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\(\left(a^2+\dfrac{b^2}{3}+\dfrac{c^2}{5}\right)\left(1+3+5\right)\ge\left(a+b+c\right)^2\)
\(\Rightarrow3\sqrt{a^2+\dfrac{b^2}{3}+\dfrac{c^2}{5}}\ge a+b+c\)
\(\Rightarrow P\ge\dfrac{2}{3}\left(a+b+c\right)+3\sqrt{\dfrac{1}{a}+\dfrac{3^2}{b}+\dfrac{5^2}{c}}\)
\(\Rightarrow P\ge\dfrac{2}{3}\left(a+b+c\right)+3\sqrt{\dfrac{\left(1+3+5\right)^2}{a+b+c}}=\dfrac{2}{3}\left(a+b+c\right)+\dfrac{27}{\sqrt{a+b+c}}\)
\(\Rightarrow P\ge\dfrac{1}{2}\left(a+b+c\right)+\dfrac{27}{2\sqrt{a+b+c}}+\dfrac{27}{2\sqrt{a+b+c}}+\dfrac{1}{6}\left(a+b+c\right)\)
\(\Rightarrow P\ge3\sqrt[3]{\dfrac{27^2\left(a+b+c\right)}{2^3\left(a+b+c\right)}}+\dfrac{1}{6}.9=15\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(1;3;5\right)\)
`#3107.101107`
a)
`2/5 \sqrt{25} - 1/2 \sqrt{4}`
`= 2/5 * \sqrt{5^2} - 1/2 * \sqrt{2^2}`
`= 2/5*5 - 1/2*2`
`= 2 - 1`
`= 1`
b)
`0,5*\sqrt{0,09} + 5*\sqrt{0,81}`
`= 0,5*\sqrt{(0,3)^2} + 5*\sqrt{(0,9)^2}`
`= 0,5*0,3 + 5*0,9`
`= 0,15 + 4,5`
`= 4,65`
c)
`2/5\sqrt{25/36} - 5/2\sqrt{4/25}`
`= 2/5*\sqrt{(5^2)/(6^2)} - 5/2*\sqrt{(2^2)/(5^2)}`
`= 2/5*5/6 - 5/2*2/5`
`= 1/3 - 1`
`= -2/3`
d)
`-2 \sqrt{(-36)/(-16)} + 5 \sqrt{(-81)/(-25)}`
`= -2*\sqrt{36/16} + 5*\sqrt{81/25}`
`= -2*\sqrt{(6^2)/(4^2)} + 5*\sqrt{(9^2)/(5^2)}`
`= -2*6/4 + 5*9/5`
`= -3 + 9`
`= 6`
`A)đk:x>=0,x ne 25`
`A=9=>A=(3+2)/(3-5)=-5/2`
`B)B=(3sqrtx-15+20-2sqrtx)/(x-25)`
`=(sqrtx+5)/(x-25)`
`=1/(sqrtx-5)`
`A=B.|x-4|`
`<=>A/B=|x-4|`
`<=>\sqrtx+2=|x-4|`
`<=>\sqrtx+2=(sqrtx+2)|sqrtx-2|`
`<=>|sqrtx-2|=1`
`+)sqrtx-2=1<=>x=9(tm)`
`+)sqrtx-2=-1<=>x=1(tm)`
Vậy `S={1,9}`
a, Thay x=9 vào biểu thức A ta có
\(A=\dfrac{\sqrt{9}+2}{\sqrt{9}-5}\)
\(A=\dfrac{3+2}{3-5}=\dfrac{5}{-2}=-2,5\)
Vậy A =-2,5 khi x=9
Nguyễn Việt Lâm DƯƠNG PHAN KHÁNH DƯƠNG Mysterious Person help
Do \(a,b,c>\dfrac{25}{4}\Rightarrow\) các mẫu số đều dương
Áp dụng BĐT Cauchy:
\(M\ge3\sqrt[3]{\dfrac{abc}{\left(2\sqrt{b}-5\right)\left(2\sqrt{c}-5\right)\left(2\sqrt{a}-5\right)}}\)
\(\Rightarrow M\ge3\sqrt[3]{\dfrac{5^3.abc}{5\left(2\sqrt{b}-5\right).5\left(2\sqrt{c}-5\right).5\left(2\sqrt{a}-5\right)}}\)
Ta có: \(\left\{{}\begin{matrix}5\left(2\sqrt{a}-5\right)\le\dfrac{\left(5+2\sqrt{a}-5\right)^2}{4}=a\\5\left(2\sqrt{b}-5\right)\le\dfrac{\left(5+2\sqrt{b}-5\right)^2}{4}=b\\5\left(2\sqrt{c}-5\right)\le\dfrac{\left(5+2\sqrt{c}-5\right)^2}{4}=c\end{matrix}\right.\)
\(\Rightarrow M\ge3\sqrt[3]{\dfrac{5^3.abc}{abc}}=3.5=15\)
\(\Rightarrow M_{min}=15\) khi \(a=b=c=25\)
Bạn áp dụng BĐT \(xy\le\dfrac{\left(x+y\right)^2}{4}\)
Dấu "=" xảy ra khi x=y
Hơn nữa, cũng áp dụng để tìm dấu "=" cuối bài, ta có \(5=2\sqrt{a}-5\Rightarrow2\sqrt{a}=10\Rightarrow a=25\), đó là lý do tại sao biết đẳng thức xảy ra tại a=b=c=25
Câu 1
a, \(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\) ( ĐKXĐ: \(x\ge0;x\ne25\))
=\(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-\dfrac{5}{\sqrt{x}+5}\)
=\(\dfrac{\sqrt{x}\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-\dfrac{10\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-\dfrac{5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
=\(\dfrac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
=\(\dfrac{x-10\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
=\(\dfrac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
=\(\dfrac{\sqrt{x}-5}{\sqrt{x}+5}\)
b, Với \(x\ge0;x\ne25\) để \(A< 0\) thì \(\sqrt{x}-5\) < 0 ( Vì \(\sqrt{x}+5\) > 0 )
<=> x < 25
Bạn tham khảo:
Cho \(a,b,c>\dfrac{25}{4}.\)Tìm GTNN của \(Q=\dfrac{a}{2\sqrt{b}-5}+\dfrac{b}{2\sqrt{c}-5}+\dfrac{c}{2\sqrt{a}-5}\) - Hoc24