a) \(\left(6x^3+3x^2+4x+2\right):\left(3x^2+2\right)\)

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

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

\(=2x+1\)

b) \(\left(2x^3-22x^2-5x^2+60x+55x-150\right):\left(x-5\right)\)

\(=\left[\left(2x^3-22x^2+60x\right)-\left(5x^2-55x+150\right)\right]:\left(x-5\right)\)

\(=\left[2x\left(x^2-11x+30\right)-5\left(x^2-11x+30\right)\right]:\left(x-5\right)\)

\(=\left[\left(2x-5\right)\left(x^2-11x+30\right)\right]:\left(x-5\right)\)

\(=\left[\left(2x-5\right)\left(x^2-5x-6x+30\right)\right]:\left(x-5\right)\)

\(=\left[\left(2x-5\right)\left(x-5\right)\left(x-6\right)\right]:\left(x-5\right)\)

\(=\left(2x-5\right)\left(x-6\right)\)

\(=2x^2-17x+30\)

d) \(\left(x^5+4x^3+3x^2-5x+15\right):\left(x^3-x+3\right)\)

\(=\left(x^5+5x^3+3x^2-x^3-5x+15\right):\left(x^3-x+3\right)\)

\(=\left[\left(x^5-x^3+3x^2\right)+\left(5x^3-5x^2+15\right)\right]:\left(x^3-x+3\right)\)

\(=\left[x^2\left(x^3-x+3\right)+5\left(x^3-x^2+3\right)\right]:\left(x^3-x+3\right)\)

\(=\left[\left(x^2+5\right)\left(x^3-x+3\right)\right]:\left(x^3-x+3\right)\)

\(=x^2+5\)

1. (-2x - 1)(x² - x - 3) - (x + 2)(x + 1)²

= -2x³ + 2x² + 6x - x² + x + 3 - (x + 2)(x² + 2x + 1)

= -2x³ + x² + 7x + 3 - x³ - 2x² - x - 2x² - 2x - 2

= -3x³ - 3x² + 4x + 1

2. (x + 2)(x - 1) - (x - 3)(x + 2) = 3

=> (x + 2)(x - 1 - x + 3) = 3

=> (x + 2).0 = 3

...(xem lại đề)

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

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

\(\Leftrightarrow2\left(x+2\right)=3\)

\(\Leftrightarrow x+2=\frac{3}{2}\)

\(\Leftrightarrow x=\frac{3}{2}-2\)

\(\Leftrightarrow x=-\frac{1}{2}\)

\(a^2+b^2+c^2+d^2+e^2\ge a\left(b+c+d+e\right)\)

\(\Leftrightarrow4a^2+4b^2+4c^2+4d^2+4e^2\ge4ab+4ac+4ad+4ae\)

\(\Leftrightarrow4a^2+4b^2+4c^2+4d^2+4e^2-4ab-4ac-4ad-4ae\ge0\)

\(\Leftrightarrow\left(a^2-4ab+4b^2\right)+\left(a^2-4ac+4c^2\right)+\left(a^2-4ad+4d^2\right)+\left(a^2-4ae+4e^2\right)\ge0\)

\(\Leftrightarrow\left(a-2b\right)^2+\left(a-2c\right)^2+\left(a-2d\right)^2+\left(a-2e\right)^2\ge0\)( luôn đúng )

Vậy ...

Có nhiều cách biểu diễn:

VD

\(VT-VP=\frac{\left(a-b-c\right)^2+\left(a-d-e\right)^2+\left(b-c\right)^2+\left(d-e\right)^2}{2}\) (còn rất nhiều ...)

Áp dụng BĐT Cauchy cho 2 số không âm:

\(4a^2+9b^2\ge2\sqrt{4a^2.9b^2}=2.6ab=12ab\)

\(9b^2+25c^2\ge2\sqrt{9b^2.25c^2}=2.15bc=30bc\)

\(4a^2+25c^2\ge2\sqrt{4a^2.25c^2}=2.10ac=20ac\)

Cộng từng vế của các BĐT trên:

\(2\left(4a^2+9b^2+25c^2\right)\ge2\left(6ab+10ac+15bc\right)\)

\(\Rightarrow4a^2+9b^2+25c^2\ge6ab+10ac+15bc\)

(Dấu "="\(\Leftrightarrow a=b=c=0\))

\(\text{BĐT}\Leftrightarrow\frac{\left(4a-3b-5c\right)^2+3\left(3b-5c\right)^2}{4}\ge0\) (đúng)

Đẳng thức xảy ra khi \(\hept{\begin{cases}4a=3b+5c\\3b=5c\end{cases}}\Leftrightarrow\hept{\begin{cases}4a=6b\\4a=10c\end{cases}}\Leftrightarrow a=\frac{3}{2}b=\frac{5}{2}c\)

Không chắc chỗ dấu bằng cho lắm:)

Áp dụng BĐT Cauchy cho 2 số không âm:

\(a^2+b^2\ge2\sqrt{a^2b^2}=2ab\)

\(b^2+1\ge2\sqrt{b^2}=2b\)

\(a^2+1\ge2\sqrt{a^2}=2a\)

Cộng từng vế của các BĐT trên:

\(2\left(a^2+b^2+1\right)\ge2\left(ab+a+b\right)\)

\(\Rightarrow a^2+b^2+1\ge ab+a+b\)

(Dấu "="\(\Leftrightarrow a=b=1\))

\(VT-VP=\frac{\left(2a-b-1\right)^2+3\left(b-1\right)^2}{4}\ge0\)