Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-8\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)\(=\left(x^2+7x+11-1\right)\left(x^2+7x+11+1\right)-8\)
\(=\left(x^2+7x+11\right)^2-9\)
\(=\left(x^2+7x+11-3\right)\left(x^2+7x+11+3\right)=\left(x^2+7x+8\right)\left(x^2+7x+14\right)\)
PTĐTTNT
a, x3+4x2-24x+24
b,x8+x4+1
PTĐTTNT: x8 + x + 1
PTĐTTNT:
a) x(x+1)2 + x(x-5) - 5(x+1)
b) 25x2 - 4y2 - 9 + 12y
1.PTĐTTNT
a, x^2-2xy-25-y^2
b, x( x-1)+y (1-x)
c, 7x+7y-(x-y)
d, x^4+y^4
2, Chứng minh rằng:
a, x^2-5x+3≥0
b, -x^2+3x-4<0 với mọi x
1. PTĐTTNT:
a, 4x2 - 4x - 3
b, x3 - x2 - 4
c, 64x4 + y4
Cảm ơn mọi người nhiều nhé!
PTĐTTNT:
x2 - x + 1
PTĐTTNT:
xn + xn+3 -xn
I : PTĐTTNT
\(\left(x^2-3\right)^2+16\)
help me
PTĐTTNT
x11 + x + 1