - 11 + x = 4

=> 2x = 4 - 4

=> 2x = 0

=> x = 0

3) 40 - 4x² = (-6)²

=> 40 -  4x² = 36

=> 4x² = 40 - 36

=> 4x² = 4

=> x² = 1

=> x = \(\pm\)1

4) (-50) + 10x² = (-25) x |-2|

=> -50 + 10x² = -50

=> 10x² = -50 + 50

=> 10x² = 0

=> x² = 0

=> x = 0

5) |x + 1| = 2020

=> \(\orbr{\begin{cases}x+1=2020\\x+1=-2020\end{cases}}\)

=> \(\orbr{\begin{cases}x=2019\\x=-2021\end{cases}}\)

6) (x + 1)⁵ + 8 = 0 (xem lại đề)

7) (-20) + x³ : 16 = -24

=> x³ : 16 = -24 + 20

=> x³ : 16 = -4

=> x³ = -4 . 16

=> x³ = -64 = (-4)³

=> x = -4

9) x¹⁴ = x¹⁷

=> x¹⁴ - x¹⁷ = 0

=> x¹⁴(1 - x³) = 0

=> \(\orbr{\begin{cases}x^{14}=0\\1-x^3=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

10) (-36) + (1 - x)² = 0

=> (1 - x)² = 36

=> (1 - x)² = 6²

=> \(\orbr{\begin{cases}1-x=6\\1-x=-6\end{cases}}\)

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