Ta có : \(\frac{x-1}{2}=\frac{x+1}{3}\)
<=> \(3\left(x-1\right)=2\left(x+1\right)\)
<=> \(3x-3=2x+2\)
<=> \(3x-2x=2+3\)
<=> x = 5
a, \(\frac{x-1}{2}=\frac{x+1}{3}\)
=> (x-1)3 = 2(x+1)
=> 3x - 3 = 2x + 2
=> 3x - 2x = 2 + 3
=> x = 5
b, \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}=\frac{99}{100}< 1\) (ĐPCM)
a, 3.(x - 1) =2. ( x+ 1)
<=>3.x-3.1=2.x+2.1
<=>3x-2x=2+3
<=>x=5
b, A= 1/1.2+1/2.3+1/3.4+...+1/99.100
=1-1/2+1/2-1/3+1/3-...+1/99-1/100
=1-1/100
=99/100
+ Vì 99/100<1 nên =>1/1.2=1/2.3+1/3.4+...+1/99.100
Vậy......
a. x-1/2 = x+1/3
=>(x-1)3=2(x+1)
=> 3x-3=2x+2
=> 3x-2x=2+3
=> x = 5
b. A = 1/1.2+1/2.3+..+1/99.100
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ..+ 1/99 - 1/100
= 1 - 1/100 = 99/ 100 < 1