2.a) (ko phân tích được, bạn coi lại nhé)

b) phần còn lại của chứng minh là gì thế bạn?

a.

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

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

b.

\(x^3-9x^2+6x+16=\left(x^3-7x^2-8x\right)-\left(2x^2-14x-16\right)\)

\(=x\left(x^2-7x-8\right)-2\left(x^2-7x-8\right)\)

\(=\left(x-2\right)\left(x^2-7x-8\right)=\left(x-2\right)\left(x^2+x-8x-8\right)\)

\(=\left(x-2\right)\left[x\left(x+1\right)-8\left(x+1\right)\right]=\left(x-2\right)\left(x+1\right)\left(x-8\right)\)

c.

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+10+2\right)-24\)

\(=\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)-24\)

\(=\left(x^2+7x+10\right)^2-4\left(x^2+7x+10\right)+6\left(x^2+7x+10\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+10-4\right)+6\left(x^2+7x+10-4\right)\)

\(=\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

Bài 2:

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

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

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

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

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

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

Dấu '=' xảy ra khi \(\begin{cases}x^2-x=0\\ x-1=0\end{cases}\Rightarrow x=1\)

b: \(B=\left(x-5\right)\left(x-3\right)\left(x+2\right)\left(x+4\right)+2022\)

\(=\left(x-5\right)\left(x+4\right)\left(x-3\right)\left(x+2\right)+2022\)

\(=\left(x^2-x-20\right)\left(x^2-x-6\right)+2022\)

\(=\left(x^2-x-6\right)^2-14\left(x^2-x-6\right)+49+1973=\left(x^2-x-6+7\right)^2+1973\)

\(=\left(x^2-x+1\right)^2+1973\)

Ta có: \(x^2-x+1=\left(x-\frac12\right)^2+\frac34\ge\frac34\forall x\)

=>\(\left(x^2-x+1\right)^2\ge\frac{9}{16}\forall x\)

=>\(\left(x^2-x+1\right)^2+1973\ge\frac{9}{16}+1973\forall x\)

=>B>=31577/16∀x

Dấu '=' xảy ra khi \(x-\frac12=0\)

=>\(x=\frac12\)

Bài 2:

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

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

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

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

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

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

Dấu '=' xảy ra khi \(\begin{cases}x^2-x=0\\ x-1=0\end{cases}\Rightarrow x=1\)

b: \(B=\left(x-5\right)\left(x-3\right)\left(x+2\right)\left(x+4\right)+2022\)

\(=\left(x-5\right)\left(x+4\right)\left(x-3\right)\left(x+2\right)+2022\)

\(=\left(x^2-x-20\right)\left(x^2-x-6\right)+2022\)

\(=\left(x^2-x-6\right)^2-14\left(x^2-x-6\right)+49+1973=\left(x^2-x-6+7\right)^2+1973\)

\(=\left(x^2-x+1\right)^2+1973\)

Ta có: \(x^2-x+1=\left(x-\frac12\right)^2+\frac34\ge\frac34\forall x\)

=>\(\left(x^2-x+1\right)^2\ge\frac{9}{16}\forall x\)

=>\(\left(x^2-x+1\right)^2+1973\ge\frac{9}{16}+1973\forall x\)

=>B>=31577/16∀x

Dấu '=' xảy ra khi \(x-\frac12=0\)

=>\(x=\frac12\)

Bài 2:

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

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

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

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

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

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

Dấu '=' xảy ra khi \(\begin{cases}x^2-x=0\\ x-1=0\end{cases}\Rightarrow x=1\)

b: \(B=\left(x-5\right)\left(x-3\right)\left(x+2\right)\left(x+4\right)+2022\)

\(=\left(x-5\right)\left(x+4\right)\left(x-3\right)\left(x+2\right)+2022\)

\(=\left(x^2-x-20\right)\left(x^2-x-6\right)+2022\)

\(=\left(x^2-x-6\right)^2-14\left(x^2-x-6\right)+49+1973=\left(x^2-x-6+7\right)^2+1973\)

\(=\left(x^2-x+1\right)^2+1973\)

Ta có: \(x^2-x+1=\left(x-\frac12\right)^2+\frac34\ge\frac34\forall x\)

=>\(\left(x^2-x+1\right)^2\ge\frac{9}{16}\forall x\)

=>\(\left(x^2-x+1\right)^2+1973\ge\frac{9}{16}+1973\forall x\)

=>B>=31577/16∀x

Dấu '=' xảy ra khi \(x-\frac12=0\)

=>\(x=\frac12\)

a/ = x⁴ + 36x² +81 +12x³ + 108x + 18x² + x⁴ +1 + 4x² + 2x² - 4x  - 4x³ -16 = 2x⁴ + 8x³ + 60x² + 104x + 66 = 2(x⁴ + 4x³ + 30x² + 52x + 33)

Đặt \(A=\left(x+3\right)^4+\left(x+1\right)^4-16\)

\(=\left\lbrack\left(x+2\right)+1\right\rbrack^4+\left\lbrack\left(x+2\right)-1\right\rbrack^4-16\)

Đặt b=x+2

=>\(A=\left(b+1\right)^4+\left(b-1\right)^4-16\)

\(=\left(b^2+2b+1\right)^2+\left(b^2-2b+1\right)^2-16\)

\(=\left(b^2+1\right)^2+4b\left(b^2+1\right)+4b^2+\left(b^2+1\right)^2-4b\left(b^2+1\right)+4b^2-16\)

\(=2\left(b^2+1\right)^2+8b^2-16\)

\(=2\left\lbrack\left(b^2+1\right)^2+4b^2-8\right\rbrack\)

\(=2\left\lbrack b^4+2b^2+1+4b^2-8\right\rbrack=2\left(b^4+6b^2-7\right)\)

\(=2\left(b^2+7\right)\left(b^2-1\right)=2\left(b^2+7\right)\left(b-1\right)\left(b+1\right)\)

\(=2\left\lbrack\left(x+2\right)^2+7\right\rbrack\left(x+2-1\right)\left(x+2+1\right)=2\left(x+1\right)\left(x+3\right)\left\lbrack\left(x+2\right)^2+7\right\rbrack\)

Đặt \(A=\left(x+3\right)^4+\left(x+1\right)^4-16\)

\(=\left\lbrack\left(x+2\right)+1\right\rbrack^4+\left\lbrack\left(x+2\right)-1\right\rbrack^4-16\)

Đặt b=x+2

=>\(A=\left(b+1\right)^4+\left(b-1\right)^4-16\)

\(=\left(b^2+2b+1\right)^2+\left(b^2-2b+1\right)^2-16\)

\(=\left(b^2+1\right)^2+4b\left(b^2+1\right)+4b^2+\left(b^2+1\right)^2-4b\left(b^2+1\right)+4b^2-16\)

\(=2\left(b^2+1\right)^2+8b^2-16\)

\(=2\left\lbrack\left(b^2+1\right)^2+4b^2-8\right\rbrack\)

\(=2\left\lbrack b^4+2b^2+1+4b^2-8\right\rbrack=2\left(b^4+6b^2-7\right)\)

\(=2\left(b^2+7\right)\left(b^2-1\right)=2\left(b^2+7\right)\left(b-1\right)\left(b+1\right)\)

\(=2\left\lbrack\left(x+2\right)^2+7\right\rbrack\left(x+2-1\right)\left(x+2+1\right)=2\left(x+1\right)\left(x+3\right)\left\lbrack\left(x+2\right)^2+7\right\rbrack\)

mình biết nội quy rồi nên đưng đăng nội quy

ai chơi bang bang 2 kết bạn với mình

mình có nick có 54k vàng đang góp mua pika 

ai kết bạn mình cho

a) \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)-4=\left(x^2+6x+5\right)\left(x^2+6x+8\right)-4\)

Đặt \(t=x^2+6x+5\)

\(PT=t\left(t+3\right)-4=t^2+3t-4=\left(t-1\right)\left(t+4\right)\)

Thay t: \(PT=\left(x^2+6x+5-1\right)\left(x^2+6x+5+4\right)=\left(x^2+6x+4\right)\left(x^2+6x+9\right)=\left(x^2+6x+4\right)\left(x+3\right)^2\)

b)  Đặt \(t=\left(2x+1\right)^2\)

\(PT=t^2-3t+2=\left(t^2-3t+\dfrac{9}{4}\right)-\dfrac{1}{4}=\left(t+\dfrac{3}{2}\right)^2-\dfrac{1}{4}=\left(t+1\right)\left(t+2\right)\)

Thay t:

\(PT=\left[\left(2x+1\right)^2+1\right]\left[\left(2x+1\right)^2+2\right]=\left[4x^2+4x+2\right]\left[4x^2+4x+3\right]=2\left[2x^2+2x+1\right]\left[4x^2+4x+3\right]\)