Áp dụng AM-GM có:

\(2a^2+2b^2\ge4ab\)

\(8b^2+\dfrac{1}{2}c^2\ge4bc\)

\(8a^2+\dfrac{1}{2}c^2\ge4ac\)

Cộng vế với vế \(\Rightarrow VT\ge4\left(ab+bc+ac\right)=4\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}ab+bc+ac=1\\a=b=\dfrac{c}{4}\end{matrix}\right.\)\(\Rightarrow a=b=\dfrac{1}{3};c=\dfrac{4}{3}\)

Đk: \(x\ge\dfrac{1}{2}\)

Bpt\(\Leftrightarrow\left(x^2+2x\sqrt{2x-1}+2x-1\right)-\left[4\left(2x-1\right)+4\sqrt{2x-1}+1\right]\ge0\)

\(\Leftrightarrow\left(x+\sqrt{2x-1}\right)^2-\left(2\sqrt{2x-1}+1\right)^2\ge0\)

\(\Leftrightarrow\left(x-\sqrt{2x-1}-1\right)\left(x+3\sqrt{2x-1}+1\right)\ge0\) (1)

Vì \(x\ge\dfrac{1}{2}\Rightarrow x+3\sqrt{2x-1}+1>0\)

Từ (1) \(\Rightarrow x-\sqrt{2x-1}-1\ge0\)

\(\Leftrightarrow\sqrt{2x-1}\le x-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-1\ge0\\x-1\ge0\\2x-1\le\left(1-x\right)^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x\in R\backslash\left(2-\sqrt{2};2+\sqrt{2}\right)\end{matrix}\right.\)\(\Rightarrow x\ge2+\sqrt{2}\)

Vậy...

Với \(n\in N;n>0\) có:

\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}\left(n+1-n\right)}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

Áp dụng vào P có:
\(P=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{2016}}-\dfrac{1}{\sqrt{2017}}\)

\(=1-\dfrac{1}{\sqrt{2017}}\)

\(\Rightarrow a^2+b=1^2+2017=2018\)

Ý A

\(V_{nc}=10\left(l\right)\Rightarrow m=10\left(kg\right)\)

Gọi số nước đang sôi cần đổ là \(V_1=a\left(l\right)\Rightarrow m_1=a\left(kg\right)\)

Gọi số nước ở 20⁰C cần đổ là \(V_2=b\left(l\right)\Rightarrow m_2=b\left(kg\right)\)

\(\Rightarrow a+b=10\left(kg\right)\) \(\Leftrightarrow a=10-b\)

Có \(Q_{tỏa}=Q_{thu}\)

\(\Leftrightarrow m_1.c\left(100-40\right)=m_2.c\left(40-20\right)\)

\(\Leftrightarrow60a=20b\) \(\Leftrightarrow3a=b\)

\(\Leftrightarrow3\left(10-b\right)=b\Leftrightarrow b=7,5\) \(\Rightarrow a=2,5\)

\(\Rightarrow V_1=2,5\left(l\right);V_2=7,5\left(l\right)\)

Ý A

a)\(\left(x-3\right)\left(x+3\right)\left(x+2\right)-\left(x-1\right)\left(x^2-3\right)-5x\left(x+4\right)^2-\left(x-5\right)^2\)

\(=\left(x^2-9\right)\left(x+2\right)-\left(x^3-3x-x^2+3\right)-5x\left(x^2+8x+16\right)-\left(x^2-10x+25\right)\)

\(=x^3+2x^2-9x-18-x^3+x^2+3x-3-5x^3-40x^2-80x-x^2+10x-25\)

\(=-5x^3-38x^2-76x-46\)

b)\(2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x+5\right)^2-\left(x-1\right)^2\)

\(=2x\left(x^2-8x+16\right)-\left(x+5\right)\left(x^2-4\right)+2\left(x^2+10x+25\right)-\left(x^2-2x+1\right)\)

\(=2x^3-16x^2+32x-\left(x^3+5x^2-4x-20\right)+2x^2+20x+50-x^2+2x-1\)

\(=x^3-20x^2+58x+69\)

c)\(\left(x+5\right)^2-4x\left(2x+3\right)^2-\left(2x-1\right)\left(x+3\right)\left(x-3\right)\)

\(=x^2+10x+25-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)\)

\(=x^2+10x+25-16x^3-48x^2-36x-\left(2x^3-x^2-18x+9\right)\)

\(=-18x^3-46x^2-8x+16\).

Không phải làm như nàu đâu, pt ban đầu có hai nghiệm pb khi pt (1) có hai nghiệm pb \(\ge-\dfrac{1}{2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\Delta>0\\-\dfrac{1}{2}\le x_1< x_2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(m-4\right)^2+12>0\left(lđ\right)\\\left(x_1+\dfrac{1}{2}\right)\left(x_2+\dfrac{1}{2}\right)\ge0\end{matrix}\right.\)

\(\Leftrightarrow x_1x_2+\dfrac{1}{2}\left(x_1+x_2\right)+\dfrac{1}{4}\ge0\)

\(\Leftrightarrow\dfrac{-1}{3}+\dfrac{1}{2}.\dfrac{m-4}{3}+\dfrac{1}{4}\ge0\)

\(\Leftrightarrow m\ge\dfrac{9}{2}\)