Violympic toán 9 - Hỏi đáp

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Mình vừa giải một bài tương tự với câu hỏi của bạn, Đây là phần bài làm của mình bạn nhé!

\(P=\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1-\frac{2x+\sqrt{x}}{\sqrt{x}}\)

\(P=\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1-\left(2\sqrt{x}+1\right)\)

\(P=\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1-2\sqrt{x}-1\)

\(P=\frac{x^2-2x\sqrt{x}+2x-\sqrt{x}}{x-\sqrt{x}+1}\)

E không chắc đâu ạ

\(VT=\sqrt{ab.ac}+\sqrt{ab.bc}+\sqrt{ac.bc}\)

\(VT\le\frac{1}{2}\left(ab+ac+ab+bc+ac+bc\right)\)

\(VT\le ab+bc+ca\le\frac{1}{3}\left(a+b+c\right)^2\)

Dấu "=" xảy ra khi \(a=b=c\)

ĐK:\(x\ge0\)

\(x\sqrt{x}-2\sqrt{x}-x=0\Leftrightarrow\sqrt{x}\left(x-2-\sqrt{x}\right)=0\Leftrightarrow\sqrt{x}\left(x-2\sqrt{x}+\sqrt{x}-2\right)=0\Leftrightarrow\sqrt{x}\left[\sqrt{x}\left(\sqrt{x}-2\right)+\left(\sqrt{x}-2\right)\right]=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\\\sqrt{x}+1=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}=2\\\sqrt{x}=-1\left(loai\right)\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\left(tm\right)\\x=4\left(tm\right)\end{matrix}\right.\)

Vậy S={0;4}

ĐKXĐ: \(x\ge\frac{2}{3}\)

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

\(\Leftrightarrow\frac{4x-8}{\sqrt{4x+1}+3}+\frac{3x-6}{\sqrt{3x-2}+2}=0\)

\(\Leftrightarrow\left(x-2\right)\left(\frac{2}{\sqrt{4x+1}+3}+\frac{2}{\sqrt{3x-2}+2}\right)=0\)

\(\Leftrightarrow x=2\)

\(\dfrac{\left(\sqrt{X}+\sqrt{Y}\right)\left(1+\sqrt{XY}\right)+\left(\sqrt{X}-\sqrt{Y}\right)\left(1-\sqrt{XY}\right)}{1-XY}\cdot\dfrac{1-XY}{1-XY+\sqrt{X}+\sqrt{Y}+2\sqrt{XY}}=\dfrac{\sqrt{X}+X\sqrt{Y}+\sqrt{Y}+Y\sqrt{X}+\sqrt{X}-X\sqrt{Y}-\sqrt{Y}+Y\sqrt{X}}{1-XY}\cdot\dfrac{1-XY}{XY+X+Y+1}=\dfrac{2\sqrt{X}\left(1+Y\right)}{\left(1+Y\right)\left(X+1\right)}=\dfrac{2\sqrt{X}}{X+1}\)

\(\frac{x^3}{y+1}+\frac{y+1}{4}+\frac{1}{2}\ge3\sqrt[3]{\frac{x^3\left(y+1\right)}{8\left(y+1\right)}}=\frac{3}{2}x\)

Tương tự: \(\frac{y^3}{x+1}+\frac{x+1}{4}+\frac{1}{2}\ge\frac{3}{2}y\)

Cộng vế với vế:

\(B+\frac{x+y+2}{4}+1\ge\frac{3}{2}\left(x+y\right)\)

\(\Rightarrow B\ge\frac{5}{4}\left(x+y\right)-\frac{3}{2}\ge\frac{5}{4}.2\sqrt{xy}-\frac{3}{2}=1\)

Dấu "=" xảy ra khi \(x=y=1\)

M= \(\sqrt{x}-1+\sqrt{y-x}\)(đk : \(y\ge x\ge0\)

Áp dụng bđt cosi vs hai số dương có:

\(\sqrt{x}=1.\sqrt{x}\le\frac{x+1}{2}\)

\(\sqrt{y-x}\le\frac{y-x+1}{2}\)

=> \(\sqrt{x}+\sqrt{y-x}\le\frac{x+1}{2}+\frac{y-x+1}{2}\) <=> \(\sqrt{x}-1+\sqrt{y-x}\le\frac{x}{2}+\frac{1}{2}+\frac{y}{2}-\frac{x}{2}+\frac{1}{2}-1\)

<=> \(M\le\frac{y}{2}\) (1)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}1=y-x\\x=1\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)

Từ (1) => M=1 tại y=2,x=1

Vậy maxM=1 <=> x=1,y=2

\(A=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2-4\sqrt{xy}}{\sqrt{x}-\sqrt{y}}+\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}\)(đk: \(x,y>0\) ,\(x\ne y\))

=\(\frac{x+y+2\sqrt{xy}-4\sqrt{xy}}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\) (vì x,y>0)

=\(\frac{x-2\sqrt{xy}+y}{\sqrt{x}-\sqrt{y}}+\sqrt{x}-\sqrt{y}\)

= \(\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{x}-\sqrt{y}}+\sqrt{x}-\sqrt{y}=\sqrt{x}-\sqrt{y}+\sqrt{x}-\sqrt{y}=2\sqrt{x}-2\sqrt{y}\)

Vậy A= \(2\sqrt{x}-2\sqrt{y}\)

\(A=11^3+12^3+...+1945^3\)

Ta có: \(A=11^3+12^3+...+1945^3\)

\(=\left(12^3+14^3+...+1944^3\right)+\left(11^3+13^3+...+1945^3\right)\)

Do dãy \(11;13;...;1945\) có \(\frac{1945-11}{2}+1=968\) số hạng

\(\Rightarrow \left(11^3+13^3+...+1945^3\right)⋮2\) mà \(\left(12^3+14^3+...+1944^3\right)⋮2\)

\(\Rightarrow A⋮2\left(1\right)\)

Mặt khác:

\(A=\left(11^3+1945^3\right)+\left(12^3+1944^3\right)+...+\left(977^3+979^3\right)+978^3\)

\(=967.1956^3+978^3⋮3\)

\(\Rightarrow A⋮3\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow A⋮6\)

\(=\frac{1}{2}\left(\sqrt[3]{40+16\sqrt{13}}+\sqrt[3]{40-16\sqrt{13}}\right)\)

\(=\frac{1}{2}\left(\sqrt[3]{\left(\sqrt{13}+1\right)^3}+\sqrt[3]{\left(1-\sqrt{13}\right)^3}\right)\)

\(=\frac{1}{2}\left(\sqrt{13}+1+1-\sqrt{13}\right)=1\)