Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ĐKXĐ: \(x\ge-1\)
Đặt \(\sqrt{x+1}=y\ge0\)
\(\Rightarrow4x^2+12xy=27y^2\)
\(\Leftrightarrow\left(2x-3y\right)\left(2x+9y\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3y=2x\\9y=-2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3\sqrt{x+1}=2x\left(x\ge0\right)\\9\sqrt{x+1}=-2x\left(x\le0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}9\left(x+1\right)=4x^2\left(x\ge0\right)\\81\left(x+1\right)=4x^2\left(x\le0\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{81-9\sqrt{97}}{8}\end{matrix}\right.\)
\(A^2=9-x+2\sqrt{\left(2x+5\right)\left(4-3x\right)}\ge9-x\ge9-\frac{4}{3}=\frac{23}{3}\)
\(\Rightarrow A\ge\sqrt{\frac{23}{3}}\Rightarrow a+b=26\)
ĐK: x>0
\(bpt\Leftrightarrow\hept{\begin{cases}x\ge0\\6x^2-13x-15=0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge0\\x=3;x=\frac{-5}{6}\end{cases}\Leftrightarrow}x=3\Rightarrow y=\pm2}\)
\(\Leftrightarrow\frac{4}{\sqrt{x}}\ge\frac{\left(\sqrt{2x+17}-\sqrt{2x+1}\right)\left(\sqrt{2x+17}+\sqrt{2x+1}\right)}{\sqrt{2x+17}+\sqrt{2x+1}}\)
\(\Leftrightarrow\frac{4}{\sqrt{x}}\ge\frac{16}{\sqrt{2x+17}+\sqrt{2x+1}}\)
\(\Leftrightarrow\sqrt{2x+17}+\sqrt{2x+1}\ge4\sqrt{x}\)
\(\Leftrightarrow\left(\sqrt{2x+17}+\sqrt{2x+1}\right)^2\ge16x\)
\(\Leftrightarrow\sqrt{\left(2x+17\right)\left(2x+1\right)}\ge6x-9\)
\(\Leftrightarrow x\in\left\{\frac{3}{2},4\right\}\)
Theo đk, ta có tập nghiệm của bpt là S= \(\left\{0;4\right\}\)
1)ĐK:\(x\in\left[-3;\frac{6}{5}\right]\)
pt\(\Leftrightarrow3\left(x^2-x+2\right)-3\left[\sqrt{6-5x}-\left(x-2\right)\right]+\left[3\sqrt{x+3}-\left(x+5\right)\right]=0\)
\(\Leftrightarrow\left(x^2-x+2\right)\left(\frac{3}{\sqrt{6-5x}+x-2}+\frac{1}{3\sqrt{x+3}+x+5}+3\right)=0\)
\(\Leftrightarrow x^2\)-x+2=0(do(...)>0)
\(\Leftrightarrow x=-2\)hoặc \(x=1\)(t/m)
ÁD BĐT Bunhiacopxki:
\(\left(a+b+c\right)\left[\frac{a}{\left(b+c\right)^2}+\frac{b}{\left(c+a\right)^2}+\frac{c}{\left(a+b\right)^2}\right]\ge\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)^2\)
Lại có:\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3\)
\(=\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3\ge\frac{9}{2\left(a+b+c\right)}-3=\frac{3}{2}\)
\(\Rightarrow VT\ge\left(\frac{3}{2}\right)^2\)=\(\frac{9}{4}\)(đpcm)
Dấu''='' xảy ra\(\Leftrightarrow a=b=c=\frac{1}{3}\)
\(-x^2+4x-5=-\left(x-2\right)^2-1< 0;\forall x\)
Do đó BPT tương đương:
\(x^2-2\left(2m-3\right)x+4m-3>0\)
Do \(a=1>0\) nên để tập nghiệm BPT là R
\(\Leftrightarrow\Delta'=\left(2m-3\right)^2-\left(4m-3\right)< 0\)
\(\Leftrightarrow4m^2-16m+12< 0\)
\(\Rightarrow1< m< 3\Rightarrow b-3a=3-3.1=0\)
ĐKXĐ: \(x\ge\frac{2}{3}\)
\(\Leftrightarrow x^3-1+2x-1-\sqrt{3x-2}+x+1-\sqrt{x+3}\le0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)+\frac{4x^2-7x+3}{2x-1+\sqrt{3x-2}}+\frac{x^2+x-2}{x+1+\sqrt{x+3}}\le0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)+\frac{\left(x-1\right)\left(4x-3\right)}{2x-1+\sqrt{3x-2}}+\frac{\left(x-1\right)\left(x+2\right)}{x+1+\sqrt{x+3}}\le0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1+\frac{4x-3}{2x-1+\sqrt{3x-2}}+\frac{x+2}{x+1+\sqrt{x+3}}\right)\le0\)
\(\Leftrightarrow x-1\le0\) (ngoặc đằng sau luôn dương)
\(\Rightarrow x\le1\Rightarrow\frac{2}{3}\le x\le1\Rightarrow\left\{{}\begin{matrix}a=2\\b=3\\c=1\end{matrix}\right.\) \(\Rightarrow a+b=5\)