\(Q=x^2+5y^2+4xy-2x-8y+2015\)       

\(=\left(x^2+4xy+4y^2\right)-\left(2x+4y\right)+1+y^2-4y+4+2010\)

\(=\left(x+2y\right)^2-2\left(x+2y\right)+1+\left(y-2\right)^2+2010\)

\(=\left(x+2y-1\right)^2+\left(y-2\right)^2+2010\ge2010\forall x;y\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}x+2y-1=0\\y-2=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-3\\y=2\end{cases}}\)

Vậy GTNN của Q là 2010 khi \(x=-3,y=2\)

Gọi M là trung điểm BC

+) vecto AI=vecto IG=vecto GM

+) vecto AI=1/3vecto AM=1/3(vecto CM-vecto CA)=2/3vecto CB-1/3vecto CA

+) vecto AK=1/5vecto AB=1/5vecto CB-1/5vectoCA

+) vecto CK=vecto CA+vecto AK=vecto CA+1/5vecto AB

=vecto CA+1/5vecto CB-1/5vecto CA=1/5vecto CB+4/5vecto CA

+)vecto CI=vecto CA+vecto AI= vecto CA+1/3vecto AM

=vecto CA+1/3vecto AC+1/6vecto CB=2/3vecto CA+1/6vecto CB

b/

+) vecto CI =2/3vecto CA+1/6vecto CB=5(4/30vecto CA+1/30vecto CB)

+) vecto CK=6(4/30vecto CA+1/30vecto CB)

do đó 1/5vecto CI=1/6vecto CK

Nên C,I,K thẳng hàng.

Có:\(x+y=30\Rightarrow\left(x+y\right)^2=900\Rightarrow x^2+y^2+2xy=900\Rightarrow x^2+y^2=900-2.216=468\)(Vì xy=216)

Lại có: \(\left(x-y\right)^2=x^2+y^2-2xy=468-2.216=0\Rightarrow x-y=0\)

\(A=x^2-y^2=\left(x+y\right)\left(x-y\right)=30.0=0\)

Lộn : y = -1.5

2^2 - 4y = 10

4 - 4y = 10

4y = 10 - 4

4y = 6

y = 6 : 4

y = 1,5

Đúng không hả!!

Đặt A=\(n^6+n^4-2n^2=n^2\left(n^4+n^2-2\right)\)

\(=n^2\left(n^4-n^2+2n^2-2\right)=n^2\left[n^2\left(n^2-1\right)+2\left(n^2-1\right)\right]\)

\(=n^2\left(n^2-1\right)\left(n^2+2\right)\)

- Nếu n = 2k (k thuộc Z) thì \(A=\left(2k\right)^2\left[\left(2k\right)^2-1\right]\left[\left(2k\right)^2+2\right]\)

\(=4k^2\left(4k^2-1\right)\left(4k^2+2\right)=8k^2\left(4k^2-1\right)\left(2k^2+1\right)⋮8\) 

- Nếu n = 2k + 1 thì \(A=\left(2k+1\right)^2\left[\left(2k+1\right)^2-1\right]\left[\left(2k+1\right)^2+2\right]\)

\(=\left(4k^2+4k+1\right)\left(4k^2+4k\right)\left(4k^2+4k+3\right)\)

\(=4k\left(k+1\right)\left(4k^2+4k+1\right)\left(4k^2+4k+3\right)\)

=>\(A⋮4.2\left(4k^2+4k+1\right)\left(4k^2+4k+3\right)=8\left(4k^2+4k+1\right)\left(4k^2+4k+3\right)⋮8\) (vì k(k+1) là tích 2 số nguyên liên tiếp)

Từ 2 trường hợp trên thì A chia hết cho 8 với mọi n (1)

- Nếu n chia hết cho 3 thì A chia hết cho 3

- Nếu n không chia hết cho 3

Vì n² là số chính phương => n² chia 3 dư 1 (vì n không chia hết cho 3) =>n² + 2 chia hết cho 3

Ta có: \(A=n^2\left(n^2-1\right)\left(n^2+2\right)=n\left(n-1\right)\left(n+1\right)n\left(n^2+2\right)\)

Mà n(n-1)(n+1) là tích 3 số nguyên liên tiếp =>n(n-1)(n+1) chia hét cho 3

=>\(A⋮3.3.n=9n⋮9\)

Từ 2 trường hợp trên A chia hết cho 9 với mọi n (2)

Mà (8,9) = 1 (3)

Từ (1),(2),(3) => \(A⋮72\left(đpcm\right)\)

\(x+y=1\Rightarrow x=1-y\)        

\(A=x^3+y^3+xy\)

\(=\left(x+y\right)\left(x^2-xy+y^2\right)+xy\)

\(=x^2+y^2\) (vì x + y = 1)

\(=\left(1-y\right)^2+y^2\)

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

\(=2\left(y^2-y+\frac{1}{4}\right)+\frac{1}{2}=2\left(y-\frac{1}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\forall y\)

Dấu "=" xảy ra khi: \(y-\frac{1}{2}=0\Rightarrow y=\frac{1}{2}\Rightarrow x=1-y=\frac{1}{2}\)

Vậy GTNN của A là \(\frac{1}{2}\)khi \(x=y=\frac{1}{2}\)

\(A=x^3+y^3+xy=\left(x+y\right)\left(x^2-xy+y^2\right)+xy\)

\(=x^2-xy+y^2+xy=x^2+y^2\ge\frac{\left(x+y\right)^2}{2}=\frac{1}{2}\)

Nên min A là \(\frac{1}{2}\) khi \(x=y=\frac{1}{2}\)

\(x^3+x^2+4=0\)

\(\Leftrightarrow x^3+2x^2-x^2-2x+2x+4=0\)

\(\Leftrightarrow x^2\left(x+2\right)-x\left(x+2\right)+2\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left[\left(x^2-2.\frac{1}{2}x+\frac{1}{4}\right)+\frac{7}{4}\right]=0\)

\(\Leftrightarrow\left(x+2\right)\left[\left(x-\frac{1}{2}\right)^2+\frac{7}{4}\right]=0\)

\(\text{Vì}\)\(\left[\left(x-\frac{1}{2}\right)^2+\frac{7}{4}\right]>0\)

\(\text{Nên x + 2 = 0 }\Leftrightarrow x=-2\)

\(\text{Vậy x = -2}\)

     \(x^3+x^2+4=0\)

\(\Rightarrow x^3+2x^2-x^2-2x+2x+4=0\)

\(\Rightarrow x^2\left(x+2\right)-x\left(x+2\right)+2\left(x+2\right)=0\)

\(\Rightarrow\left(x+2\right)\left(x^2-x+2\right)=0\)

Mà \(x^2-x+2=\left(x-\frac{1}{2}\right)^2+\frac{7}{4}>0\forall x\)

Do đó: \(x+2=0\Rightarrow x=-2\)

a) \(x^3+2x^2y+xy^2-4xz^2=x\left(x^2+2xy+y^2-4z^2\right)=x\left[\left(x+y\right)^2-\left(2z\right)^2\right]\)

\(=x\left(x+y-2z\right)\left(x+y+2z\right)\)

b)\(-8x^3+12x^2y-6xy^2+y^3=y^3+3.y.\left(2x\right)^2-3.y^2.2x-\left(2x\right)^3\)\(=\left(y-2x\right)^3\)

c)\(6x^2+7x-5=2x\left(3x+5\right)-\left(3x+5\right)=\left(3x+5\right)\left(2x-1\right)\)

d)\(x^4+64y^4=\left(x^2\right)^2+2.x^2.8y^2+\left(8y^2\right)^2-16x^2y^2=\left(x^2+8y^2\right)-\left(4xy\right)^2\)

\(=\left(x^2+8y^2-4xy\right)\left(x^2+8y^2+4xy\right)\)

e)\(x\left(2-x\right)-x+2=x\left(2-x\right)+\left(2-x\right)=\left(2-x\right)\left(x+1\right)\)

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

h)\(3x^2-6xy+3y^2-12z^2=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)

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

g)\(x^3-3x^2-9x+27=x^2\left(x-3\right)-9\left(x-3\right)=\left(x-3\right)\left(x^2-9\right)\)\(=\left(x-3\right)^2\left(x+3\right)\)

B2: \(x^3-5x=0\Rightarrow x\left(x^2-5\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x^2-5=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x^2=5\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{5}\end{cases}}}\)\(\Rightarrow\orbr{\begin{cases}x=0\\x^2=5\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\\orbr{\begin{cases}x=\sqrt{5}\\x=-\sqrt{5}\end{cases}}\end{cases}}\)