\(A=x^3+1+x-x^3+1+1994=x+2+1994=x+1996\)

Khi x=-1995 thì A=1

a: \(=\dfrac{1}{4}\cdot\dfrac{2^2}{5^2}=\dfrac{1}{25}\)

b: \(=\left(\dfrac{5}{2}\cdot\dfrac{4}{5}\right)^3=2^3=8\)

\(\left(a\right):\left(\dfrac{-1}{2}\right)^2.\left(\dfrac{2}{5}\right)^2=\left(\dfrac{-1}{2}.\dfrac{2}{5}\right)^2=\left(-\dfrac{1}{5}\right)^2=\dfrac{1}{25}\\ \left(b\right):\left(\dfrac{5}{2}\right)^3.\left(\dfrac{4}{5}\right)^3=\left(\dfrac{5}{2}.\dfrac{4}{5}\right)^3=2^3=8\)

a: \(=-23+23+15+18=33\)

b: \(=119-9-511+301=110-210=-100\)

d: \(=815+\left[95-815-45\right]=50\)

2/5 x 7/4 - 2/5 x 3/7

= 2/5 x (7/4 - 3/7)

= 2/5 x (49/28 - 12/28)

= 2/5 x 37/28

= 37/70

\(=\dfrac{2}{5}\left(\dfrac{7}{4}-\dfrac{3}{7}\right)\\ =\dfrac{2}{5}\left(\dfrac{49-12}{28}\right)\\ =\dfrac{2}{5}.\dfrac{37}{28}\\ =\dfrac{37}{70}\)

a: Xét ΔAHB vuông tại H cso HE là đường cao

nên \(AE\cdot AB=AH^2\left(1\right)\)

Xét ΔAHC vuông tại H có HF là đươngc ao

nên \(AF\cdot AC=AH^2\left(2\right)\)

Từ (1) và (2) suy ra \(AE\cdot AB=AF\cdot+AC\)

b: \(AE\cdot EB+AF\cdot FC=HE^2+HF^2=AH^2\)

thiếu ý C ạ

Lx bn ey

nó k có ảnh ý:))

\(a,x-35=120\\ x=120+35\\ x=155\\ b,118-x=310-217\\ 118-x=93\\ x=25\\ c,x+61=156-82\\ x+61=74\\ x=74-61\\ x=13\\ d,x-305=814-712\\ x-305=102\\ x=407\\ e,2x-138=72\\ 2x=72+138\\ 2x=210\\ x=105\\ g,x+35=515:5\\ x+35=103\\ x=68\)

Với a;b > 0 ; AD BĐT Cô-si ; ta được : \(a^4+1+1+1\ge4a\)  \(\Rightarrow a^4+3\ge4a\)  

\(\Rightarrow\dfrac{1}{a^4+4}\le\dfrac{1}{4a+1}\) . CMTT : \(\dfrac{1}{b^4+4}\le\dfrac{1}{4b+1}\)  

Suy ra : \(A=\dfrac{1}{a^4+4}+\dfrac{1}{b^4+4}\le\dfrac{1}{4a+1}+\dfrac{1}{4b+1}\)

Mặt khác : \(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{a}+1\ge\dfrac{\left(1+1+1+1+1\right)^2}{4a+1}\)  ( B.C.S)

\(\Rightarrow\dfrac{4}{a}+1\ge\dfrac{25}{4a+1}\Rightarrow\dfrac{1}{25}\left(\dfrac{4}{a}+1\right)\ge\dfrac{1}{4a+1}\)

CMTT : \(\dfrac{1}{4b+1}\le\dfrac{1}{25}\left(\dfrac{4}{b}+1\right)\)

Suy ra : \(A\le\dfrac{4}{25}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\dfrac{2}{25}=\dfrac{4}{25}.2+\dfrac{2}{25}=\dfrac{2}{5}\)

" = " \(\Leftrightarrow a=b=1\)

= 78 năm

0,78 thế kỉ= 78 năm