99x99+99
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Ta có:99x99:99
=99x(99:99)
=99x1
=99
ai tích mình mình tích lại cho
99x99=99x9+99+99x9+99x.....
= 99x99=99 x 10+99x.......
= 99x99=99x100
= 99x99= 99x(99-10)
= > ... = 89
99x99=99x9+99+99x...
suy ra 99x99=99x(9+1+...)
suy ra 99x99=99x(10+...)
suy ra 99=10+...
suy ra ...=99-10
suy ra ....=89
thong minh bo duoi dau co dung khong ban khe thay doi khong ha ban ge chua minh do ban 1+1=maythui deo biet
Ta có :
\(A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+.................+\dfrac{1}{99.99}+\dfrac{1}{100.100}\)
Ta thấy :
\(\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
.............................
\(\dfrac{1}{99.99}< \dfrac{1}{98.99}\)
\(\dfrac{1}{100.100}< \dfrac{1}{99.100}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..................+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...........+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow A< 1-\dfrac{1}{100}=\dfrac{99}{100}\)
\(\Rightarrow A< \dfrac{99}{100}\)
\(A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+.....+\dfrac{1}{99.99}+\dfrac{1}{100.100}\)
\(A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)
\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A< 1-\dfrac{1}{100}\)
\(A< \dfrac{99}{100}\)
\(A< B\)
99.(100-1)+99
=9900-99+99
=9900
99.99+99
=99.99+99.1
=99.(99+1)
=99.100
=9900