|x+1.2|+|x+2.3|+...+|x+49.50|=50^x
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VK
12 tháng 6 2018
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{49.50}\right)x=\frac{49}{50}\)
\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)x=\frac{49}{50}\)
\(\left(1-\frac{1}{50}\right)x=\frac{49}{50}\)
\(\frac{49}{50}x=\frac{49}{50}\)
\(x=\frac{\frac{49}{50}}{\frac{49}{50}}\)
\(x=1\)
Vậy \(x=1\)
3 tháng 2 2019
\(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+...+\left|x+99\right|=100x\)
\(\left|x+1\right|\ge0;\left|x+2\right|\ge0;...;\left|x+99\right|\ge0\)
\(\Rightarrow100x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow x+1+x+2+x+3+...+x+99=100x\)
\(\Rightarrow99x+1+2+3+...+99=100x\)
\(\Rightarrow99x+4950=100x\)
\(\Rightarrow-x=-4950\)
\(\Rightarrow x=4950\)
\(\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+\left|x+\frac{1}{3\cdot4}\right|+...+\left|x+\frac{1}{49\cdot50}\right|=50x\)
\(\left|x+\frac{1}{1\cdot2}\right|\ge0;\left|x+\frac{1}{2\cdot3}\right|\ge0;...;\left|x+\frac{1}{49\cdot50}\right|\ge0\)
\(\Rightarrow50x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow x+\frac{1}{1\cdot2}+x+\frac{1}{2\cdot3}+...+x+\frac{1}{49\cdot50}\)
\(\Rightarrow49x+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=50x\)
\(\Rightarrow49x+\frac{49}{50}=50x\)
tu lam
4 tháng 2 2019
\(a;\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+..............+\left|x+99\right|=100x^{\left(1\right)}\)
Ta có \(\left|x+1\right|\ge0;\left|x+2\right|\ge0;\left|x+3\right|\ge0;.............;\left|x+99\right|\ge0\)
\(\Rightarrow VT\ge0\Rightarrow VP\ge0\Rightarrow100x\ge0\Rightarrow x\ge0\)
Với \(x\ge0\).Từ (1) \(\Rightarrow x+1+x+2+x+3+..................+x+99=100x\)
\(\Rightarrow\left(x+x+x+........+x\right)+\left(1+2+3+..........+99\right)=100x\)
\(\Rightarrow99x+4950=100x\)
\(\Rightarrow x=4950\)(t/m đk x > = 0)
\(\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+.........+\left|x+\frac{1}{49.50}\right|=50x^{(∗)}\)
\(\left|x+\frac{1}{1.2}\right|\ge0;\left|x+\frac{1}{2.3}\right|\ge0;............;\left|x+\frac{1}{49.50}\right|\ge0\)
\(\Rightarrow VT\ge0\Rightarrow VP\ge0\Rightarrow50x\ge0\Rightarrow x\ge0\)
Với x > = 0 .Từ (*) \(\Rightarrow x+\frac{1}{1.2}+x+\frac{1}{2.3}+............+x+\frac{1}{49.50}=50x\)
\(\Rightarrow\left(x+x+x+.......+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+...........+\frac{1}{49.50}\right)=50x\)
\(\Rightarrow49x+\left(1-\frac{1}{50}\right)=50x\)
\(\Rightarrow49x+\frac{49}{50}=50x\)
\(\Rightarrow x=\frac{49}{50}\)(t/m đk \(x\ge0\))
NA
20 tháng 3 2022
\(x\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ x\cdot\left(1-\dfrac{1}{50}\right)=1\\ \dfrac{49}{50}x=1\\ x=1:\dfrac{49}{50}\\ x=\dfrac{50}{49}\)
VD
20 tháng 3 2022
\(x.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)=1\\ \Rightarrow x.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ \Rightarrow x.\left(1-\dfrac{1}{50}\right)=1\\ \Rightarrow x.\dfrac{49}{50}=1\\ \Rightarrow x=1:\dfrac{49}{50}\\ \Rightarrow x=\dfrac{50}{49}\)
TH
4
8 tháng 5 2017
1.Tính
\(E=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(E=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(E=\frac{1}{1}-\frac{1}{50}\)
\(E=\frac{49}{50}\)
Câu 2 mình không biết, xin lỗi nha
DL
24 tháng 8 2019
1/1.2+1/3.4+1/5.6+...+1/49.50
=1/1-1/2+1/3-1/4+...+1/49-1/50
=1/1+1/2+1/3+1/4+...+1/49+1/50-2(1/2+1/4+1/6+...+1/50)
=1/1+1/2+1/3+1/4+...+1/49+1/50-(1/1+1/2+1/3+1/4+...+1/25)
=1/26+1/27+...+1/50=1/26+1/27+...+1/50(đpcm)
b. 1/1-1/2+1/3-1/4+...+1/99-1/100=99/100
7/12=175/300; 5/6=10/12=250/300; 99/100=297/300
(hình như khúc này đề bài sai hả bạn) bạn tự tính ra nhé
bài 2: a.x+1/10+x/12+x/14+...x+1/20
(x+x+x...+x)+(1/10+1/12+...+1/20)
ko có kết quả sao tìm x được bạn:[
b.x+1/2000+x+2/1999=x+3/1998+x+4/1997
x+1/2000+x+2/1999=x+3/1998+x+4/1997
(x+1/2000+1)+(x+2/1999+1)=(x+3/1998+1)+(x+4/1997+1)
x+2002/2000+x+2002/1999=x+2002/1998+x+2002/1997
x+2002(1/2000+1/1999)=(x+2002)(1/1998+1/1997)
=>(1/2000+1/1999)=(1/1998+1/1997)
x+2002(1/2000+1/1999)-(x+2002)(1/1998+1/1997)=0
(x+2002)(1/2000+1/1999-1/1998-1/1997)=0
(x+2002).0=0
(x+2002)=0
x =0-2002=-2002
Chúc bạn học tốt.
TL
1
15 tháng 4 2020
đặt A = 1.2. + 2.3 + 3.4 + ... + 49.50
3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 49.50.3
3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 49.50.(51 - 48)
3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 49.50.51 - 48.49.50
3A = 49.50.51
A = 41650
Thay vào ta được
41650 + 1/2x = 40642
=> 1/2x = 1008
=> x = 2016
KG
Cho A=1/1.2 + 1/2.3 + + 1/ 3.4+...+1/49.50 ; B = 1.2+2.3+3.4+4.5+5.6+...+49.50
Tính 50 mủ 2 A – B/17
0
LT
3
5 tháng 5 2019
\(\frac{1}{2}-\left(\frac{2}{3}x-\frac{1}{3}\right)=\frac{2}{3}\)
\(\frac{2}{3}x-\frac{1}{3}=\frac{1}{2}-\frac{2}{3}\)
\(\frac{2}{3}x-\frac{1}{3}=\frac{-1}{6}\)
\(\frac{2}{3}x=\frac{-1}{6}+\frac{1}{3}\)
\(\frac{2}{3}x=\frac{1}{6}\)
\(x=\frac{1}{6}:\frac{2}{3}\)
\(x=\frac{1}{4}\)
~ Hok tốt ~
KN
5 tháng 5 2019
\(\frac{3}{x+5}=15\%\)
\(\Leftrightarrow\frac{3}{x+5}=\frac{15}{100}\)
\(\Leftrightarrow\frac{3}{x+5}=\frac{3}{20}\)
\(\Leftrightarrow x+5=20\)
\(\Leftrightarrow x=20-5\)
\(\Leftrightarrow x=15\)
AL
9 tháng 8 2016
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=\left(\frac{1}{1}-\frac{1}{2}\right)+...+\left(\frac{1}{49}-\frac{1}{50}\right)\)
\(=\frac{1}{1}-\frac{1}{50}\)
\(=\frac{49}{50}\)