\(Q=x^2\left(4-3x\right)=\dfrac{4}{9}.\dfrac{3}{2}x.\dfrac{3}{2}x\left(4-3x\right)\)

\(Q\le\dfrac{1}{27}.\dfrac{4}{9}.\left(\dfrac{3x}{2}+\dfrac{3x}{2}+4-3x\right)^3=\dfrac{256}{243}\)

\(Q_{maxx}=\dfrac{256}{243}\) khi \(\dfrac{3x}{2}=4-3x\Leftrightarrow x=\dfrac{8}{9}\)

A = \(\frac{3x}{2}+\frac{2}{x-1}=3.\frac{x-1}{2}+\frac{2}{x-1}+\frac{3}{2}\)\(\ge2\sqrt{3}+\frac{3}{2}\)

\(\Rightarrow\)min A = \(2\sqrt{3}+\frac{3}{2}\Leftrightarrow x=\frac{2}{\sqrt{3}}+1\)(thỏa mãn)

B = \(x+\frac{3}{3x-1}=\frac{1}{3}\left(3x-1+\frac{9}{3x-1}+1\right)\)\(\ge\frac{1}{3}\left(2\sqrt{9}+1\right)=\frac{7}{3}\)

\(\Rightarrow\)min B = \(\frac{7}{3}\Leftrightarrow x=\frac{4}{3}\)

\(A\) \(=\) \(3x^2\left(8-x^2\right)\le3\frac{\left(x^2+8-x^2\right)^2}{4}=48\)

\(\Rightarrow\) maxA = 48 \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)(thỏa mãn)

\(B=\) \(4x\left(8-5x\right)\)\(=\frac{4}{5}.5x\left(8-5x\right)\le\frac{4}{5}.\frac{\left(5x+8-5x\right)^2}{4}=\frac{64}{5}\)

\(\Rightarrow\)max B = \(\frac{64}{5}\Leftrightarrow x=\frac{4}{5}\)(thỏa mãn)

C = \(4\left(x-1\right)\left(8-5x\right)=\frac{4}{5}.\left(5x-5\right)\left(8-5x\right)\)\(\le\frac{4}{5}.\frac{\left(5x-5+8-5x\right)^2}{4}=\frac{9}{5}\)

\(\Rightarrow\)max C = \(\frac{9}{5}\)\(\Leftrightarrow x=\frac{13}{10}\)(thỏa mãn)

D = \(x\left(3-\sqrt{3}\right)\)(quá dễ rồi)

\(A=\dfrac{x^2+x-2+x^2-x-2-4}{x\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x-3\right)}{2\left(x+2\right)}=\dfrac{2\left(x-2\right)\left(x+2\right)\left(x-3\right)}{2\left(x-2\right)\left(x+2\right)^2}=\dfrac{x-3}{x+2}\\ A\le0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\\x+2< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3\le0\\x+2>0\end{matrix}\right.\end{matrix}\right.\Rightarrow-2< x< 3;x\ne0\left(ĐKXD\right)\)

Ai trả lời giùm em với

Câu 1:ĐkXĐ \(x\ge-\frac{1}{4}\)

\(\left(2\sqrt{x+2}-\sqrt{4x+1}\right)\left(2x+3+\sqrt{4x^2+9x+2}\right)=7\)(theo đề ở dưới)

Nhân liên hợp ta có

\(\left(4\left(x+2\right)-4x-1\right)\left(2x+3+\sqrt{4x^2+9x+2}\right)=7\left(2\sqrt{x+2}+\sqrt{4x+1}\right)\)<=>\(2x+3+\sqrt{4x^2+9x+2}=2\sqrt{x+2}+\sqrt{4x+1}\)(1)

Đặt \(2\sqrt{x+2}+\sqrt{4x+1}=t\left(t\ge0\right)\)

=> \(t^2=8x+9+4\sqrt{4x^2+9x+2}\)

=> \(\frac{t^2-8x-9}{4}=\sqrt{4x^2+9x+2}\)

Khi đó (1)

<=> \(2x+3+\frac{t^2-8x-9}{4}=t\)

<=> \(\frac{3}{4}+\frac{t^2}{4}=t\)

=> \(\left[{}\begin{matrix}t=1\\t=3\end{matrix}\right.\)(tm)

+ \(t=1\) => \(\sqrt{4x^2+9x+2}=-2x-2\)

Mà \(x\ge-\frac{1}{4}\)

=> pt vô nghiệm

+ t=3 => \(\sqrt{4x^2+9x+2}=-2x\)

=> \(\left\{{}\begin{matrix}x\le0\\9x+2=0\end{matrix}\right.\)

=> \(x=-\frac{2}{9}\)(tmĐKXĐ)

Vậy x=-2/9

Câu 3:

\(\frac{1}{a+bc}+\frac{1}{b+ac}=\frac{1}{a+b}\)

<=> \(\frac{\left(a+b\right)\left(c+1\right)}{\left(a+bc\right)\left(b+ac\right)}=\frac{1}{a+b}\)

<=> \(\left(a+b\right)^2\left(c+1\right)=ab\left(c^2+1\right)+c\left(a^2+b^2\right)\)

<=> \(2abc+a^2+b^2+ab=abc^2\)

<=> \(\left(a^2+b^2+2ba\right)=ab\left(c^2-2c+1\right)\)

<=> \(\left(a+b\right)^2=ab\left(c-1\right)^2\)

=> ab>0 , ab là bình phương của số hữu tỉ

=> \(c-1=\frac{a+b}{\sqrt{ab}}\)

=> \(c+1=\frac{a+b}{\sqrt{ab}}+2=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{ab}}\)

Khi đó

\(\frac{c-3}{c+1}=1-\frac{4}{c+1}=1-\frac{4\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)^2}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)^2}\)

Mà \(\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{a-b}=\frac{a+b-2\sqrt{ab}}{a-b}\)là số hữu tỉ do ab là bình phương của số hữu tỉ

=> \(\frac{c-3}{c+1}\)là bình phương của số hữu tỉ(ĐPCM)

Câu 4 : Anh có cách này tạm được

Xét \(B\le10\)

Quy đồng chuyển vế ta có:\(\left(a+b+c+3\right)\left[\left(ab+bc+ac\right)+2\left(a+b+c\right)+3\right]\le10\left(a+1\right)\left(b+1\right)\left(c+1\right)\)

<=>\(\Sigma a^2\left(b+c\right)+\Sigma2a^2+7\left(ab+bc+ac\right)+9\left(a+b+c\right)+9\le10\left(abc+\Sigma a+\Sigma ab+1\right)\)

<=> \(\Sigma a^2\left(b+c\right)+2\left(a^2+b^2+c^2\right)\le7abc+3\left(ab+bc+ac\right)+1+a+b+c\)

Mà \(\left\{{}\begin{matrix}a^2\le a\\b^2\le b\\c^2\le c\end{matrix}\right.\)do \(0\le a,b,c\le1\)

=> \(a\left(b+c\right)+b\left(a+c\right)+c\left(a+b\right)+2\left(a+b+c\right)\le7abc+3\left(ab+bc+ac\right)+\Sigma a+1\)

<=> \(a+b+c\le7abc+ab+bc+ac+1\)

Lại có \(7abc\ge-abc\)

=> \(a+b+c\le-abc+ab+bc+ac+1\)

<=> \(\left(1-a\right)\left(1-b\right)\left(1-c\right)\ge0\)luôn đúng với mọi \(0\le a,b,c\le1\)

=> ĐPCM

Dấu bằng xảy ra khi \(\left\{{}\begin{matrix}a^2=a\\b^2=b\\c^2=c\\\left(a-1\right)\left(b-1\right)\left(c-1\right)=0\end{matrix}\right.\)

=> \(\left(a,b,c\right)=\left(1,1,0\right);\left(0,0,1\right),...\)và các hoán vị

a/

\(\frac{3x-4}{x-2}-1>0\Leftrightarrow\frac{2x-2}{x-2}>0\Rightarrow\left[{}\begin{matrix}x>2\\x< 1\end{matrix}\right.\)

b/

\(\frac{2x-5}{2-x}+1\le0\Rightarrow\frac{x-3}{2-x}\le0\Rightarrow\left[{}\begin{matrix}x\ge3\\x< 2\end{matrix}\right.\)

c/

\(\frac{x^2+x-3}{x^2-4}-1\le0\Rightarrow\frac{x+1}{x^2-4}\le0\Rightarrow\frac{x+1}{\left(x-2\right)\left(x+2\right)}\le0\Rightarrow\left[{}\begin{matrix}x< -2\\-1\le x< 2\end{matrix}\right.\)

d/

\(\frac{4x^2-8x+6+x^2-x-6}{2\left(x^2-x-6\right)}>0\Rightarrow\frac{x\left(5x-9\right)}{2\left(x+2\right)\left(x-3\right)}>0\Rightarrow\left[{}\begin{matrix}x>3\\0< x< \frac{9}{5}\\x< -2\end{matrix}\right.\)

e/

\(\frac{x^2+3x+2}{2x+3}-\frac{2x-5}{4}\ge0\Rightarrow\frac{4x^2+12x+8-\left(2x-5\right)\left(2x+3\right)}{4\left(2x+3\right)}\ge0\)

\(\Rightarrow\frac{28x+23}{4\left(2x+3\right)}\ge0\Rightarrow\left[{}\begin{matrix}x\ge-\frac{23}{28}\\x< -\frac{3}{2}\end{matrix}\right.\)

bạn xem câu hỏi số 905663 nhé

Đề kì vậy bạn. Sao vế trái không có \(y\) vậy?