Tìm max \(12\sqrt{6-x}+5\sqrt{x-5}\) với \(6\ge x\ge5\\ \)
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27 tháng 8 2021
`sqrt{4x+20}-3sqrt{5+x}+4/3sqrt{9x+15}=6(x>=-5)`
`<=>sqrt{4(x+5)}-3sqrt{x+5}+4/3sqrt{9(x+5)}=6`
`<=>2sqrt{x+5}-3sqrt{x+5}+4sqrt{x+5}=6`
`<=>3sqrt{x+5}=6`
`<=>sqrt{x+5}=2`
`<=>x+5=4`
`<=>x=-1(tm)`
Vậy `x=-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 \(-2\le x\le3\)
LD
19 tháng 6 2019
a.
\(A=\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{9}}\)
\(=\frac{\sqrt{3}-\sqrt{1}}{3-1}+\frac{\sqrt{5}-\sqrt{3}}{5-3}+\frac{\sqrt{7}-\sqrt{5}}{7-5}+\frac{\sqrt{9}-\sqrt{7}}{9-7}\)
\(=\frac{\sqrt{9}-\sqrt{7}+\sqrt{7}-\sqrt{5}+\sqrt{5}-\sqrt{3}+\sqrt{3}-\sqrt{1}}{2}\)
\(=\frac{3-1}{2}=1\)
b.
\(B=2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
\(=2\sqrt{80\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
\(=8\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}=0\)
c.
\(C=\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}-\sqrt{6}\)
\(=\frac{15\sqrt{6}-15}{6-1}+\frac{4\sqrt{6}+8}{6-4}-\frac{36+12\sqrt{6}}{9-6}-\sqrt{6}\)
\(=\frac{15\sqrt{6}-15}{5}+\frac{4\sqrt{6}+8}{2}-\frac{36+12\sqrt{6}}{3}-\sqrt{6}\)
\(=3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}-\sqrt{6}\)
\(=-11\)
20 tháng 8 2019
d)D=\(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\)( \(x\ge2\))
=\(\sqrt{x+2\sqrt{2}.\sqrt{x-2}}+\sqrt{x-2\sqrt{2}.\sqrt{x-2}}\)
=\(\sqrt{\left(x-2\right)+2\sqrt{2}.\sqrt{x-2}+2}+\sqrt{\left(x-2\right)-2\sqrt{2}.\sqrt{x-2}+2}\)
=\(\sqrt{\left(\sqrt{x-2}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{x-2}-\sqrt{2}\right)^2}\)
=\(\sqrt{x-2}+\sqrt{2}+\left|\sqrt{x-2}-\sqrt{2}\right|\)(1)
TH1: \(2\le x\le4\)
Từ (1)<=> \(\sqrt{x-2}+\sqrt{2}-\sqrt{x-2}+\sqrt{2}\)
=\(2\sqrt{2}\)
TH2. x\(>4\)
Từ (1) <=> \(\sqrt{x-2}+\sqrt{2}-\sqrt{2}+\sqrt{x-2}\)=\(2\sqrt{x-2}\)
Vậy \(\left[{}\begin{matrix}2\le x\le4\\x>4\end{matrix}\right.< =>\left[{}\begin{matrix}D=2\sqrt{2}\\D=2\sqrt{x-2}\end{matrix}\right.\)
TC
27 tháng 8 2017
Gọi 1/4 số a là 0,25 . Ta có :
a . 3 - a . 0,25 = 147,07
a . (3 - 0,25) = 147,07 ( 1 số nhân 1 hiệu )
a . 2,75 = 147,07
a = 147,07 : 2,75
a = 53,48
TN
28 tháng 8 2017
Ta c/m BĐT mạnh hơn \(\frac{1}{x^5-x^2+3xy+6}+\frac{1}{y^5-y^2+3yz+6}+\frac{1}{z^5-z^2+3zx+6}\le\frac{1}{3}\)
Áp dụng BĐT AM-GM ta có:
\(x^5+x+1\ge3x^2\)và \(2x^2+2\ge4x\)
\(\Rightarrow x^5-x^2+6\ge3x+3\)
\(\Rightarrow\frac{1}{x^5-x^2+3xy+6}\le\frac{1}{3(x+xy+1)}\)
\(P\le\frac{1}{3(x+xy+1)}+\frac{1}{3(y+yz+1)}+\frac{1}{3(z+zx+1)}=\frac{1}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 8 2021
\(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}+1\right)=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}+1\right)\)
\(=\left|\sqrt{3}-1\right|\left(\sqrt{3}+1\right)=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=3-1=2\)
\(\dfrac{x-25}{\sqrt{x}-5}-\dfrac{x+4\sqrt{x}+4}{\sqrt{x}+2}=\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\sqrt{x}-5}-\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}+2}\)
\(=\sqrt{x}+5-\left(\sqrt{x}+2\right)=5-2=3\)
23 tháng 8 2021
a: Ta có: \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}+\sqrt{2}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
=3-1
=2
b: Ta có: \(\dfrac{x-25}{\sqrt{x}-5}-\dfrac{x+4\sqrt{x}+4}{\sqrt{x}+2}\)
\(=\sqrt{x}+5-\sqrt{x}-2\)
=3
16 tháng 5 2021
`A=(2\sqrtx-9)(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)(3-sqrtx)(x>=0,x ne 4, x ne 9)`
`=(2\sqrtx-9)(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)(sqrtx-3)`
`=(2sqrtx-9-x+9+2x-3sqrtx-2)/(x-5sqrtx+6)`
`=(x-sqrtx-2)/(x-5sqrtx+6)`
`=((\sqrtx+1)(sqrtx-2))/((sqrtx-2)(sqrtx-3))`
`=(sqrtx+1)/(sqrtx-3)`
16 tháng 5 2021
`A=(2\sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)/(3-sqrtx)(x>=0,x ne 4, x ne 9)`
`=(2\sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)/(sqrtx-3)`
`=(2sqrtx-9-x+9+2x-3sqrtx-2)/(x-5sqrtx+6)`
`=(x-sqrtx-2)/(x-5sqrtx+6)`
`=((\sqrtx+1)(sqrtx-2))/((sqrtx-2)(sqrtx-3))`
`=(sqrtx+1)/(sqrtx-3)`
AT
12 tháng 7 2021
\(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}}{x-1}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-2\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)
\(A< \dfrac{3}{5}\Rightarrow\dfrac{3}{5}-A>0\Rightarrow\dfrac{3}{5}-\dfrac{\sqrt{x}-3}{\sqrt{x}-1}>0\)
\(\Rightarrow\dfrac{3\left(\sqrt{x}-1\right)-5\left(\sqrt{x}-3\right)}{5\left(\sqrt{x}-1\right)}>0\Rightarrow\dfrac{12-2\sqrt{x}}{5\left(\sqrt{x}-1\right)}>0\)
\(\Rightarrow\dfrac{2}{5}.\dfrac{6-\sqrt{x}}{\sqrt{x}-1}>0\Rightarrow\dfrac{6-\sqrt{x}}{\sqrt{x}-1}>0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}6-\sqrt{x}>0\\\sqrt{x}-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}6-\sqrt{x}< 0\\\sqrt{x}-1< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}1< x< 36\\\left\{{}\begin{matrix}x>36\\x< 1\end{matrix}\right.\left(l\right)\end{matrix}\right.\)
\(\Rightarrow1< x< 36\)
12 tháng 7 2021
\(=>A=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{x-2\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)
để \(A< \dfrac{3}{5}< =>\dfrac{\sqrt{x}-3}{\sqrt{x}-1}< \dfrac{3}{5}\)
\(< =>\dfrac{5\left(\sqrt{x}-3\right)-3\left(\sqrt{x}-1\right)}{5\left(\sqrt{x}-1\right)}< 0\)
\(< =>\dfrac{2\sqrt{x}-12}{5\left(\sqrt{x}-1\right)}< 0\)
\(=>\left\{{}\begin{matrix}\left[{}\begin{matrix}2\sqrt{x}-12>0\\5\left(\sqrt{x}-1\right)< 0\end{matrix}\right.\\\left[{}\begin{matrix}2\sqrt{x}-12< 0\\5\left(\sqrt{x}-1\right)>0\end{matrix}\right.\end{matrix}\right.\)\(=>\left\{{}\begin{matrix}\left[{}\begin{matrix}x>36\\x< 1\end{matrix}\right.\\\left[{}\begin{matrix}x< 36\\x>1\end{matrix}\right.\end{matrix}\right.=>1< x< 36\left(tm\right)\)
theo bđt cauchy schwarz ta có
\(12\sqrt{6-x}\le\dfrac{12^2+6-x}{2}\)
\(5\sqrt{x-5}\le\dfrac{25+x-5}{2}\)
\(\Rightarrow12\sqrt{6-x}+5\sqrt{x-5}\le\dfrac{144+6-x+25+x-5}{2}=85\)
vậy Amax=85
Max A = 85 khi x=?