- Thể tích đống cát Bác An đang có là:

            \(V_{cát}=\dfrac{1}{3}\pi R^2h=\dfrac{1}{3}\pi.\left(\dfrac{6}{2}\right)^2.2=6\pi\left(m^3\right)\)

- Vậy Bác An cần mua bổ sung: \(30-6\pi\approx11,16\) m³ cát.

\(\dfrac{ab}{\sqrt{5a^2+10ab+10b^2}}=\dfrac{ab}{\sqrt{5\left[b^2+\left(a+b\right)^2\right]}}=\dfrac{ab}{\sqrt{\left[b^2+\left(a+b\right)^2\right]\left(1^2+2^2\right)}}\le^{Bunhiacopxki}\dfrac{ab}{b.1+\left(a+b\right).2}=\dfrac{ab}{2a+3b}\)

\(\Rightarrow\dfrac{ab}{\sqrt{5a^2+10ab+10b^2}}\le\dfrac{ab}{2a+3b}\left(1\right)\)

Áp dụng bất đẳng thức Bunhiaxcopki dạng phân thức:

\(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{5^2}{2a+3b}\)

\(\Rightarrow\dfrac{25}{2a+3b}\le\dfrac{2}{a}+\dfrac{3}{b}\Rightarrow\dfrac{ab}{2a+3b}\le\dfrac{2b+3a}{25}\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow\dfrac{ab}{\sqrt{5a^2+10ab+10b^2}}\le\dfrac{2b+3a}{25}\left(3\right)\)

Tương tự: 

\(\dfrac{bc}{\sqrt{5b^2+10bc+10c^2}}\le\dfrac{2c+3b}{25}\left(4\right)\)

\(\dfrac{ca}{\sqrt{5c^2+10ca+10a^2}}\le\dfrac{2a+3c}{25}\left(5\right)\)

\(\left(3\right)+\left(4\right)+\left(5\right)\Rightarrow M\le\dfrac{a+b+c}{5}\)

Ta có: \(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)=3.3=9\)

\(\Rightarrow a+b+c\le3\)

\(\Rightarrow M\le\dfrac{a+b+c}{5}\le\dfrac{3}{5}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Vậy \(MaxM=\dfrac{3}{5}\)

Đây là box Toán nhé bạn, phiền bạn chuyển bài này sang box Văn nhé.

36) A supermarket has just been built near the city center.

37) Miss Nga said that she enjoyed reading books in her free time.

38) Unless Tom hurry up, he will be late for the train.

39) I haven't seen Mr. John for a week.

\(Đk:x\ge\dfrac{3}{2}\Rightarrow x>0\)

\(x^3-4x^2+5x-1-\sqrt{2x-3}=0\)

\(\Leftrightarrow2x^3-8x^2+10x-2-2\sqrt{2x-3}=0\)

\(\Leftrightarrow\left(2x^3-8x^2+8x\right)+\left[\left(2x-3\right)-2\sqrt{2x-3}+1\right]=0\)

\(\Leftrightarrow2x\left(x-2\right)^2+\left(\sqrt{2x-3}-1\right)^2=0\)

Ta có: \(\left\{{}\begin{matrix}2x\left(x-2\right)^2\ge0\left(x>0\right)\\\left(\sqrt{2x-3}-1\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow2x\left(x-2\right)^2+\left(\sqrt{2x-3}-1\right)^2\ge0\)

Do đó: \(\left\{{}\begin{matrix}2x\left(x-2\right)^2=0\\\left(\sqrt{2x-3}-1\right)^2=0\end{matrix}\right.\Leftrightarrow x=2\)

Thử lại ta có x=2 là nghiệm duy nhất của phương trình đã cho.

Anh kiểm tra lại lời giải của anh lại nhé.

\(S=a^2+\dfrac{1}{a^2}=\left(\dfrac{1}{16}a^2+\dfrac{1}{a^2}\right)+\dfrac{15}{16}a^2\ge2.\sqrt{\dfrac{1}{16}a^2.\dfrac{1}{a^2}}+\dfrac{15}{16}.2^2=\dfrac{17}{4}\)

Dấu "=" xảy ra khi \(a=2\)

Vậy \(MinS=\dfrac{17}{4}\)

Câu 6:

\(A=2x^2+4xy+3y^2-2x+2y+2\)

\(\dfrac{A}{2}=x^2+2xy+\dfrac{3}{2}y^2-x+y+1\)

\(=\left(x^2+y^2+\dfrac{1}{4}+2xy-x-y\right)+\left(\dfrac{1}{2}y^2+2y+\dfrac{3}{4}\right)\)

\(=\left(x^2+y^2+2xy\right)-\left(x+y\right)+\dfrac{1}{4}+\dfrac{1}{2}\left(y^2+4y+\dfrac{3}{2}\right)\)

\(=\left(x+y-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\left[\left(y+2\right)^2-\dfrac{5}{2}\right]\)

\(\ge\dfrac{1}{2}.\left(-\dfrac{5}{2}\right)=\dfrac{-5}{4}\)

\(\Rightarrow A\ge\dfrac{-5}{2}\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x+y-\dfrac{1}{2}=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=-2\end{matrix}\right.\)

Vậy \(MinA=\dfrac{-5}{2}\)

1) That present was given to me on my 20^th birthday.

2) He asked me what my name was.

1) If you work late tonight, I'll pay you well.

2) If you recycle waste paper now, you will help save money and labor.

3) If the air were pure, they would feel healthy.

4) If they didn't speak French to her, she could help them.

5) If the weather weren't bad, we would enjoy our holiday.