Tìm min của: \(x+\sqrt{2x-5}\) với \(x\ge\dfrac{5}{2}\)
Giúp mk vs ạ mk xin cảm ơn.
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11 tháng 1 2022
a: =>2x>-6
hay x>-3
e: =>(5-x)/x<0
=>0<x<5
h: \(\Leftrightarrow\dfrac{x+5-x-3}{x+3}< 0\)
\(\Leftrightarrow x+3< 0\)
hay x<-3
g: \(\Leftrightarrow\dfrac{2x+7}{x+4}>0\)
\(\Leftrightarrow\left[{}\begin{matrix}x>-\dfrac{7}{2}\\x< -4\end{matrix}\right.\)
MM
1
4 tháng 4 2019
\(\sqrt{2x+1}-\sqrt{5-x}+x-6=0\)
\(\Leftrightarrow\left(\sqrt{2x+1}-3\right)+\left(1-\sqrt{5-x}\right)+x-4=0\)
\(\Leftrightarrow\frac{2\left(x-4\right)}{\sqrt{2x+1}+3}+\frac{x-4}{\sqrt{5-x}+1}+x-4=0\)
\(\Leftrightarrow\left(x-4\right)\left(\frac{2}{\sqrt{2x+1}+3}+\frac{1}{\sqrt{5-x}+1}+1\right)=0\)
\(\Leftrightarrow x=4\)
17 tháng 1 2023
\(1,\dfrac{4x-3}{x-5}=\dfrac{29}{3}\left(ĐKXĐ:x\ne5\right)\)
\(\Rightarrow3\left(4x-3\right)=29\left(x-5\right)\)
\(\Leftrightarrow12x-9=29x-145\)
\(\Leftrightarrow12x-9-29x+145=0\)
\(\Leftrightarrow-17x+136=0\)
\(\Leftrightarrow-17x=-136\)
\(\Leftrightarrow x=8\left(tm\right)\)
Vậy \(S=\left\{8\right\}\)
\(2,\dfrac{2x-1}{5-3x}=2\left(ĐKXĐ:x\ne\dfrac{5}{3}\right)\)
\(\Rightarrow2x-1=2\left(5-3x\right)\)
\(\Leftrightarrow2x-1=10-6x\)
\(\Leftrightarrow2x-1-10+6x=0\)
\(\Leftrightarrow8x-11=0\)
\(\Leftrightarrow8x=11\)
\(\Leftrightarrow x=\dfrac{11}{8}\left(tm\right)\)
Vậy \(S=\left\{\dfrac{11}{8}\right\}\)
\(3,\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\left(ĐKXĐ:x\ne1\right)\)
\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{2\left(x-1\right)}{x-1}+\dfrac{x}{x-1}\)
\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{2x-2}{x-1}+\dfrac{x}{x-1}\)
\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{3x-2}{x-1}\)
\(\Rightarrow4x-5=3x-2\)
\(\Leftrightarrow4x-5-3x+2=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\left(tm\right)\)
Vậy \(S=\left\{3\right\}\)
\(4,\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\left(ĐKXĐ:x\ne\dfrac{1}{2};x\ne-5\right)\)
\(\Leftrightarrow\dfrac{\left(2x+5\right)\left(x+5\right)}{2x\left(x+5\right)}-\dfrac{2x^2}{2x\left(x+5\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2+15x+25}{2x\left(x+5\right)}-\dfrac{2x^2}{2x\left(x+5\right)}=0\)
\(\Leftrightarrow\dfrac{15x+25}{2x\left(x+5\right)}=0\)
\(\Rightarrow15x+25=0\)
\(\Leftrightarrow15x=-25\)
\(\Leftrightarrow x=\dfrac{-5}{3}\left(tm\right)\)
Vậy \(S=\left\{\dfrac{-5}{3}\right\}\)
Y
17 tháng 1 2023
\(1,\dfrac{4x-3}{x-5}=\dfrac{29}{3}\)
\(\Leftrightarrow\dfrac{3\left(4x-3\right)-29\left(x-5\right)}{3\left(x-5\right)}=0\)
\(\Leftrightarrow12x-9-29x+145=0\)
\(\Leftrightarrow-17x=-136\)
\(\Leftrightarrow x=8\)
\(2,\dfrac{2x-1}{5-3x}=2\)
\(\Leftrightarrow\dfrac{2x-1-2\left(5-3x\right)}{5-3x}=0\)
\(\Leftrightarrow2x-1-10+6x=0\)
\(\Leftrightarrow8x=11\)
\(\Leftrightarrow x=\dfrac{11}{8}\)
\(3,\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\)
\(\Leftrightarrow\dfrac{4x-5-2\left(x-1-x\right)}{x-1}=0\)
\(\Leftrightarrow4x-5-2x+2+2x=0\)
\(\Leftrightarrow4x=3\)
\(\Leftrightarrow x=\dfrac{3}{4}\)
\(4,\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\)
\(\Leftrightarrow\dfrac{\left(2x+5\right)\left(x+5\right)-2x^2}{2x\left(x+5\right)}=0\)
\(\Leftrightarrow2x^2+10x+5x+25-2x^2=0\)
\(\Leftrightarrow15x=-25\)
\(\Leftrightarrow x=-\dfrac{5}{3}\)
11 tháng 6 2021
`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:
`=>` \(\left[ \begin{array}{l}x \ge 6\\\dfrac52 \le x \le 3\end{array} \right.\)
20 tháng 11 2022
\(B=\dfrac{xy}{xy}+\dfrac{\left(x-y\right)x}{x\left(x-y\right)}-\dfrac{y\left(x-y\right)}{y\left(x-y\right)}=1\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 3 2021
ĐKXĐ: \(-2\le x\le3\)
\(\dfrac{\sqrt{-x^2+x+6}}{2x+5}-\dfrac{\sqrt{-x^2+x+6}}{x-4}\ge0\)
\(\Leftrightarrow\sqrt{-x^2+x+6}\left(\dfrac{1}{2x+5}-\dfrac{1}{x-4}\right)\ge0\)
\(\Leftrightarrow\dfrac{\left(-x-9\right)\sqrt{x^2+x+6}}{\left(2x+5\right)\left(x-4\right)}\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x^2+x+6=0\\\dfrac{-x-9}{\left(2x+5\right)\left(x-4\right)}\ge0\end{matrix}\right.\) \(\Leftrightarrow-2\le x\le3\)
Hoặc có thể biện luận như sau:
Ta có: \(\left\{{}\begin{matrix}2x+5>0;\forall x\in\left[-2;3\right]\\x-4< 0;\forall x\in\left[-2;3\right]\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{-x^2+x+6}}{2x+5}\ge0\\\dfrac{\sqrt{-x^2+x+6}}{x-4}\le0\end{matrix}\right.\) ; \(\forall x\in\left[-2;3\right]\)
Do đó nghiệm của BPT là \(-2\le x\le3\)
Với \(x\ge\dfrac{5}{2}\)có: \(A=x+\sqrt{2x-5}\ge\dfrac{5}{2}+0=\dfrac{5}{2}\)
Dấu '=' xảy ra \(\Leftrightarrow x=\dfrac{5}{2}\)
\(\Rightarrow A_{min}=\dfrac{5}{2}\)
đúng như mk dự đoán chớ mk thủ hết cách rk mà có dc à