Lõm !!!!

Câu 1 tự lm.

Câu 2:

Ta có: \(f\left(x\right)=x^{1994}+x^{1993}+1\)

= \(\left(x^{1994}-x^2\right)+\left(x^{1993}-x\right)+\left(x^2+x+1\right)\)

= \(x^2\left(x^{1992}-1\right)+x\left(x^{1992}-1\right)+\left(x^2+x+1\right)\)

= \(\left[\left(x^3\right)^{664}-\left(1^3\right)^{664}\right]\left(x^2+x\right)+\left(x^2+x+1\right)\)

= \(\left(x^3-1^3\right)\left(x^{1989}+x^{1986}+...+x^3+1\right)+\left(x^2+x+1\right)\)

= \(\left(x-1\right)\left(x^2+x+1\right)\left(x^{1989}+x^{1986}+..+1\right)+\left(x^2+x+1\right)\)

= \(\left(x^2+x+1\right)\left[\left(x-1\right)\left(x^{1989}+..+1\right)+1\right]\)

Vì \(x^2+x+1\) \(⋮\) \(x^2+x+1\)

=> \(f\left(x\right)\) \(⋮\) \(x^2+x+1\) hay số dư trong phép chia là 0

123456789+123456789=246913578

Áp dụng BĐT Svac-sơ vào P

Ta có: P = \(\dfrac{1}{x+\sqrt{yz}}+\dfrac{1}{y+\sqrt{zx}}+\dfrac{1}{z+\sqrt{xy}}\)

\(\ge\) \(\dfrac{9}{x+\sqrt{yz}+y+\sqrt{xz}+z+\sqrt{xy}}\)

\(\ge\) \(\dfrac{9}{x+y+z+\dfrac{x+y}{2}+\dfrac{y+z}{2}+\dfrac{z+x}{2}}\) ( BĐT cosi)

= \(\dfrac{9}{2\left(x+y+z\right)}\)

Hjhj! Đến đoạn này thì chịu!

Ok! Lm lại:

\(\overline{a5}^2=\left(10a+5\right)^2=100a\left(a+1\right)+25\)

Áp dụng vào ta đươc:

15² = ( 10 + 5)² = 100.1.2 + 25 = 225

25² = (2.10+5)² = 100.2.3 + 25 = 625

75² = ( 7.10+5)² = 100.7.8 + 25 = 5625

115² = ( 10.11 + 5)² = 100.12.11 + 25 = 13225