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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 \(x=-\frac{1921}{2205}\)
\(\Leftrightarrow2x+10\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=2\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{47\cdot49}\right)-7x\)
\(\Leftrightarrow2x+10\cdot\dfrac{49}{50}=2\left(1-\dfrac{1}{49}\right)-7x\)
\(\Leftrightarrow9x=-\dfrac{1921}{245}\)
hay x=-1921/2205
\(\frac{3}{1.2}+\frac{3}{2.3}+....+\frac{3}{x.\left(x+1\right)}=\frac{11}{4}\)
\(3.\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{x.\left(x+1\right)}\right)=\frac{11}{4}\)
\(3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{11}{4}\)
\(3.\left(1-\frac{1}{x+1}\right)=\frac{11}{4}\)
\(1-\frac{1}{x+1}=\frac{11}{4}:3=\frac{11}{12}\)
\(\frac{1}{x+1}=1-\frac{11}{12}=\frac{1}{12}\)
\(\Rightarrow x+1=12\)
\(\Rightarrow x=12-1\)
\(\Rightarrow x=11\)
a)\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)
=\(\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+\frac{4}{4!}-\frac{1}{4!}+...+\frac{100}{100!}-\frac{1}{100!}\)
=\(1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
=\(1-\frac{1}{100!}< 1\)
\(\Rightarrow\)\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}< 1\)
b)\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
=\(\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
=\(\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)\)=\(1+1-\frac{1}{99}-\frac{1}{100}\)
=\(2-\frac{1}{99}-\frac{1}{100}< 2\)
\(\Rightarrow\)\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}< 2\)
A=4/3+9/8+16/15+..............+4064256/4064255
A=1+1/3+1+1/8+1/15+...............+1/4064255
A=(1+1+...+1)+(1/3+1/8+...+1/406255) (có 2015 số 1)
A=2015+(1/1.3+1/2.4+...........+1/2015.2017)
A=2015+1/2(1/1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+1/5-1/7+....+1/2012-1/2014+1/2013-1/2015+1/2014-1/2016+1/2015-1/2017)
A=2015+1/2(1+1/2-1/2016-1/2017)
A=2015,749504
k cho mình nhé mình k lại cho
Đặt \(A=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+...+\frac{1}{2013.2015}+\frac{1}{2014.2016}< \frac{3}{4}\)
\(\Leftrightarrow A=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2013.2015}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2014.2016}\right)\)
\(\Leftrightarrow A=\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\right)+\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2014}-\frac{1}{2016}\right)\)
\(\Leftrightarrow A=\left(1-\frac{1}{2015}\right)+\left(\frac{1}{2}-\frac{1}{2016}\right)\)
\(\Leftrightarrow A=\frac{2014}{2015}+\frac{1007}{2016}\)
\(\Leftrightarrow A=1,5\)
Đổi \(\frac{3}{4}=0,75\)
Vì 0,75 < 1,5
Nên ko thể CM
Bài này mà cũng hỏi thì đừng có thi nữa. đợi vài ngày sau có đáp án nhé.
sửa đề câu 1 :
\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)
\(=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{100-1}{100!}\)
\(=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=1-\frac{1}{100!}< 1\)
sửa đề câu 2
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)
