Tìm x khi biết x:☺
\(\left(\frac{x-4}{2017}\right)+\left(\frac{x-3}{2018}\right)+\left(\frac{x-2}{2019}\right)+\left(\frac{x-1}{2020}\right)=4\)
gip mik với mik cần gấppppppppppppppppp
a.\(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+_{\frac{x-11}{12}}\)
b.\(\frac{x-1}{2020}+\frac{x-2}{2019}-\frac{x-3}{2018}=\frac{x-4}{2017}\)
c.\(\left(\frac{3}{4}x+3\right)-\left(\frac{2}{3}x-4\right)-\left(\frac{1}{6}x+1\right)=\left(\frac{1}{3}x+4\right)-\left(\frac{1}{3}x-3\right)\)
giúp mk vs mk đg cần gấp
Thank you
a) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
=> \(\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)
=> \(\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}+\frac{x+1}{12}=0\)
=> \(\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)
=> x + 1 = 0
=> x = -1
b) \(\frac{x-1}{2020}+\frac{x-2}{2019}-\frac{x-3}{2018}=\frac{x-4}{2017}\)
=> \(\left(\frac{x-1}{2020}-1\right)+\left(\frac{x-2}{2019}-1\right)-\left(\frac{x-3}{2018}-1\right)=\left(\frac{x-4}{2017}-1\right)\)
=> \(\frac{x-2021}{2020}+\frac{x-2021}{2019}-\frac{x-2021}{2018}=\frac{x-2021}{2017}\)
=> \(\left(x-2021\right)\left(\frac{1}{2020}+\frac{1}{2019}-\frac{1}{2018}-\frac{1}{2017}\right)=0\)
=> x - 2021 = 0
=> x = 2021
c) \(\left(\frac{3}{4}x+3\right)-\left(\frac{2}{3}x-4\right)-\left(\frac{1}{6}x+1\right)=\left(\frac{1}{3}x+4\right)-\left(\frac{1}{3}x-3\right)\)
=> \(\frac{3}{4}x+3-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}x+3\)
=> \(-\frac{1}{12}x+6=7\)
=> \(-\frac{1}{12}x=1\)
=> x = -12
Giải các phương trình sau:
a) \(\left(\frac{x-2}{x-1}\right)^2-5\left(\frac{x+2}{x+1}\right)^2+4\left(\frac{x^2-4}{x^2-1}\right)=1\)
b) \(\left(\frac{x-1}{x}\right)^2+\left(\frac{x-1}{x-2}\right)^2=\frac{40}{9}\)
c) \(x.\frac{4-x}{x+2}.\left(\frac{8-2x}{x+2}\right)=3\)
d) \(\frac{1}{3x-2020}+\frac{1}{4x-2018}+\frac{1}{5x-2017}=\frac{1}{12x-2019}\)
Tìm x biết \(\frac{\left(2019-x\right)^2+\left(2019-x\right)\left(x-2020\right)}{\left(2019-x\right)^2-\left(2019-x\right)\left(x-2020\right)}\)\(\frac{+\left(x-2020\right)^2}{+\left(x-2020\right)^2}\)\(=\frac{19}{49}\)
Tìm x biết
a) \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
b) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(a)\) Ta có :
\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)
\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)
\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
Lại có :
\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)
\(\Rightarrow\)\(x=2019\)
Vậy \(x=2019\)
Chúc bạn học tốt ~
\(b)\) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(1-\frac{2}{x+1}=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(\frac{2}{x+1}=1-\frac{2017}{2019}\)
\(\Leftrightarrow\)\(\frac{2}{x+1}=\frac{2}{2019}\)
\(\Leftrightarrow\)\(x+1=2019\)
\(\Leftrightarrow\)\(x=2019-1\)
\(\Leftrightarrow\)\(x=2018\)
Vậy \(x=2018\)
Chúc bạn học tốt ~
Tìm x, biết:
a) \(x-\left\{\left[-x-\left(x+3\right)\right]-\left[\left(x+2018\right)-\left(x+2019\right)\right]+21\right\}=2020\)
b) \(\frac{1}{x}-\left\{\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2018.2020}\right\}=\frac{1}{2019}\)
Help me! ( tặng 2 - 3 tk )
a) Ta có:
\(x-\left\{\left[-x-\left(x+3\right)\right]-\left[\left(x+2018\right)-\left(x+2019\right)\right]+21\right\}\)
\(=x-\left\{\left[-x-x-3\right]-\left[x+2018-x-2019\right]+21\right\}\)
\(=x-\left\{\left[-2x-3\right]-\left[2018-2019\right]+21\right\}\)
\(=x+2x+-3+1-21\)
\(=3x-23\)
=> \(3x-23=2020\)
\(3x=2020+23=2043\)
=> \(x=2043:3=681\)
Nhầm
\(=x-\left\{-2x-3+1+21\right\}\\ =x+2x+3-1-21\)
\(=3x-17\\ =>3x-17=2020\\ 3x=2020+17=2037\\ x=2037:3=679\)
tìm x biết
\(\frac{\left(2019-x^2\right)+\left(2019-x\right)\left(x-2020\right)+\left(x-2020\right)^2}{\left(2019-x\right)^2-\left(2019-x\right)\left(x-2020\right)+\left(x-2020^2\right)}\) = \(\frac{19}{49}\)
Cho \(f\left(x\right)=\frac{x^3}{1-3x+3x^2}.\) Tính \(A=f\left(\frac{1}{2020}\right)+f\left(\frac{2}{2020}\right)+...+f\left(\frac{2018}{2020}\right)+f\left(\frac{2019}{2020}\right).\)
Cho \(f\left(x\right)=\frac{x^3}{1-3x+3x^2}.\) Tính \(A=f\left(\frac{1}{2020}\right)+f\left(\frac{2}{2020}\right)+...+f\left(\frac{2018}{2020}\right)+f\left(\frac{2019}{2020}\right).\)
Xét \(f\left(x\right)+f\left(1-x\right)=\frac{x^3}{1-3x+3x^2}+\frac{\left(1-x\right)^3}{1-3\left(1-x\right)+3\left(1-x\right)^2}\)
\(=\frac{x^3}{1-3x+3x^2}+\frac{1-3x+3x^2-x^3}{1-3+3x+3-6x+3x^2}\)
\(=\frac{x^3}{1-3x+3x^2}+\frac{1-3x+3x^2-x^3}{1-3x+3x^2}\)
\(=\frac{1-3x+3x^2}{1-3x+3x^2}=1\)
Thay vào ta tính được:
\(A=\left[f\left(\frac{1}{2020}\right)+f\left(\frac{2019}{2020}\right)\right]+...+\left[f\left(\frac{1009}{2020}\right)+f\left(\frac{1011}{2020}\right)\right]+f\left(\frac{1010}{2020}\right)\)
\(A=1+...+1+f\left(\frac{1010}{2020}\right)\) (với 1009 số 1)
\(A=1009+f\left(\frac{1}{2}\right)=1009+\frac{\left(\frac{1}{2}\right)^3}{1-3\cdot\frac{1}{2}+3\cdot\left(\frac{1}{2}\right)^2}\)
\(A=1009+\frac{1}{2}=\frac{2019}{2}\)
Vậy \(A=\frac{2019}{2}\)
hello ae xin chào
\(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x......x\left(1-\frac{1}{2017}\right)x\left(1-\frac{1}{2018}\right)\)
=\(\frac{1}{2}x\frac{2}{3}x...x\frac{2017}{2018}\)
=\(\frac{1}{2018}\)
bạn trừ ra là đc
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{2017}\right)\cdot\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2016}{2017}\cdot\frac{2017}{2018}\)
\(=\frac{1\cdot2\cdot3\cdot....\cdot2016\cdot2017}{2\cdot3\cdot4\cdot....\cdot2017\cdot2018}\)
\(=\frac{1}{2018}\)
dấu lớn hơn vì 23/12 lớn hơn 2008/2009