Tìm x: \(\left(\frac{10}{1.2}+\frac{10}{2.3}+...+\frac{10}{49.50}\right)+2x=\frac{4}{1.3}+\frac{4}{3.5}+...+\frac{4}{47.49}-7x\)
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30 tháng 1 2022
\(\Leftrightarrow2x+10\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=2\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{47\cdot49}\right)-7x\)
\(\Leftrightarrow2x+10\cdot\dfrac{49}{50}=2\left(1-\dfrac{1}{49}\right)-7x\)
\(\Leftrightarrow9x=-\dfrac{1921}{245}\)
hay x=-1921/2205
DT
1
15 tháng 6 2016
\(x-\frac{\frac{x}{2}-\frac{3+x}{4}}{2}=\frac{2x-\frac{10-7x}{3}}{2}-\left(x+1\right)\)
<=>\(2x-\frac{x}{2}+\frac{3+x}{4}=2x-\frac{10-7x}{3}-2\left(x+1\right)\)
<=>\(24x-6x+9+3x=24x-40+28x-24x-24\)
<=>\(21x+9=28x-64\)
<=>\(-7x=-73\)
<=>x=73/7
23 tháng 3 2018
mk làm câu c cho nó dễ
c)1/1.2+1/2.3+...+1/x.(x+1)=2009/2010
=1-1/2+1/2-1/3+...+1/x-1/x+1=2009/2010
=1-1/x+1=2009/2010
=1/x+1=1-2009/2010
=1/x+1=1/2010
=) x+1=2010
x =2010-1
x =2009
DD
Đoàn Đức Hà
Giáo viên
18 tháng 5 2021
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}=\frac{99}{100}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(B=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{101-99}{99.101}\)
\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(B=1-\frac{1}{101}=\frac{100}{101}\)
DD
Đoàn Đức Hà
Giáo viên
18 tháng 5 2021
\(C=\frac{3^2}{10}+\frac{3^2}{40}+\frac{3^2}{88}+...+\frac{3^2}{340}\)
\(C=3\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\right)\)
\(C=3\left(\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{20-17}{17.20}\right)\)
\(C=3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(C=3\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{27}{20}\)
\(D=\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+...+\frac{7}{99.101}\)
\(D=\frac{7}{2}B=\frac{7}{2}.\frac{100}{101}=\frac{350}{101}\)
8 tháng 1 2020
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
11 tháng 3 2020
1) \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
<=> \(\frac{21x}{24}-\frac{100\left(x-9\right)}{24}=\frac{80x+6}{24}\)
<=> 21x - 100x + 900 = 80x + 6
<=> -79x - 80x = 6 - 900
<=> -159x = -894
<=> x = 258/53
Vậy S = {258/53}
2) \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x+1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
<=> \(\frac{3\left(4x^2+4x+1\right)}{15}-\frac{5\left(x^2+2x+1\right)}{15}=\frac{7x^2-14x-5}{15}\)
<=> 12x2 + 12x + 3 - 5x2 - 10x - 5 = 7x2 - 14x - 5
<=> 7x2 + 2x - 7x2 + 14x = -5 + 2
<=> 16x = 3
<=> x = 3/16
Vậy S = {3/16}
11 tháng 3 2020
3) 4(3x - 2) - 3(x - 4) = 7x+ 10
<=> 12x - 8 - 3x + 12 = 7x + 10
<=> 9x - 7x = 10 - 4
<=> 2x = 6
<=> x = 3
Vậy S = {3}
4) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
<=> \(\frac{x^2+14x+40}{12}+\frac{3\left(x^2+2x-8\right)}{12}=\frac{4\left(x^2+8x-20\right)}{12}\)
<=> x2 + 14x + 40 + 3x2 + 6x - 24 = 4x2 + 32x - 80
<=> 4x2 + 20x - 4x2 - 32x = -80 - 16
<=> -12x = -96
<=> x = 8
Vậy S = {8}
Ta có: \(\left(\frac{10}{1.2}+\frac{10}{2.3}+...+\frac{10}{49.50}\right)+2x=\frac{4}{1.3}+\frac{4}{3.5}+...+\frac{4}{47.49}-7x\) (1)
Xét vế trái ta có:
\(\left(\frac{10}{1.2}+\frac{10}{2.3}+...+\frac{10}{49.50}\right)+2x\)
\(=10.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)\)
\(=10.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)+2x\)
\(=10.\left(1-\frac{1}{50}\right)+2x\)
\(=10.\frac{49}{50}+2x\)
\(=\frac{49}{5}+2x\) (2)
Xét vế phải ta có:
\(\frac{4}{1.3}+\frac{4}{3.5}+...+\frac{4}{47.49}-7x\)
\(=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{47.49}\right)-7x\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\right)-7x\)
\(=2.\left(1-\frac{1}{49}\right)-7x\)
\(=2.\frac{48}{49}-7x\)
\(=\frac{96}{49}-7x\) (3)
Từ (1), (2) và (3) => \(\frac{49}{5}+2x=\frac{96}{49}-7x\)
\(\Rightarrow2x+7x=\frac{96}{49}-\frac{49}{5}\)
\(\Rightarrow9x=\frac{480}{245}-\frac{2401}{245}\)
\(\Rightarrow9x=-\frac{1921}{245}\)
\(\Rightarrow x=-\frac{1921}{245}:9=-\frac{1921}{2205}\)
Vậy \(x=-\frac{1921}{2205}\)
Chúc bạn học tốt!
Ta có:\(\left(10-\frac{10}{2}+\frac{10}{2}-\frac{10}{3}+...+\frac{10}{49}-\frac{10}{50}\right)+2x=\left(2-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+...+\frac{2}{47}-\frac{2}{49}\right)-7x\)
\(\left(10-\frac{10}{50}\right)+2x=\left(2-\frac{2}{49}\right)-7x\)
\(\frac{49}{5}+2x=\frac{96}{49}-7x\)
\(7x+2x=\frac{96}{49}-\frac{49}{5}\)
\(9x=-\frac{1921}{245}\)
\(x=-\frac{1921}{245}:9\)
\(x=-\frac{1921}{2205}\)
Vậy \(x=-\frac{1921}{2205}\)