\(\left(x^4-3x^2+9\right)\left(x^2+3\right)-\left(3-x^2\right)^3=x^6+27-\left(27-27x^2+9x^4-x^6\right).\)

                                                                                     =\(x^6+27-27+27x^2-9x^4+x^6\)

                                                                                     =2x⁶+27x²-9x⁴

\(C=3x^2-5x-8=3\left(x^2-\frac{5}{3}x+\frac{25}{36}\right)-\frac{121}{12}=3\left(x-\frac{5}{6}\right)^2-\frac{121}{12}\ge-\frac{121}{12}\)

Dấu "=" xảy ra khi x=5/6

Vậy Cmin=-121/12 khi x=5/6

\(C=3x^2-5x-8=3\left(x^2-2x.\frac{5}{6}+\frac{25}{36}\right)-\frac{25}{12}-8\)

\(=3\left(x-\frac{5}{6}\right)^2-\frac{121}{12}\)\(\Rightarrow C\ge-\frac{121}{12}\)(Do \(3\left(x-\frac{5}{6}\right)^2\ge0\))

Vậy \(Min_C=-\frac{121}{12}\). Dấu "=" xảy ra <=> \(x=\frac{5}{6}.\)

\(1,24^2-2,48\times0,24+0,24^2\)

\(=1,24-2\times1,24\times0,24+0,24^2\)

\(=\left(1,24-0,24\right)^2\)

\(=1\)

\(89^2-18.89+9^2\)

\(=89^2-2.9.89+9^2\)

\(=\left(89-9\right)^2\)

\(=80^2=6400\)

Mũ 3 ngoài ngoặc chứ đúng k

chỉnh đề:

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

\(=\left(x^2-1\right)^3-\left(x^2-1\right)^3=0\)

p/s: bài này áp dụng HĐT:  (a - b)(a² + ab + b²) = (a - b)³

\(5\left(x+2\right)\left(x-2\right)-\frac{1}{2}\left(6-8x^2\right)+17\)

\(=5\left(x^2-4\right)-\left(3-4x^2\right)+17\)

\(=5x^2-20-3+4x^2+17\)

\(=9x^2-6\)

C1:  \(B=x^3+3xy+y^3\)

\(=\left(x+y\right)^3-3xy\left(x+y\right)+3xy\)

\(=\left(x+y\right)^3-3xy\left(x+y-1\right)\)

Thay \(x+y=1\)ta được:

\(B=1^3-3xy\left(1-1\right)=1\)

C2: \(x+y=1\)\(\Rightarrow\)\(x=1-y\)

\(B=x^3+3xy+y^3=\left(1-y\right)^3+3\left(1-y\right)y+y^3\)

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

mk chỉnh lại đề: \(D=7x+2y\)   (thử \(x=1;y=3\)thì D k chia hết cho 13)

\(4x+3y\)\(⋮\)\(13\)

\(\Rightarrow\)\(5\left(4x+3y\right)\)\(⋮\)\(13\)

\(13\left(x+y\right)\)\(⋮\)\(13\)

suy  ra:  \(5\left(4x+3y\right)-13\left(x+y\right)\)\(⋮\)\(13\)

\(\Leftrightarrow\)\(20x+15y-13x-13y\)\(⋮\)\(13\)

\(\Leftrightarrow\)\(7x+2y\)\(⋮\)\(13\)

Vậy D chia hết cho 13

n chia 9 dư 5 nên n có dạng:  \(n=9k+5\)   \(\left(n\in N\right)\)

Ta có:  \(n^2=\left(9k+5\right)^2=81k^2+90k+25=9\left(9k^2+10k+2\right)+7\)

Ta thấy:  \(9\left(9k^2+10k+2\right)\)\(⋮\)\(9\);  7 không chia hết cho 9

Vậy  \(n^2\)chia 9 dư 7