8 x 2.019 x 125 - 250 x 2.018 x 4

= (8 x 125) x 2.019 - (250 x 4) x 2.018

= 1.000 x 2.019 - 1.000 x 2.018

=2.019.000 - 2.018.000

= 1.000

nhớ tick nhe

\(5x\left(x-2018\right)-x+2018=0\)

\(5x\left(x-2018\right)-\left(x-2018\right)=0\)

\(\left(x-2018\right)\left(5x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2018=0\\5x-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2018\\x=\frac{1}{5}\end{cases}}\)

Vậy.........

x = 2018

x = 1/5

t i c k nha

\(5x\left(x-2018\right)-x+2018=0\)

\(\Leftrightarrow5x\left(x-2018\right)-\left(x-2018\right)=0\)

\(\Leftrightarrow\left(x-2018\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2018=0\\5x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2018\\x=\frac{1}{5}\end{cases}}\)

( 1 + x ) + ( 2 + x ) + ( 3 + x ) + ...+ ( 100 + x ) = 2018

\(\Rightarrow\)( 1 + 2 + 3 + ... + 100 ) + ( x + x + x + ... + x ) = 2018

\(\Rightarrow\){( 1 + 100 ) . [( 100 - 1 ) : 1 + 1 ] : 2 } + ( x + x + ... + x ) = 2018

\(\Rightarrow\)5050 + x . 100 = 2018

\(\Rightarrow\)  x100 = 2018 - 5050 = -3032

\(\Rightarrow\)x = -3032 : 100 = -30,32

vậy x = -30,32

mà nè sai thì xin lỗi đề hơi có vấn đề nếu sai thì sorry nha !!!

a) ( 1/2-1/3-1/6).(1/2+2/3+3/4+...+2017/2018) + 3/4.x = 9/10

0.(1/2+2/3+3/4+...+2017/2018) + 3/4.x = 9/10

0+3/4.x = 9/10

3/4.x = 9/10

x = 9/10: 3/4

x = 6/5

b) x + ( 3/1.3+3/3.5+...+3/13.15) = 11/5

x + 3/2. ( 1-1/3 + 1/3 - 1/5 + ...+ 1/13 - 1/15) = 11/5

x + 3/2. ( 1-1/15) = 11/5

x + 3/2.14/15 = 11/5

x + 7/5 = 11/5

x = 11/5 - 7/5

x = 4/5

..... là gì?

a) ( 1/2-1/3-1/6).(1/2+2/3+3/4+...+2017/2018) + 3/4.x = 9/10

0.(1/2+2/3+3/4+...+2017/2018) + 3/4.x = 9/10

0+3/4.x = 9/10

3/4.x = 9/10

x = 9/10: 3/4

x = 6/5

b) x + ( 3/1.3+3/3.5+...+3/13.15) = 11/5

x + 3/2. ( 1-1/3 + 1/3 - 1/5 + ...+ 1/13 - 1/15) = 11/5

x + 3/2. ( 1-1/15) = 11/5

x + 3/2.14/15 = 11/5

x + 7/5 = 11/5

x = 11/5 - 7/5

x = 4/5

Ta có số số hạng của dãy:                        (2018-1) :1+1=2018                                Tổng gtri biểu thức:                                 =(1+2018).2018:2=2019.2018:2=2037171

Gợi ý :

Bài 1 : Cộng thêm 1 vào 3 phân thức đầu, trừ cho 3 ở phân thức thứ 4, có nhân tử chung là (x+2020)

Bài 2 : Trừ mỗi phân thức cho 1, chuyển vế và có nhân tử chung là (x-2021)

Bài 3 : Phân thức thứ nhất trừ đi 1, phân thức hai trù đi 2, phân thức ba trừ đi 3, phân thức bốn trừ cho 4, phân thức 5 trừ cho 5. Có nhân tử chung là (x-100)

bài 3

\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15.\)

=>\(\frac{x-90}{10}-1+\frac{x-76}{12}-2+\frac{x-58}{14}-3+\frac{x-36}{16}-4+\frac{x-15}{17}-5=0\)

=>\(\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)

=>\(\left(x-100\right).\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)

=>(x-100)=0 do \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)

=> x=100

\(\frac{x+2}{2018}+\frac{x+3}{2017}+\frac{x+4}{2016}+\frac{x+2036}{6}=0\)

\(\Leftrightarrow\frac{x+2}{2018}+1+\frac{x+3}{2017}+1+\frac{x+4}{2016}+1+\frac{x+2038}{6}-3=0\)

\(\Leftrightarrow\frac{x+2020}{2018}+\frac{x+2020}{2017}+\frac{x+2020}{2016}+\frac{x+2020}{6}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+\frac{1}{6}\right)=0\)

có : \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+\frac{1}{6}\ne0\)

\(\Leftrightarrow x+2020=0\)

\(\Leftrightarrow x=-2020\)

\(\frac{x-3}{2018}+\frac{x-2}{2019}=\frac{x-2019}{2}+\frac{x-2018}{3}\)

\(\Leftrightarrow\frac{x-3}{2018}-1+\frac{x-2}{2019}-1=\frac{x-2019}{2}-1+\frac{x-2018}{3}-1\)

\(\Leftrightarrow\frac{x-2021}{2018}+\frac{x-2021}{2019}=\frac{x-2021}{2}+\frac{x-2021}{3}\)

bài 3 thì lần lượt trừ đi 1; 2; 3; 4; 5

\(a,A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)-2018\)

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)-2018\)

Đặt \(x^2+5x=a\)

\(\Rightarrow A=\left(a-6\right)\left(a+6\right)-2018=a^2-2054\)

\(\Rightarrow A_{min}=2054\Leftrightarrow a=0\)

\(\Rightarrow x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\)

\(\Leftrightarrow x\in\left\{0;-5\right\}\)

\(b,B=\left(x-1\right)\left(x-4\right)\left(x-5\right)\left(x-8\right)+2018.\)

\(=\left(x^2-9x+8\right)\left(x^2-9x+20\right)+2018\)

Đặt \(x^2-9x+14=a\)

\(\Rightarrow B=\left(a-6\right)\left(a+6\right)+2018\)

\(=a^2-36+2018=a^2+1982\)

\(\Rightarrow B_{min}=1982\Leftrightarrow a^2=0\Rightarrow a=0\)

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

\(\Rightarrow x^2-2x-7x+14=0\)

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

\(\Rightarrow\left(x-2\right)\left(x-7\right)=0\)

\(\Rightarrow x\in\left\{2;7\right\}\)