\(\frac{1+2.3^6}{2^3.3^6-2^3.5^3}-\frac{1+3^6}{8\left(9^3-125\right)}-\frac{5^3}{18^3-10^3}\)

\(=\frac{1+2.3^6}{2^3\left(3^6-5^5\right)}-\frac{1+3^6}{2^3\left[\left(3^2\right)^3-5^3\right]}-\frac{5^3}{\left(2.3^2\right)^3-\left(2.5\right)^3}\)

\(=\frac{1+2.3^6}{2^3\left(3^6-5^3\right)}-\frac{1+3^6}{2^3\left(3^6-5^3\right)}-\frac{5^3}{2^3\left(3^6-5^3\right)}\)

\(=\frac{\left(1+2.3^6\right)-\left(1+3^6\right)-5^3}{2^3\left(3^6-5^2\right)}\)

\(=\frac{3^6-5^3}{2^3\left(3^6-5^3\right)}\)

\(=\frac{1}{8}\)

mình mới học lớp 7 sorry nha

𝑝=−2856279824648840

a: \(A=\left(2x-5\right)^2-4x\left(x-5\right)\)

\(=4x^2-20x+25-4x^2+20x\)

=25

b: \(B=\left(4-3x\right)\left(4+3x\right)+\left(3x+1\right)^2\)

\(=16-9x^2+9x^2+6x+1\)

=6x+17

c: \(C=\left(x+1\right)^3-x\left(x^2+3x+3\right)\)

\(=x^3+3x^2+3x+1-x^3-3x^2-3x\)

=1

d: \(D=\left(2021x-2020\right)^2-2\left(2021x-2020\right)\left(2020x-2021\right)+\left(2020x-2021\right)^2\)

\(=\left(2021x-2020-2020x+2021\right)^2\)

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

\(=x^2+2x+1\)

Bài 2: 

c: \(=x^2\left(x-3\right)-4\left(x-3\right)\)

\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)

ta có (a-1)² ≥ 0 ∀a

<=> a²-2a+1 ≥ 0

<=>a²+4a-2a+1 ≥ 4a (cộng cả 2 vế va 4a)

<=> a²+2a+1 ≥ 4a

<=> (a+1)² ≥ 4a

CM tương tự ta đc

(b+1)² ≥ 4b

(c+1)² ≥ 4c

Nhân các vế với nhau ta có

[(a+1)²+(b+1)² +(c+1)² ]² ≥ 4a.4b.4c

<=> [(a+1)²+(b+1)² +(c+1)² ]² ≥64abc

<=> [(a+1)²+(b+1)² +(c+1)² ]² ≥64 (vì abc =1)

<=> (a+1)²+(b+1)² +(c+1)² ≥8 (đpcm)

link bài giải đây ạ => http://bblink.com/ghyht