Sử dụng BPT tích
\(\frac{\sqrt{x-1}+6-3x}{\sqrt{x-1}+3-x}\ge\frac{1}{2}\)
Những câu hỏi liên quan
Bài 1. Tìm điều kiện các BPT sau
a, sqrt{20-x}sqrt{3x-6}+1
b, frac{sqrt{9-x^2}}{x-1}frac{1}{sqrt{x}}+1
c, x+frac{x+1}{sqrt{x-4}}2-frac{2}{x^2-25}
d, sqrt{x}sqrt{-x}
e, 3x+frac{4}{sqrt{x-5}}le9+frac{x}{x-6}
f, frac{x+2}{10+3x^2}ge7+frac{4}{left(3x+9right)^2}
g, frac{sqrt{x+2}}{sqrt{x-2}}+frac{1}{left(x-4right)left(x+6right)}lefrac{3}{sqrt{8-x}}
h, frac{sqrt{x+6}}{left|xright|-sqrt{x+6}}gesqrt{16-2x}
Đọc tiếp
Bài 1. Tìm điều kiện các BPT sau
a, \(\sqrt{20-x}>\sqrt{3x-6}+1\)
b, \(\frac{\sqrt{9-x^2}}{x-1}>\frac{1}{\sqrt{x}}+1\)
c, \(x+\frac{x+1}{\sqrt{x-4}}>2-\frac{2}{x^2-25}\)
d, \(\sqrt{x}>\sqrt{-x}\)
e, \(3x+\frac{4}{\sqrt{x-5}}\le9+\frac{x}{x-6}\)
f, \(\frac{x+2}{10+3x^2}\ge7+\frac{4}{\left(3x+9\right)^2}\)
g, \(\frac{\sqrt{x+2}}{\sqrt{x-2}}+\frac{1}{\left(x-4\right)\left(x+6\right)}\le\frac{3}{\sqrt{8-x}}\)
h, \(\frac{\sqrt{x+6}}{\left|x\right|-\sqrt{x+6}}\ge\sqrt{16-2x}\)
giải bpt
\(\left(\sqrt{x+4}-1\right)\sqrt{x+2}\ge\frac{x^3+4x^2+3x-2\left(x+3\right)\sqrt[3]{2x+3}}{\left(\sqrt[3]{2x+3}-3\right)\left(\sqrt{x+4}+1\right)}\)
3 tháng 6 2020 lúc 22:26
giải bpt
\(\frac{\sqrt{x-3}}{\sqrt{2x-1}-1}\ge\frac{1}{\sqrt{x+3}-\sqrt{x-3}}\)
ĐKXĐ: \(x\ge3\)
Khi đó \(\sqrt{2x-1}\ge\sqrt{5}>1\Rightarrow\sqrt{2x-1}-1>0\)
Đồng thời \(\sqrt{x+3}>\sqrt{x-3}\) \(\forall x\Rightarrow\sqrt{x+3}-\sqrt{x-3}>0\)
Do đó BPT tương đương:
\(\sqrt{x-3}\left(\sqrt{x+3}-\sqrt{x-3}\right)\ge\sqrt{2x-1}-1\)
\(\Leftrightarrow\sqrt{x^2-9}-x+3\ge\sqrt{2x-1}-1\)
\(\Leftrightarrow\sqrt{x^2-9}\ge x-4+\sqrt{2x-1}\)
Do \(x-4+\sqrt{2x-1}\ge3-4+\sqrt{5}>0;\forall x\ge3\) nên BPT tương đương:
\(x^2-9\ge x^2-8x+16+2x-1+2\left(x-4\right)\sqrt{2x-1}\)
\(\Leftrightarrow\left(x-4\right)\sqrt{2x-1}-3\left(x-4\right)\le0\)
\(\Leftrightarrow\left(x-4\right)\left(\sqrt{2x-1}-3\right)\le0\)
\(\Leftrightarrow\left(x-4\right)\left(\frac{2x-1-9}{\sqrt{2x-1}+3}\right)\le0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)\le0\Leftrightarrow4\le x\le5\)
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3 tháng 4 2020 lúc 14:58
Giải bpt
a) \(\frac{3}{\sqrt{x-2}-1}\ge\frac{5}{\sqrt{x-2}-3}\)
b) \(x\sqrt{x-3}-\frac{\sqrt{x-3}}{2-x}\le0\)
c) \(\frac{2\sqrt{x-1}-4}{\sqrt{4-x^2}-1}\ge2-\sqrt{x-1}\)
a/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ge2\\x\ne\left\{3;11\right\}\end{matrix}\right.\)
Đặt \(\sqrt{x-2}=t\ge0\)
\(\Rightarrow\frac{3}{t-1}\ge\frac{5}{t-3}\)
\(\Leftrightarrow\frac{3}{t-1}-\frac{5}{t-3}\ge0\)
\(\Leftrightarrow\frac{3t-9-5t+5}{\left(t-1\right)\left(t-3\right)}\ge0\)
\(\Leftrightarrow\frac{-2t-4}{\left(t-1\right)\left(t-3\right)}\ge0\)
\(\Leftrightarrow\frac{t+2}{\left(t-1\right)\left(t-3\right)}\le0\)
\(\Leftrightarrow1< t< 3\)
\(\Rightarrow1< \sqrt{x-2}< 3\)
\(\Leftrightarrow1< x-2< 9\Rightarrow3< x< 11\)
b/
ĐKXĐ: \(x\ge3\)
- Với \(x=3\) BPT thỏa mãn
- Với \(x>3\Rightarrow\sqrt{x-3}>0\) BPT tương đương
\(x-\frac{1}{2-x}\le0\Leftrightarrow x+\frac{1}{x-2}\le0\)
\(\Leftrightarrow\frac{x^2-2x+1}{x-2}\le0\)
\(\Leftrightarrow\frac{\left(x-1\right)^2}{x-2}\le0\Rightarrow\) không tồn tại x thỏa mãn
Vậy BPT có nghiệm duy nhất \(x=3\)
c/
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge1\\4-x^2\ge0\\\sqrt{4-x^2}\ne1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\-2\le x\le2\\x\ne\pm\sqrt{3}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}1\le x\le2\\x\ne\sqrt{3}\end{matrix}\right.\)
BPT tương đương:
\(\frac{2\left(\sqrt{x-1}-2\right)}{\sqrt{4-x^2}-1}+\sqrt{x-1}-2\ge0\)
\(\Leftrightarrow\left(\sqrt{x-1}-2\right)\left(\frac{2}{\sqrt{4-x^2}-1}+1\right)\ge0\)
Do \(x\le2\Rightarrow\sqrt{x-1}\le1\Rightarrow\sqrt{x-1}-2< 0\)
BPt tương đương:
\(\frac{2}{\sqrt{4-x^2}-1}+1\le0\)
\(\Leftrightarrow\frac{1+\sqrt{4-x^2}}{\sqrt{4-x^2}-1}\le0\)
\(\Leftrightarrow\sqrt{4-x^2}-1< 0\) (do \(1+\sqrt{4-x^2}>0\) \(\forall x\))
\(\Leftrightarrow\sqrt{4-x^2}< 1\Leftrightarrow x^2>3\Rightarrow x>\sqrt{3}\)
Vậy nghiệm của BPT đã cho là: \(\sqrt{3}< x\le2\)
giải bpt
1.\(\frac{1}{x+2}\ge\frac{x+2}{3x-5}\)
2.\(\sqrt{-x^2+4x-3}\le x-1\)
help!
\(\frac{1}{x+2}-\frac{x+2}{3x-5}\ge0\)
\(\Leftrightarrow\frac{-x^2-x-9}{\left(x+2\right)\left(3x-5\right)}\ge0\)
\(\Leftrightarrow\left(x+2\right)\left(3x-5\right)< 0\) (do \(-x^2-x-9< 0;\forall x\))
\(\Rightarrow-2< x< \frac{5}{3}\)
2/ ĐKXĐ: \(1\le x\le3\)
\(\Leftrightarrow-x^2+4x-3\le\left(x-1\right)^2\)
\(\Leftrightarrow2x^2-6x+4\ge0\Rightarrow\left[{}\begin{matrix}x\ge2\\x\le1\end{matrix}\right.\)
Kết hợp ĐKXĐ: \(\left[{}\begin{matrix}x=1\\2\le x\le3\end{matrix}\right.\)
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Giải bpt
\(\frac{x+2}{\sqrt{2x+3}-\sqrt{x+1}}\ge\sqrt{2x^2+5x+3}+1\)
giải bpt:
1. \(\frac{\sqrt{-3x^2+x+4}+2}{x}< 2\)
2. \(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}\ge2\sqrt{x^2-5x+4}\)
3. \(\sqrt{x^2-8x+15}+\sqrt{x^2+2x-15}\le\sqrt{4x^2-18x=18}\)
4. 4(x+1)2 \(\ge\) (2x +10)( 1- \(\sqrt{3+2x}\))2
5. \(\sqrt{1+x}-\sqrt{1-x}\ge x\)
giải BPT :
a. \(\sqrt[3]{x+6}+\sqrt{x-1}\ge x^2-1\)
b.2\(\sqrt[3]{x+4}+\sqrt{2x+7}+x^2+8x+13\)
c.\(4x^3+5x^2+1\ge\sqrt{3x+1}-3x\)
giúp với ạ
1) \(\frac{\sqrt{2\left(X^2-16\right)}}{\sqrt{X-3}}+\sqrt{X-3}>\frac{7-X}{\sqrt{X-3}}\)
2) \(\frac{1}{\sqrt{2X^2+3X-5}}\ge\frac{1}{2X-1}\)
3) \(\frac{1-\sqrt{1-4X^2}}{X}< 3\)
4) \(\frac{\sqrt{3X+1}-X}{2X-1}< 1\)