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\(x\ge9\Rightarrow x+9\ge18\Rightarrow\sqrt{x+9}\ge3\sqrt{2}\)
nguyễn thị thanh huyền
b/ ĐKXĐ: \(\left[{}\begin{matrix}x\ge-\frac{2}{3}\\x\le-1\end{matrix}\right.\)
Đặt \(3x^2+5x+2=t\ge0\)
\(\Leftrightarrow\sqrt{t+5}-\sqrt{t}>1\)
\(\Leftrightarrow\sqrt{t+5}>\sqrt{t}+1\)
\(\Leftrightarrow t+5>t+1+2\sqrt{t}\)
\(\Leftrightarrow\sqrt{t}< 2\Rightarrow t< 4\)
\(\Rightarrow3x^2+5x+2< 4\)
\(\Leftrightarrow3x^2+5x-2< 0\) \(\Rightarrow-2< x< \frac{1}{3}\)
Kết hợp ĐKXĐ ta được nghiệm của BPT:
\(\left[{}\begin{matrix}-2< x\le-1\\-\frac{2}{3}\le x< \frac{1}{3}\end{matrix}\right.\)
a) \(\sqrt{5x+3}=3x-7\)\(\Leftrightarrow\left\{{}\begin{matrix}5x+3=\left(3x-7\right)^2\\3x-7\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x+3=9x^2-42x+49\\x\ge\dfrac{7}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}9x^2-47x+46=0\\x\ge\dfrac{7}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{47+\sqrt{553}}{18}\\x=\dfrac{47-\sqrt{553}}{18}\end{matrix}\right.\\x\ge\dfrac{7}{3}\end{matrix}\right.\)\(\Leftrightarrow\dfrac{47+\sqrt{553}}{18}\).
b) \(\sqrt{3x^2-2x-1}=3x+1\)\(\Leftrightarrow\left\{{}\begin{matrix}3x^2-2x-1=\left(3x+1\right)^2\\3x+1\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x^2+8x+2=0\\x\ge\dfrac{-1}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-1\end{matrix}\right.\\x\ge-\dfrac{1}{3}\end{matrix}\right.\)\(\Leftrightarrow x=-\dfrac{1}{3}\).
a) Đặt \(a=\sqrt[3]{1+\sqrt{x}};b=\sqrt[3]{1-\sqrt{x}}\)
\(\Rightarrow a^3+b^3=2\) kết hợp với đề bài
\(\Rightarrow\left\{{}\begin{matrix}a^3+b^3=2\\a+b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)\left(a^2-ab+b^2\right)=2\\a+b=2\end{matrix}\right.\)
................
a/ \(-1\le x\le1\)
\(\Leftrightarrow\frac{2x}{\sqrt{1+x}+\sqrt{1-x}}-x\ge0\)
\(\Leftrightarrow x\left(\frac{2}{\sqrt{1+x}+\sqrt{1-x}}-1\right)\ge0\)
Do \(0< \sqrt{1+x}+\sqrt{1-x}\le\sqrt{2\left(1+x+1-x\right)}=2\)
\(\Rightarrow\frac{2}{\sqrt{1+x}+\sqrt{1-x}}\ge1\Rightarrow\frac{2}{\sqrt{1+x}+\sqrt{1-x}}-1\ge0\)
\(\Rightarrow x\ge0\)
Vậy nghiệm của BPT là \(0\le x\le1\)
b/ \(\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{\left(x-1\right)\left(x-3\right)}\ge2\sqrt{\left(x-1\right)\left(x-4\right)}\)
- Với \(x=1\) thỏa mãn
- Với \(x\ge4\Leftrightarrow\sqrt{x-2}+\sqrt{x-3}\ge2\sqrt{x-4}\)
\(\Leftrightarrow\sqrt{x-2}-\sqrt{x-4}+\sqrt{x-3}-\sqrt{x-4}\ge0\)
\(\Leftrightarrow\frac{2}{\sqrt{x-2}+\sqrt{x-4}}+\frac{1}{\sqrt{x-3}+\sqrt{x-4}}\ge0\) (luôn đúng)
- Với \(x< 1\Rightarrow\sqrt{2-x}+\sqrt{3-x}\ge2\sqrt{4-x}\)
Tương tự bên trên ta có BPT luôn sai
Vậy nghiệm của BPT đã cho là \(\left[{}\begin{matrix}x=1\\x\ge4\end{matrix}\right.\)
ĐKXĐ: \(x>\dfrac{1}{5}\)
\(1-3x^2< \left(x+2\right)\sqrt[]{5x-1}+5x-1\)
\(\Leftrightarrow3x^2+5x-2+\left(x+2\right)\sqrt{5x-1}\ge0\)
\(\Leftrightarrow\left(x+2\right)\left(3x-1\right)+\left(x+2\right)\sqrt{5x-1}>0\)
\(\Leftrightarrow\left(x+2\right)\left(3x-1+\sqrt{5x-1}\right)>0\)
\(\Leftrightarrow3x-1+\sqrt{5x-1}>0\)
\(\Leftrightarrow\sqrt{5x-1}>1-3x\)
TH1: \(\left\{{}\begin{matrix}x\ge\dfrac{1}{5}\\1-3x< 0\end{matrix}\right.\) \(\Leftrightarrow x>\dfrac{1}{3}\)
TH2: \(\left\{{}\begin{matrix}x\le\dfrac{1}{3}\\5x-1>9x^2-6x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{3}\\9x^2-11x+2< 0\end{matrix}\right.\) \(\Rightarrow\dfrac{2}{9}< x\le\dfrac{1}{3}\)
Kết luận: \(x>\dfrac{2}{9}\)