có ai biết chỉ giúp mình với

mình cũng ko biết làm

a=b và a,b đều lớn hơn c

lớp trưởng mà k lam đc thì thôi chớ

Ta có:

\(A=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+...+\frac{1}{198.101}\)

\(=\frac{2}{\left(2.3\right).2}+\frac{2}{\left(6.5\right).2}+\frac{2}{\left(10.7\right).2}+...+\frac{2}{\left(198.101\right).2}\)

\(=\frac{2}{2.\left(3.2\right)}+\frac{2}{6.\left(5.2\right)}+\frac{2}{10.\left(7.2\right)}+...+\frac{2}{198.\left(101.2\right)}\)

\(=\frac{2}{2.6}+\frac{2}{6.10}+\frac{2}{10.14}+...+\frac{2}{198.202}\)

\(=\frac{4}{2.6}:2+\frac{4}{6.10}:2+\frac{4}{10.14}:2+...+\frac{4}{198.202}:2\)

\(=\left(\frac{4}{2.6}+\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{198.202}\right):2\)

\(=\left(\frac{1}{2}-\frac{1}{202}\right):2\)

\(=\frac{50}{202}=\frac{25}{101}\)

Vậy \(A=\frac{25}{101}\)

frac,left,right là gì vậy ?

\(1,=\left(x^2+3\right)\left(2x-5\right):\left(2x-5\right)=x^2+3\left(A\right)\\ 2,\)

Vì MNPQ là hbh nên MP//QN \(\Rightarrow\widehat{M}+\widehat{N}=180^0\Rightarrow\widehat{N}=\dfrac{180^0-26^0}{2}=77^0\)

Mà MNPQ là hbh nên \(\widehat{Q}=\widehat{N}=77^0\left(B\right)\)

A = 1.2 + 2.3 + 3.4 + ... + 2017.2018

⇒ 3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 2017.218.(2019 - 2016)

= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 2017.2018.2019 - 2016.2017.2018

= 2017.2018.2019

= 2017.2018.2019

B = 2018³/3 ⇒ 3B = 2018³

Ta có:

2017.2019 = (2018 - 1).(2018 + 1)

= 2018² - 1²

= 2018.2018 - 1 < 2018.2018

⇒ 2017.2018.2019 < 2018.2018.2018

⇒ 3A < 3B

⇒ A < B

10+32+54+76+98=10+(32+98)+(54+76)=10+130+130=270

54+90+36+12+78=(54+36)+(12+78)+90=90+90+90=270

74+18+92+30+56=(74+56)+(18+92)+30=130+110+30=270

Vậy 10+32+54+76+98=54+90+36+12+78=74+18+92+30+56

10+32+54+76+98 =270 (1)

54+90+36+12+78=270 (2)

74+18+92+30+56 = 270 (3)

→ (1) = (2) =(3)

Tick nha !!!!!!!!