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vì \(\left(2^x+\dfrac{1}{3}\right)^4\) có mũ chẵn là 4 +> \(\left(2^x+\dfrac{1}{3}\right)^4\) > hoặc bằng 0 . Vậy GTNN của \(\left(2^x+\dfrac{1}{3}\right)^4\)= 0 .
vi GTNN cua \(\left(2^x+\dfrac{1}{3}\right)^4\)=> \(\left(2^x+\dfrac{1}{3}\right)^4\)-1 =0 -1=-1
vay GTNN cua \(\left(2^x+\dfrac{1}{3}\right)^4\)-1 =-1
b, vi \(\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^{2018}\) co mu chan la 2018 => \(\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^{2018}\) . hoặc bằng 0
Vậy GTLN của \(\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^{2018}\) = 0 .Vì \(\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^{2018}\) = 0 =>
\(\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^{2018}\) +3=0+3=3
Vậy GTLN của \(\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^{2018}\)+3=3
\(A=\left|2021-x\right|+\dfrac{1}{2}\left|4040-2x\right|\)
\(A=\left|2021-x\right|+\left|2020-x\right|\)
\(A=\left|2021-x\right|+\left|x-2020\right|\ge\left|2021-x+x-2020\right|=1\)
\(A_{min}=1\) khi \(2020\le x\le2021\)
\(A=\left(\dfrac{2020}{2021}xy^5z\right).\left(\dfrac{2020}{2021}x^3yz^2\right).\left(-\dfrac{2020}{2021}\right)^0\)
\(a)A=\dfrac{2020.2021.2020}{2021.2020.2021}.\left(x.x^3\right).\left(y^5.y\right).\left(z.z^2\right)\Leftrightarrow A=\dfrac{2020}{2021}x^4.y^6.z^3\)
\(b)A=\dfrac{2020}{2021}x^4.y^6.z^3\)
\(\Rightarrow\text{A có hệ số là:}\dfrac{2020}{2021}\)
\(\text{Phần biến là:}\left(x,y,z\right)\)
\(c)\text{Xét A ta có:}\dfrac{2020}{2021}< 0;x^4,y^6\text{ luôn }< 0\)
\(\Rightarrow\dfrac{2020}{2021}x^4.y^6>0\Rightarrow\text{ Nếu }z< 0\Rightarrow A\le0\text{ và z có số mũ là:3}\)
\(\text{Chẳng hạn:}\left(-\right).\left(-\right).\left(-\right)=\left(-\right).< 0\Rightarrow z\text{ phải }\ge0\text{ thì }A\ge0\)
\(\Rightarrow Z\in N\)
3 câu này bạn áp dụng cái này nhé.
`a^2 >=0 forall a`.
`|a| >=0 forall a`.
`1/a` xác định `<=> a ne 0`.
a: P=(x+30)^2+(y-4)^2+1975>=1975 với mọi x,y
Dấu = xảy ra khi x=-30 và y=4
b: Q=(3x+1)^2+|2y-1/3|+căn 5>=căn 5 với mọi x,y
Dấu = xảy ra khi x=-1/3 và y=1/6
c: -x^2-x+1=-(x^2+x-1)
=-(x^2+x+1/4-5/4)
=-(x+1/2)^2+5/4<=5/4
=>R>=3:5/4=12/5
Dấu = xảy ra khi x=-1/2
a)
`(2x-1)(x+2/3)=0`
\(< =>\left[{}\begin{matrix}2x-1=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
b)
\(\dfrac{x+4}{2019}+\dfrac{x+3}{2020}=\dfrac{x+2}{2021}+\dfrac{x+1}{2022}\)
\(< =>\dfrac{x+4}{2019}+1+\dfrac{x+3}{2020}+1=\dfrac{x+2}{2021}+1+\dfrac{x+1}{2022}+1\)
\(< =>\dfrac{x+2023}{2019}+\dfrac{x+2023}{2020}=\dfrac{x+2023}{2021}+\dfrac{x+2023}{2022}\)
\(< =>\left(x+2023\right)\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\right)=0\)
\(< =>x+2023=0\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\ne0\right)\\ < =>x=-2023\)
a, Với mọi giá trị của x;y ta có:
\(\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2\ge0\)
\(\Rightarrow\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2-10\ge-10\)
Hay \(C\ge-10\)với mọi giá trị của x;y
Để \(C=-10\) thì \(\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2-10=-10\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x+1\right)^2=0\\\left(y-\dfrac{1}{3}\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy................
b, Với mọi giá trị của x ta có:
\(\left(2x-1\right)^2+3\ge3\Rightarrow\dfrac{5}{\left(2x-1\right)^2+3}\ge\dfrac{5}{3}\)
Hay \(D\ge\dfrac{5}{3}\) với mọi giá trị của x.
Để \(D=\dfrac{5}{3}\) thì \(\dfrac{5}{\left(2x-1\right)^2+3}=\dfrac{5}{3}\)
\(\Rightarrow\left(2x-1\right)^2=0\Rightarrow x=\dfrac{1}{2}\)
Vậy..................
Chúc bạn học tốt!!!
\(C=\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2-10\)
\(\left(x+1\right)^2\ge0;\left(y-\dfrac{1}{3}\right)^2\ge0\)
\(C_{MIN}\Rightarrow\left(x+1\right)^2_{MIN};\left(y-\dfrac{1}{3}\right)^2_{MIN}\)
\(\left(x+1\right)^2_{MIN}=0;\left(y-\dfrac{1}{3}\right)^2_{MIN}=0\)
\(\Rightarrow C_{MIN}=0+0-10=-10\)
\(D=\dfrac{5}{\left(2x-1\right)^2+3}\)
\(D_{MAX}\Rightarrow\left(2x-1\right)^2+3_{MIN}\)
\(\left(2x-1\right)^2\ge0\)
\(\left(2x-1\right)^2+3_{MIN}\Rightarrow\left(2x-1\right)^2_{MIN}=0\)
\(\Rightarrow\left(2x-1\right)^2+3_{MIN}=0+3=3\)
\(\Rightarrow D_{MAX}=\dfrac{5}{3}\)
\(A=\left(x+\dfrac{2}{3}\right)^2+\dfrac{1}{2}\)
\(\left(x+\dfrac{2}{3}\right)^2\ge0\forall x\in R\)
\(A=\left(x+\dfrac{2}{3}\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}\)
Dấu "=" xảy ra khi:
\(\left(x+\dfrac{2}{3}\right)^2=0\Rightarrow x=-\dfrac{2}{3}\)
\(B=\dfrac{2}{\left(x-\dfrac{1}{2}\right)^2+2}\)
\(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\in R\)
\(\left(x-\dfrac{1}{2}\right)^2+2\ge2\)
\(B=\dfrac{2}{\left(x-\dfrac{1}{2}\right)^2+2}\le1\)
Dấu "=" xảy ra khi:
\(\left(x-\dfrac{1}{2}\right)^2=0\Rightarrow x=\dfrac{1}{2}\)
