Violympic toán 9 - Hỏi đáp

Bình phương 2 vế , ta được :

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

\(\Leftrightarrow x^4+x^2+1+2x^3-2x^2-2x=x^4+4x^3+4x^2-2x^3-8x^2-8x+2x^2+8x+8\)

\(\Leftrightarrow x^4+2x^3-x^2-2x+1=x^4+2x^3-2x^2+8\)

\(\Leftrightarrow x^2-2x-7=0\)

\(\Leftrightarrow\left(x-1\right)^2=8\)

\(\Leftrightarrow x=\pm\sqrt{8}+1\)

Vậy ...

*Không ghi lại đề nhé ĐK: x≥0;x≠4

\(C=\left[\frac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\cdot\frac{\left(\sqrt{x}+2\right)^2}{4}=\frac{4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\frac{\left(\sqrt{x}+2\right)^2}{4}=\frac{\sqrt{x}+2}{\sqrt{x}-2}\)

a/Kẻ \(OH\perp AC,OK\perp BC.\)

Ta có :\(OA=OB,OH\text{//}BC\)

\(\Rightarrow OH\) là đường trung bình.

\(\Rightarrow BC=2OH=12\left(cm\right)\)

Ttự, ta có:\(AC=2OK=2.8=16\left(cm\right)\)

Ta có:\(AB=\sqrt{AC^2+BC^2}\)

\(=\sqrt{16^2+12^2}=20\left(cm\right).\)

\(3-2P=\frac{x}{x+2\sqrt{yz}}+\frac{y}{y+2\sqrt{xz}}+\frac{z}{z+2\sqrt{xy}}\)

\(3-2P\ge\frac{x}{x+y+z}+\frac{y}{x+y+z}+\frac{z}{x+y+z}=1\)

\(\Rightarrow2P\le2\Rightarrow P\le1\)

Dấu "=" xảy ra khi \(x=y=z\)

\(M\le\sqrt{\left(1+1\right)\left(x+y+2\right)}=\sqrt{20}=4\sqrt{5}\)

\(M_{max}=4\sqrt{5}\) khi \(\left\{{}\begin{matrix}x-2=y+4\\x+y=8\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)

1) \(P\ge\frac{x^2.2\sqrt{yz}}{yz}+\frac{y^2.2\sqrt{zx}}{zx}+\frac{z^2.2\sqrt{xy}}{xy}=\frac{2x^2}{\sqrt{yz}}+\frac{2y^2}{\sqrt{zx}}+\frac{2z^2}{\sqrt{xy}}\ge4\left(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\right)=4\left\{\left[\frac{x^2}{y+z}+\frac{1}{4}\left(y+z\right)\right]+\left[\frac{y^2}{z+x}+\frac{1}{4}\left(z+x\right)\right]+\left[\frac{z^2}{x+y}+\frac{1}{4}\left(x+y\right)\right]\right\}-2\left(x+y+z\right)\ge4\left(x+y+z\right)-2\left(x+y+z\right)=2\)

Dấu "=" xảy ra <=> \(x=y=z=\frac{1}{3}\)

2) \(A=\left[\frac{1}{x^2+y^2}+4\left(x^2+y^2\right)\right]+\left(\frac{1}{xy}+16xy\right)-4\left(x+y\right)^2-8xy\ge4+8-4-2.\left(x+y\right)^2=8-2.\left(x+y\right)^2\ge8-2=6\)

Dấu "=" xảy ra <=> \(x=y=\frac{1}{2}\)

\(P=\sum\frac{x^2\left(y+z\right)}{yz}\ge\sum\frac{4x^2\left(y+z\right)}{\left(y+z\right)^2}=\sum\frac{4x^2}{y+z}\ge\frac{4\left(x+y+z\right)^2}{y+z+z+x+x+y}=2\left(x+y+z\right)=2\)

\(P_{min}=2\) khi \(x=y=z=\frac{1}{3}\)

Câu 2 có dương không nhỉ? Không dương thì không làm được

\(A=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}\ge\frac{4}{x^2+y^2+2xy}+\frac{2}{\left(x+y\right)^2}=\frac{6}{\left(x+y\right)^2}\ge6\)

\(A_{min}=6\) khi \(x=y=\frac{1}{2}\)

Do \(0< a< 1\Rightarrow b>0\)

\(A=2a+\frac{b}{4a}+b^2=\frac{3a}{2}+\frac{a}{2}+\frac{b}{4a}+b^2\ge\frac{3a}{2}+3\sqrt[3]{\frac{ab^3}{8a}}=\frac{3}{2}\left(a+b\right)\ge\frac{3}{2}\)

\(A_{min}=\frac{3}{2}\) khi \(a=b=\frac{1}{2}\)

thôi t ra r

Lời giải:

\(P=\frac{\sqrt{x}-1}{x+8}=\frac{a-1}{a^2+8}\) (đặt \(\sqrt{x}=a\), \(a\geq 0)\)

\(\Rightarrow Pa^2+8P=a-1\)

\(\Leftrightarrow Pa^2-a+(8P+1)=0\)

Coi đây là pt bậc hai bậc $a$. Vì dấu "=" tồn tại nghĩa là pt luôn có nghiệm nên:

\(\Delta=1-4P(8P+1)\geq 0\)

\(\Leftrightarrow 32P^2+4P-1\leq 0\)

\(\Leftrightarrow (8P-1)(4P+1)\leq 0\Rightarrow P\leq \frac{1}{8}\)

Vậy \(P_{\max}=\frac{1}{8}\)

ĐKXĐ: \(x\ge0\)

Khi đó \(2x+1+\sqrt{x}\) hiển nhiên dương nên ko cần tìm điều kiện cho căn to nữa