a) \(A=123\left(123+154\right)+77^2\)

\(A=123^2+\left(123.154\right)+77^2=\left(123+77\right)^2=200^2=400\)

b) \(B=3^{24}-\left(2^{47}+1\right)\left(9^6-1\right)\)

\(B=3^{24}-\left(3^{12}-1\right)\left(3^{12}+1\right)\)

\(B=3^{24}-3^{24}+1=1\)

c) \(C=85^2+75^2+65^2+55^2-45^2-35^2-25^2-15^2\)

\(C=\left(85^2-15^2\right)+\left(75^2-25^2\right)+\left(65^2-35^2\right)+\left(55^2-45^2\right)\)

\(C=\left(85+15\right)\left(85-15\right)+\left(75+25\right)\left(75-25\right)+\left(65+35\right)\left(65-35\right)\left(55+45\right)\left(55-45\right)\)

\(C=100\left(60+50+40+30+20+10\right)\)

\(C=100.210=21000\)

a) \(16\text{x}-5\text{x}^2-3\)

\(=-5\text{x}^2+15\text{x}+x-3\)

\(=-5\text{x}\left(x-3\right)+\left(x-3\right)\)

\(=\left(x-3\right)\left(-5\text{x}+1\right)\)

b) \(x^8+3x^4+4\)

\(=\left(x^8+4\text{x}^4+4\right)-x^4\)

\(=\left(x^4+2\right)^2-x^4\)

\(=\left(x^4-x^2+2\right)\left(x^4+x^2+2\right)\)

c) \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

\(=\left(x^2+x+1\right)\text{[}\left(x^2+x+1\right)+1\text{]}-12\)

\(=\left(x^2+x+1\right)^2+\left(x^2+x+1\right)-12\)

\(=\left(x^2+x+1\right)^2-3\left(x^2+x+1\right)+4\left(x^2+x+1\right)-12\)

\(=\left(x^2+x+1\right)\left(x^2+x-2\right)+4\left(x^2+x-2\right)\)

\(=\left(x^2+x-2\right)\left(x^2+x+5\right)\)

Tên : Trần Thị Mỹ Duyên

Lớp : 7 -> 8

a) \(A=x^2-10+30=\left(x-5\right)^2+5\)

Mà \(\left(x+5\right)^2\ge0\)

=> GTNN của A = 5

b) \(B=x^2-12\text{x}+8=\left(x-6\right)^2-4\)

Mà \(\left(x-6\right)^2\ge0\)

=> GTLN = -4

\(5^{100}+5^{98}=5^{98}\left(5^2+1\right)=5^{98}.26=5^{28}.13.2\)

Vậy....

\(x^2+y^2=20\)

\(=>-\left(x^2-y^2\right)=20\)

\(=>x^2-y^2=-20\)

\(=>\left(x+y\right)\left(x-y\right)=-20\)

Mà x+y = 2 ( gt )

=> x-y = -10

*Từ

x-y = -10

x+y = 2

=> x-y+x+y = -10+ 2

=> 2x = -8

=> x = \(-4\)

Mà x+y = 2 => y = 6

Thay x= -4 , y =6 vào \(x^3+y^3\)

Có:

\(x^3+y^3=\left(-4\right)^3+6^3=\left(6-4\right)\left(\left(-4\right)^2-24+6^2\right)\))

\(=2.28=56\)

Vậy x\(^3\)+y\(^3\)=56

Gọi 3 số cần tìm là a,b,c ( a,b,c theo thứ tự tăng dần )

Ta có

b.c- a.b = 50

=> b(c-a) = 50

Mà c-a = 2 ( Do a,b,c là 3 số tự nhiên liên tiếp )

=> b = 25

a = 24

c = 26

Họ và tên : Trần Thị Mỹ Duyên

Lớp : 7 ->8

Ta có :

\(\dfrac{-18}{91}>\dfrac{-18}{90}=\dfrac{-1}{5}\)

\(\dfrac{-23}{144}< \dfrac{-23}{145}=\dfrac{-1}{5}\)

\(=>\dfrac{-18}{91}>\dfrac{-23}{114}\)