=> 1/x - 1/x+1 + 1/x+1 - 1/x+2 + 1/x+2 - 1/x+3 - 1/x = 1/2010

=> -1/x+3 = 1/2010

=> 1/x+3 = 1/-2010

=> x+3 = -2010

=> x = -2010-3 = -2013

k mk nha

1/x - 1/x+1 + 1/x+1 - 1/x+2 + 1/x+2 - 1/x+3 - 1/x = 1/2010

=> -1/x+3 = 1/2010

=> 1/x+3 = 1/-2010

=> x+3 = -2010

=> x = -2010-3 = -2013

\(\frac{1}{x.\left(x+1\right)}+\frac{1}{\left(x+1\right).\left(x+2\right)}+\frac{1}{\left(x+2\right).\left(x+3\right)}-\frac{1}{x}=\frac{1}{2010}\)

=> \(\frac{-1}{x+3}=\frac{1}{2010}\)

=> \(-x-3=2010\)

=> \(x=2007\)

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{x}\right)=\frac{1}{2010}.\)

\(\Leftrightarrow\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{x-1}{x}=\frac{1}{2010}\)

\(\Leftrightarrow\frac{1}{x}=\frac{1}{2010}\)

\(\Leftrightarrow x=2010\)

Vậy x=2010.

Tìm x, biết:

3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)

2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)

x+110 +2+111 x+112 =x+113 +x+114 

x−1030 +x−1443 +x−595 +x−1488 =0

Trả lời luôn à bạn

Phần a vs phần b tính toán thông thường thôi mà bạn, vs 1 h/s lớp 7 thì ít nhất phải làm được chứ?? :((

a) \(x-\frac{4}{5}=\frac{7}{10}-\frac{3}{4}\)

\(\Leftrightarrow x-\frac{4}{5}=\frac{-1}{20}\)

\(\Leftrightarrow x=\frac{-1}{20}+\frac{4}{5}=\frac{15}{20}=\frac{3}{4}\)

b) \(2\frac{1}{3}-x=\frac{-5}{9}+2x\)

\(\Leftrightarrow2\frac{1}{3}-\frac{-5}{9}=2x+x\)

\(\Leftrightarrow3x=\frac{7}{3}+\frac{5}{9}\)

\(\Leftrightarrow3x=\frac{26}{9}\)

\(\Leftrightarrow x=\frac{26}{9}:3=\frac{26}{27}\)

d) .............................. ( Đề bài)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}\)\(-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}\)

\(\Leftrightarrow-\frac{1}{x+3}=\frac{1}{2010}\)

\(\Leftrightarrow\frac{1}{-\left(x+3\right)}=\frac{1}{2010}\)\(\Leftrightarrow-\left(x+3\right)=2010\)

\(\Leftrightarrow-x-3=2010\) \(\Leftrightarrow-x=2010+3=2013\)

\(\Leftrightarrow x=-2013\)

Bạn tự kết luận nha!

c)

\(\frac{x+3}{2016}+\frac{x+2}{2017}=\frac{x+1}{2018}+\frac{x}{2019}\\ \Leftrightarrow\frac{x+3}{2016}+1+\frac{x+2}{2017}+1=\frac{x+1}{2018}+1+\frac{x}{2019}+1\\ \Leftrightarrow\frac{x+2019}{2016}+\frac{x+2019}{2017}-\frac{x+2019}{2018}-\frac{x+2019}{2019}=0\\ \Leftrightarrow\left(x+2019\right)\left(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}\right)=0\\ \Rightarrow x-2019=0\\ \Rightarrow x=2019\)

Bài 1:

a) x - \(\frac{4}{5}=\frac{7}{10}-\frac{3}{4}\)

=> x - \(\frac{4}{5}=-\frac{1}{20}\)

x = \(\left(-\frac{1}{20}\right)+\frac{4}{5}\)

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

Vậy x = \(\frac{3}{4}\).

b) \(2\frac{1}{3}-x=-\frac{5}{9}+2x\)

=> \(2\frac{1}{3}-\left(-\frac{5}{9}\right)=2x+x\)

=> 3x = \(\frac{7}{3}+\frac{5}{9}\)

=> 3x = \(\frac{26}{9}\)

x = \(\frac{26}{9}:3\)

x = \(\frac{26}{27}\)

Vậy x = \(\frac{26}{27}\).

Chúc bạn học tốt!

1. \(\left(2x-1\right)^3+\left(x+2\right)^3=\left(3x+1\right)^3\)

\(\Rightarrow8x^3-12x^2+6x-1+x^3+6x^2+12x+8=27x^3+27x^2+9x+1\)

\(\Rightarrow-18x^3-33x^2+9x+6=0\)\(\Rightarrow\left(x+2\right)\left(-18x^2+3x+3\right)=0\)

\(\Rightarrow\left(x+2\right)\left(2x-1\right)\left(-9x-3\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{1}{2};x=-\frac{1}{3}\end{cases}}\)

Vậy \(x=-2;x=\frac{1}{2};x=-\frac{1}{3}\)

2. \(\frac{x-1988}{15}+\frac{x-1969}{17}+\frac{x-1946}{19}+\frac{x-1919}{21}=10\)

\(\Rightarrow\left(\frac{x-1988}{15}-1\right)+\left(\frac{x-1969}{17}-2\right)+\left(\frac{x-1946}{19}-3\right)+\left(\frac{x-1919}{21}-4\right)=0\)

\(\Rightarrow\frac{x-2003}{15}+\frac{x-2003}{17}+\frac{x-2003}{19}+\frac{x-2003}{21}=0\)

\(\Rightarrow x-2003=0\)do \(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\ne0\)

Vậy \(x=2003\)

3. Đặt \(\hept{\begin{cases}2009-x=a\\x-2010=b\end{cases}}\)

\(\Rightarrow\frac{a^2+ab+b^2}{a^2-ab+b^2}=\frac{19}{49}\Rightarrow49a^2+49ab+49b^2=19a^2-19ab+19b^2\)

\(\Rightarrow30a^2+68ab+30b^2=0\Rightarrow\left(5a+3b\right)\left(3a+5b\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5a=-3b\\3a=-5b\end{cases}}\)

Với \(5a=-3b\Rightarrow5\left(2009-x\right)=-3\left(x-2010\right)\)

\(\Rightarrow-2x=-4015\Rightarrow x=\frac{4015}{2}\)

Với \(3a=-5b\Rightarrow3\left(2009-x\right)=-5\left(x-2010\right)\)

\(\Rightarrow2x=4023\Rightarrow x=\frac{4023}{2}\)

Vậy \(x=\frac{4023}{2}\)hoặc \(x=\frac{4015}{2}\)

\(\frac{1}{\left(x-1\right)x}+\frac{1}{\left(x-2\right)\left(x-1\right)}+\frac{1}{\left(x-3\right)\left(x-2\right)}+\frac{1}{\left(x-4\right)\left(x-3\right)}=\frac{x}{x^2-4x}\)

\(\Leftrightarrow\)\(\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x-2}-\frac{1}{x-1}+\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-4}-\frac{1}{x-3}=\frac{x}{x\left(x-4\right)}\)

\(\Leftrightarrow\)\(-\frac{1}{x}+\frac{1}{x-4}=\frac{1}{x-4}\)

\(\Leftrightarrow\)\(\frac{-\left(x-4\right)+x}{x\left(x-4\right)}=\frac{x}{x\left(x-4\right)}\)

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

\(\Leftrightarrow x=4\)

lo nói mk làm cách lâu chứ m cx hỏi người khác!!!!!!!!!!!