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a: 2x(x+1)-135=-200
=>2(x^2+x)=-65
=>2x^2+2x+65=0
=>x^2+x+32,5=0
=>x^2+x+0,25+32,25=0
=>(x+0,5)^2+32,25=0(vô lý)
b: 4x-5(x-1)+15=13
=>4x-5x+5=-2
=>5-x=-2
=>x=5+2=7
c: 2/3x-1/4=3/5-7/8
=>2/3x=3/5-7/8+1/4=24/40-35/40+10/40=-1/40
=>x=-1/40:2/3=-1/40*3/2=-3/80
d: 1/2(2x-3)+105/2=-137/2
=>1/2(2x-3)=-137/2-105/2=-242/2=-121
=>2x-3=-242
=>2x=-239
=>x=-239/2
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-199}{200}=-\dfrac{1}{200}\)
\(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{200^2}-1\right)\)
A là tích của 199 số âm(đặt biểu thức trên là A)
\(-A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{200^2}\right)\)
\(=\frac{3}{2^2}\cdot\frac{8}{3^2}\cdot\frac{15}{4^2}\cdot...\cdot\frac{39999}{200^2}\)
\(=\frac{1\cdot3}{2^2}\cdot\frac{2\cdot4}{3^2}\cdot\frac{3\cdot5}{4^2}\cdot...\cdot\frac{199\cdot201}{200^2}\)
Để dễ rút gọn,ta viết tử dưới dạng tích các số tự nhiên liên tiếp .
\(-A=\frac{1\cdot2\cdot3\cdot...\cdot198\cdot199}{2\cdot3\cdot4\cdot...\cdot199\cdot200}\cdot\frac{3\cdot4\cdot5\cdot...\cdot201}{2\cdot3\cdot4\cdot...\cdot199\cdot200}=\frac{1}{200}\cdot\frac{201}{2}=\frac{201}{400}>\frac{1}{2}\)
=> \(A< -\frac{1}{2}\)
a, Ta có : \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
=> \(\frac{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}}{\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}}=1\)
=> đpcm
Study well ! >_<
\(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{200}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{200}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{199}{200}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{201}{200}\)
=1/200*201/2=201/400>200/400=1/2