Violympic toán 9 - Hỏi đáp

Sử dụng BĐT Cauchy-Schwarz nhiều lần, cộng với BĐT phụ \(\dfrac{1}{xy}\ge\dfrac{4}{\left(x+y\right)^2}\), ta có:

\(8\left(x^4+y^4\right)+\dfrac{1}{xy}\ge\dfrac{8\left(x^2+y^2\right)^2}{2}+\dfrac{4}{\left(x+y\right)^2}=4\left(x^2+y^2\right)^2+4\ge4\left[\dfrac{\left(x+y\right)^2}{2}\right]^2+4=5\)

Đẳng thức xảy ra khi \(x=y=\dfrac{1}{2}\)

\(\left\{{}\begin{matrix}x;y>0\\x+y=1\end{matrix}\right.\)\(\Rightarrow0< xy=t\le\dfrac{1}{4}\)

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

\(8\left(x^4+y^4\right)+\dfrac{1}{xy}\ge5\Leftrightarrow A=8\left[\left(1-2t\right)^2-2t\right]+\dfrac{1}{t}-5\ge0\)

\(\Leftrightarrow16t^2-32t+\dfrac{1}{t}+3\ge0\)\(\Leftrightarrow16t^3-32t^2+3t+1\ge0\)

<=>\(16t^3-4t^2-28t^2+7t-4t+1\ge0\)

\(\Leftrightarrow4t^2\left(4t-1\right)-7t\left(4t-1\right)-\left(4t-1\right)\ge0\)

\(\Leftrightarrow\left(4t-1\right)\left(4t^2-7t-1\right)\ge0\)

\(\Leftrightarrow B=\left(4t-1\right)\left(8t-7-\sqrt{65}\right)\left(8t-7+\sqrt{65}\right)\ge0\)

\(0< t\le\dfrac{1}{4}\Rightarrow\)\(\left\{{}\begin{matrix}4t-1\le0\\8t-7+\sqrt{65}>0\\8t-7-\sqrt{5}< 0\end{matrix}\right.\) \(\Rightarrow B\ge0\)

mọi phép biến đổi <=> => dpcm

Đề thiếu khoảng của a,b

\(a;b\in\left[\dfrac{1}{4};2\right]\)

\(x^2+x+6=y^2\)

\(\Leftrightarrow x^2+x+6-y^2=0\)

\(\Leftrightarrow4x^2+4x+24-4x^2=0\)

\(\Leftrightarrow\left(4x^2+2x+4xy\right)+\left(2x+1+2y\right)-\left(4xy+2y+4y^2\right)+23=0\)

\(\Leftrightarrow2x\left(2x+1+2y\right)+\left(2x+1+2y\right)-2y\left(2x+1+2y\right)=-23\)

\(\Leftrightarrow\left(2x+1+2y\right)\left(2x+1-2y\right)=-23\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x+1+2y=1\\2x+1-2y=-23\end{matrix}\right.\\\left\{{}\begin{matrix}2x+1+2y=-1\\2x+1-2y=23\end{matrix}\right.\\\left\{{}\begin{matrix}2x+1+2y=-23\\2x+1-2y=1\end{matrix}\right.\\\left\{{}\begin{matrix}2x+1+2y=23\\2x+1-2y=-1\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-6\\y=6\end{matrix}\right.\\\left\{{}\begin{matrix}x=5\\y=-6\end{matrix}\right.\\\left\{{}\begin{matrix}x=-6\\y=-6\end{matrix}\right.\\\left\{{}\begin{matrix}x=5\\y=6\end{matrix}\right.\end{matrix}\right.\)

Lời giải:

Ta có: \(2x^2+5y^2=12\)

\(\Rightarrow 12-5y^2=2x^2\geq 0\)

\(\Rightarrow 5y^2\leq 12\Rightarrow y^2\leq \frac{12}{5}<4\)

\(\Rightarrow -2< y< 2\)

Vì \(y\in\mathbb{Z}\Rightarrow y\in\left\{-1;0;1\right\}\)

Nếu \(y\pm 1\Rightarrow 3x^2=12-5y^2=12-5=7\)

\(\Rightarrow x^2=\frac{7}{3}\) (vô lý do $x$ nguyên)

Nếu \(y=0\Rightarrow 3x^2=12-5y^2=12\Rightarrow x^2=4\Rightarrow x=\pm 2\)

Vậy \((x,y)=(\pm 2,0)\)

Đk: x,y >/ 0

Ta sẽ chứng minh \(\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)^2\ge2\sqrt{xy}\cdot4\sqrt{x}\cdot\sqrt{y}\) (*) luôn đúng với mọi x,

*Xét \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\), ta có:

\(\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)^2\ge2\sqrt{xy}\cdot4\sqrt{x}\cdot\sqrt{y}\Leftrightarrow0\ge0\) (luôn đúng)

Vậy (*) đúng.

*Xét \(\left\{{}\begin{matrix}x=0\\y>0\end{matrix}\right.\), ta có:

\(\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)^2\ge2\sqrt{xy}\cdot4\sqrt{x}\cdot\sqrt{y}\Leftrightarrow y^2\ge0\) (luôn đúng

Vậy (*) đúng

*Xét \(\left\{{}\begin{matrix}x>0\\y=0\end{matrix}\right.\), ta có:

\(\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)^2\ge2\sqrt{xy}\cdot4\sqrt{x}\cdot\sqrt{y}\Leftrightarrow x^2\ge0\) (luôn đúng

Vậy (*) đúng

* Xét x,y > 0

Ta có: (+) \(x+y\ge2\sqrt{xy}\) (1)

(+) \(\sqrt{x}+\sqrt{y}\ge2\sqrt{\sqrt{xy}}\Leftrightarrow\left(\sqrt{x}+\sqrt{y}\right)^2\ge4\sqrt{xy}\) (2)

Nhân (1) và (2) vế theo vế, ta được:

\(\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)^2\ge2\sqrt{xy}\cdot4\sqrt{x}\cdot\sqrt{y}\)

Vậy (*) luôn đúng với mọi x,y >/0 (đpcm)

\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\)
\(\Leftrightarrow\left(a^2+b^2\right)cd=ab\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2cd-b^2cd=abc^2+abd^2\)
\(\Leftrightarrow a^2cd-abc^2-abd^2+b^2cd=0\)
\(\Leftrightarrow ac\left(ad-bc\right)-bd\left(ad-bc\right)=0\)
\(\Leftrightarrow\left(ac-bd\right)\left(ad-bc\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}ac-bd=0\\ad-bc=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}ac=bd\\ad=bc\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{a}{b}=\dfrac{d}{c}\\\dfrac{a}{b}=\dfrac{c}{d}\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}\dfrac{a}{b}=\dfrac{d}{c}\\\dfrac{a}{b}=\dfrac{c}{d}\end{matrix}\right.\) (ĐPCM)

tương tự câu này thôi nha : https://hoc24.vn/hoi-dap/question/636899.html

ai giúp e nhanh lên đc k

\(\sqrt{x-5}-\dfrac{x-14}{3+\sqrt{x-5}}=3\)

\(\sqrt{x-5}-\dfrac{x-14}{3+\sqrt{x-5}}=3,x\in\left[5,+\infty\right]\)

\(\sqrt{x-5}-\dfrac{x-14}{3+\sqrt{x-5}}-3=0\)

\(\dfrac{\left(3+\sqrt{x-5}\right)\sqrt{x-5}-\left(x-14\right)-3\left(3+\sqrt{x-5}\right)}{3+\sqrt{x-5}}=0\)

\(3\sqrt{x-5}\sqrt{x-5}-\left(x-14\right)-3\left(3+\sqrt{x-5}\right)=0\)

\(3\sqrt{x-5}+x-5-\left(x-14\right)-9-3\sqrt{x-5}=0\)

\(x-5-x+14-9=0\)

\(0=0\)

\(x\in R,x\in\left[5,+\infty\right]\)

Lam lai nha , nay cau tha qua :(

\(\sqrt{x-5}-\dfrac{x-14}{3+\sqrt{x-5}}=3\) ( x ≥ 5 )

Dat : \(\sqrt{x-5}=a\) ( x ≥ 0 ) , ta co :

\(\sqrt{a}-\dfrac{a-9}{3+\sqrt{a}}=3\)

⇔ \(\sqrt{a}-\sqrt{a}+3=3\)

⇔ \(3=3\left(Luon-dung\right)\)

KL........

a)\(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)

\(=\dfrac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a-1}\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a\left(\sqrt{a}-1\right)}}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{\left(a-1\right)\left(a-4\right)}\)

\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{a-1-a+4}\)

\(=\dfrac{1}{\sqrt{a}}\cdot\dfrac{\sqrt{a}-2}{3}\)

\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\) ĐKXĐ: \(x>0\) \(a\ne4\) \(a\ne1\)

b)\(Q>0\)

\(\Leftrightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}>0\)

mà \(3\sqrt{a}>0\) (Kết hợp ĐKXĐ \(a>0\))

\(\Leftrightarrow\sqrt{a}-2>0\)

\(\Leftrightarrow\sqrt{a}>2\)

\(\Leftrightarrow a>4\) (thỏa mãn ĐKXĐ)

Vậy \(a>4\) thì \(Q>0\)

___♫ Chúc bạn học tốt ♫___

Lời giải:

a) Ta có:

\(Q=\frac{\sqrt{a}-(\sqrt{a}-1)}{\sqrt{a}(\sqrt{a}-1)}:\frac{(\sqrt{a}+1)(\sqrt{a}-1)-(\sqrt{a}+2)(\sqrt{a}-2)}{(\sqrt{a}-2)(\sqrt{a}-1)}\)

\(=\frac{1}{\sqrt{a}(\sqrt{a}-1)}: \frac{(a-1)-(a-4)}{(\sqrt{a}-2)(\sqrt{a}-1)}\)

\(=\frac{1}{\sqrt{a}(\sqrt{a}-1)}:\frac{3}{(\sqrt{a}-2)(\sqrt{a}-1)}\)

\(=\frac{1}{\sqrt{a}(\sqrt{a}-1)}.\frac{(\sqrt{a}-2)(\sqrt{a}-1)}{3}=\frac{\sqrt{a}-2}{3\sqrt{a}}\)

b) Để \(Q\) dương thì \(\frac{\sqrt{a}-2}{3\sqrt{a}}>0\). Mà \(3\sqrt{a}>0\) với mọi $a$ thuộc ĐKXĐ nên $Q$ dương khi \(\sqrt{a}-2>0\Leftrightarrow \sqrt{a}>2\Leftrightarrow a>4\)

Kết hợp với đkxđ suy ra để $Q$ dương thì $a>4$

mik làm hơi tắt ko bt có sai j ko :")

c) \(\dfrac{a-b}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\)

= \(\dfrac{a\sqrt{a}+a\sqrt{b}-b\sqrt{a}-b\sqrt{b}-\sqrt{a^3}+\sqrt{b^3}}{a-b}\)

= \(\dfrac{\sqrt{a^3}+\sqrt{a^2b}-\sqrt{ab^2}-\sqrt{b^3}-\sqrt{a^3}+\sqrt{b^3}}{a-b}\)

= \(\dfrac{\sqrt{a^2b}-\sqrt{ab^2}}{a-b}\)