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ĐKXĐ: \(x\ge3\)
\(\sqrt{x-1}>\sqrt{x-2}+\sqrt{x-3}\)
\(\Leftrightarrow x-1>2x-5+2\sqrt{x^2-5x+6}\)
\(\Leftrightarrow4-x>2\sqrt{x^2-5x+6}\)
\(\Leftrightarrow\left\{{}\begin{matrix}4-x\ge0\\\left(4-x\right)^2>4\left(x^2-5x+6\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le4\\3x^2-12x+8< 0\end{matrix}\right.\)
\(\Rightarrow\dfrac{6-2\sqrt{3}}{3}< x< \dfrac{6+2\sqrt{3}}{3}\)
Kết hợp ĐKXĐ \(\Rightarrow3\le x< \dfrac{6+2\sqrt{3}}{3}\)
ĐKXĐ: \(x\ge2\)
Khi đó ta có \(x^2-x+1\ge3\Rightarrow1-2\sqrt{x^2-x+1}< 0\)
Do đó BPT tương đương:
\(\sqrt{2\left(x^2+7x+3\right)}-\sqrt{x^2+x-6}-3\sqrt{x+1}\le0\)
\(\Leftrightarrow\sqrt{2x^2+14x+6}\le\sqrt{x^2+x-6}+3\sqrt{x+1}\)
\(\Leftrightarrow2x^2+14x+6\le x^2+10x+3+6\sqrt{\left(x+1\right)\left(x^2+x-6\right)}\)
\(\Leftrightarrow x^2+4x+3\le6\sqrt{\left(x+1\right)\left(x+3\right)\left(x-2\right)}\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)\le6\sqrt{\left(x+1\right)\left(x+3\right)\left(x-2\right)}\)
\(\Leftrightarrow\sqrt{\left(x+1\right)\left(x+3\right)}\le6\sqrt{x-2}\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)\le36\left(x-2\right)\)
\(\Leftrightarrow x^2-32x+75\le0\)
\(\Rightarrow16-\sqrt{181}\le x\le16+\sqrt{181}\)
`sqrt{x-2}-2>=sqrt{2x-5}-sqrt{x+1}`
`đk:x>=5/2`
`bpt<=>\sqrt{x-2}+\sqrt{x+1}>=\sqrt{2x-5}+2`
`<=>x-2+x+1+2\sqrt{(x-2)(x+1)}>=2x-5+4+4\sqrt{2x-5}`
`<=>2x-1+2\sqrt{(x-2)(x+1)}>=2x-1+4\sqrt{2x-5}`
`<=>2\sqrt{(x-2)(x+1)}>=4\sqrt{2x-5}`
`<=>sqrt{x^2-x-2}>=2sqrt{2x-5}`
`<=>x^2-x-2>=4(2x-5)`
`<=>x^2-x-2>=8x-20`
`<=>x^2-9x+18>=0`
`<=>(x-3)(x-6)>=0`
`<=>` \(\left[ \begin{array}{l}x \ge 6\\x \le 3\end{array} \right.\)
Kết hợp đkxđ:
`=>` \(\left[ \begin{array}{l}x \ge 6\\\dfrac52 \le x \le 3\end{array} \right.\)
2/ \(\left[{}\begin{matrix}x< -12\\x>12\end{matrix}\right.\)
- Với \(x< -12\Rightarrow x+\frac{12x}{\sqrt{x^2-144}}=x\left(1+\frac{12}{\sqrt{x^2-144}}\right)< 0< 35\)
\(\Rightarrow\) BPT luôn đúng
- Với \(x>12\), hai vế không âm, bình phương hai vế ta được:
\(x^2+\frac{144x^2}{x^2-144}+24\frac{x^2}{\sqrt{x^2-144}}-1225\le0\)
\(\Leftrightarrow\frac{x^4}{x^2-144}+24\frac{x^2}{\sqrt{x^2-144}}-1225\le0\)
\(\Leftrightarrow\left(\frac{x^2}{\sqrt{x^2-144}}+49\right)\left(\frac{x^2}{\sqrt{x^2-144}}-25\right)\le0\)
\(\Leftrightarrow\frac{x^2}{\sqrt{x^2-144}}-25\le0\)
\(\Leftrightarrow x^2\le25\sqrt{x^2-144}\)
\(\Leftrightarrow x^4-625x^2+90000\le0\)
\(\Leftrightarrow\left(x^2-400\right)\left(x^2-225\right)\le0\)
\(\Leftrightarrow225\le x^2\le400\)
\(\Leftrightarrow15\le x\le20\)
Vậy nghiệm của BPT là \(\left[{}\begin{matrix}x< -12\\15\le x\le20\end{matrix}\right.\)
giải bpt
a) \(x^2-3x-\sqrt{x^2-3x+5}>1\)
b) \(\sqrt[4]{x-\sqrt{x^2-1}}+4\sqrt{x+\sqrt{x^2-1}}-3< 0\)
a/ Đặt \(\sqrt{x^2-3x+5}=t>0\)
\(\Leftrightarrow t^2-5-t>1\Leftrightarrow t^2-t-6>0\)
\(\Rightarrow\left[{}\begin{matrix}t>3\\t< -2\left(l\right)\end{matrix}\right.\) \(\Rightarrow\sqrt{x^2-3x+5}>3\)
\(\Leftrightarrow x^2-3x+5>9\Leftrightarrow x^2-3x-4>0\Rightarrow\left[{}\begin{matrix}x>4\\x< -1\end{matrix}\right.\)
b/ ĐKXĐ: \(x\ge1\)
Đặt \(\sqrt[4]{x-\sqrt{x^2-1}}=t>0\Rightarrow\sqrt[4]{x+\sqrt{x^2-1}}=\frac{1}{t}\)
\(\Leftrightarrow t+\frac{4}{t^2}-3< 0\)
\(\Leftrightarrow t^3-3t^2+4< 0\)
\(\Leftrightarrow\left(t+1\right)\left(t-2\right)^2< 0\)
Do \(t>0\Rightarrow t+1>0\Rightarrow VT\ge0\Rightarrow\) BPT vô nghiệm
ĐK: \(x\ge1;x\le-2\)
\(\sqrt{x^2-1}+\sqrt{x^2-x}\le\sqrt{x^2+x-2}\)
\(\Leftrightarrow2x^2-x-1+2\sqrt{\left(x^2-1\right)\left(x^2-x\right)}\le x^2+x-2\)
\(\Leftrightarrow x^2-2x+1+2\sqrt{\left(x^2-1\right)\left(x^2-x\right)}\le0\)
\(\Leftrightarrow\left(x-1\right)^2+2\sqrt{\left(x^2-1\right)\left(x^2-x\right)}\le0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\\left(x^2-1\right)\left(x^2-x\right)=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\left(tm\right)\)
Vậy bất phương trình có nghiệm \(x=1\)