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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 \(x=-\frac{1921}{2205}\)
\(x-\frac{\frac{x}{2}-\frac{3+x}{4}}{2}=\frac{2x-\frac{10-7x}{3}}{2}-\left(x+1\right)\)
<=>\(2x-\frac{x}{2}+\frac{3+x}{4}=2x-\frac{10-7x}{3}-2\left(x+1\right)\)
<=>\(24x-6x+9+3x=24x-40+28x-24x-24\)
<=>\(21x+9=28x-64\)
<=>\(-7x=-73\)
<=>x=73/7
A=\(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.2^9.3^9}{\left(2^2\right)^6.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3+1\right)}\)
\(=\frac{2.6}{3.7}\)\(=\frac{4}{7}\)
\(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{\left(2n-1\right)\left(2n+1\right)}\)
\(=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n-1}-\frac{1}{2n+1}\right)\)
\(=2.\left(1-\frac{1}{2n+1}\right)\)
\(=2.\left(\frac{2n}{2n+1}\right)\)
\(=\frac{4n}{2n+1}\)
Tham khảo nhé~
khỏi chép lại đề ha
- 2 - 4x - 5x + \(\frac{3}{2}\)= \(\frac{7}{4}\)
\(\frac{7}{2}\)- 9x = \(\frac{7}{4}\)
-9x = \(\frac{7}{2}-\frac{7}{4}\)
-9x = \(\frac{7}{4}\)
x = \(\frac{7}{4}:\left(-9\right)\)
x = \(\frac{-7}{36}\)
- 3 - 2x - \(\frac{1}{3}=7x-\frac{1}{4}\)
-2x - 7x = \(\frac{-1}{4}-3+\frac{1}{3}\)
-9x = \(\frac{-35}{12}\)
x = \(\frac{-35}{12}:\left(-9\right)\)
x = \(\frac{35}{108}\)
- \(\frac{-15}{2}\)+ \(\frac{1}{4}\)+ 4x -2 = 1
4x = 1 + \(\frac{15}{2}-\frac{1}{4}+2\)
4x = \(\frac{41}{4}\)
x = \(\frac{41}{4}:4\)
x = \(\frac{41}{16}\)
\(\Leftrightarrow2x+10\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=2\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{47\cdot49}\right)-7x\)
\(\Leftrightarrow2x+10\cdot\dfrac{49}{50}=2\left(1-\dfrac{1}{49}\right)-7x\)
\(\Leftrightarrow9x=-\dfrac{1921}{245}\)
hay x=-1921/2205