Max y=9-sin2 3x-\(\sqrt{2}\)cos 3x
Max y= 3sin 3x -8 cos2 3x +4
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NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 6 2021
a.
\(y'=\dfrac{2-x}{2x^2\sqrt{x-1}}=0\Rightarrow x=2\)
\(y\left(1\right)=0\) ; \(y\left(2\right)=\dfrac{1}{2}\) ; \(y\left(5\right)=\dfrac{2}{5}\)
\(\Rightarrow y_{min}=y\left(1\right)=0\)
\(y_{max}=y\left(2\right)=\dfrac{1}{2}\)
b.
\(y'=\dfrac{1-3x}{\sqrt{\left(x^2+1\right)^3}}< 0\) ; \(\forall x\in\left[1;3\right]\Rightarrow\) hàm nghịch biến trên [1;3]
\(\Rightarrow y_{max}=y\left(1\right)=\dfrac{4}{\sqrt{2}}=2\sqrt{2}\)
\(y_{min}=y\left(3\right)=\dfrac{6}{\sqrt{10}}=\dfrac{3\sqrt{10}}{5}\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 6 2021
c.
\(y=1-cos^2x-cosx+1=-cos^2x-cosx+2\)
Đặt \(cosx=t\Rightarrow t\in\left[-1;1\right]\)
\(y=f\left(t\right)=-t^2-t+2\)
\(f'\left(t\right)=-2t-1=0\Rightarrow t=-\dfrac{1}{2}\)
\(f\left(-1\right)=2\) ; \(f\left(1\right)=0\) ; \(f\left(-\dfrac{1}{2}\right)=\dfrac{9}{4}\)
\(\Rightarrow y_{min}=0\) ; \(y_{max}=\dfrac{9}{4}\)
d.
Đặt \(sinx=t\Rightarrow t\in\left[-1;1\right]\)
\(y=f\left(t\right)=t^3-3t^2+2\Rightarrow f'\left(t\right)=3t^2-6t=0\Rightarrow\left[{}\begin{matrix}t=0\\t=2\notin\left[-1;1\right]\end{matrix}\right.\)
\(f\left(-1\right)=-2\) ; \(f\left(1\right)=0\) ; \(f\left(0\right)=2\)
\(\Rightarrow y_{min}=-2\) ; \(y_{max}=2\)
AH
Akai Haruma
Giáo viên
30 tháng 4 2023
Lời giải:
$A^2=9x^2(8-3x^2)=3.3x^2(8-3x^2)\leq 3.\left(\frac{3x^2+8-3x^2}{2}\right)^2=3.4^2$ (theo BĐT AM-GM)
$\Rightarrow A\leq 4\sqrt{3}$
Vậy $A_{\max}=4\sqrt{3}$. Giá trị này đạt tại $x=\frac{2}{\sqrt{3}}$
MY
1
24 tháng 5 2021
\(P=-3x^2-4x\sqrt{y}+16x--y+12\sqrt{y}+1999\)
\(=-2\left(x^2+2x\sqrt{y}+y\right)+12\left(x+\sqrt{y}\right)-18-x^2+4x-4+2021\)
\(=-2\left(x+\sqrt{y}\right)^2+12\left(x+\sqrt{y}\right)-18-\left(x-2\right)^2+2021\)
\(=-2\left(x+\sqrt{y}-3\right)^2-\left(x-2\right)^2+2021\)\(\le2021\) với mọi x và y không âm
Dấu = xảy ra <=> x=2 và y=1
Vậy maxP=2021
22 tháng 7 2019
1) \(\frac{1}{2}=\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)\(\Leftrightarrow\)\(x+y\ge8\)
\(\frac{1}{2}=\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}\)\(\Leftrightarrow\)\(xy=2\left(x+y\right)\ge16\)
\(A=\sqrt{x}+\sqrt{y}\ge2\sqrt[4]{xy}\ge2\sqrt[4]{16}=4\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=4\)
2) \(B=\sqrt{3x-5}+\sqrt{7-3x}\ge\sqrt{3x-5+7-3x}=\sqrt{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{7}{3}\end{cases}}\)
\(B=\sqrt{3x-5}+\sqrt{7-3x}\le\frac{3x-5+1+7-3x+1}{2}=2\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=2\)
8 tháng 8 2021
a) \(A=\sqrt{x-2}+\sqrt{6-x}\)
\(\Rightarrow A^2=x-2+6-x+2\sqrt{\left(x-2\right)\left(6-x\right)}\)
Ta có \(\sqrt{\left(x-2\right)\left(6-x\right)}\ge0,\forall x\)
Do đó \(A^2=4+2\sqrt{\left(x-2\right)\left(6-x\right)}\ge4\)
Mà A không âm \(\Leftrightarrow A\ge2\)
Dấu "=" \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
Áp dụng BĐT Bunhiacopxky:
\(A^2=\left(\sqrt{x-2}+\sqrt{6-x}\right)^2\le\left(x-2+6-x\right)\left(1+1\right)=4\cdot2=8\)
\(\Leftrightarrow A\le\sqrt{8}\)
Dấu "=" \(\Leftrightarrow x-2=6-x\Leftrightarrow x=4\)
Mấy bài còn lại y chang nha
Tick hộ nha
AD
1
VH
30 tháng 11 2019
a. ĐKXĐ: \(\frac{5}{3}\le x\le\frac{7}{3}\)
Áp dụng BĐT Bunhiacopxki:
\(T^2=\left(\sqrt{3x-5}+\sqrt{7-3x}\right)\)
\(\le\left(1+1\right)\left(3x-5+7-3x\right)=4\)
\(\Rightarrow T\le2\left(\text{Vì }T>0\right)\)
b.
\(x^2-25=y\left(y+6\right)\)
\(\Leftrightarrow x^2-y^2-6y-9=16\)
\(\Leftrightarrow x^2-\left(y+3\right)^2=16\)
\(\Leftrightarrow\left(x-y-3\right)\left(x+y+3\right)=16=1.16=\left(-1\right)\left(-16\right)=2.8=\left(-2\right)\left(-8\right)\)
TH1: \(\left\{{}\begin{matrix}x-y-3=1\\x+y+3=16\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{17}{2}\\y=\frac{21}{2}\end{matrix}\right.\left(l\right)\)
TH2: \(\left\{{}\begin{matrix}x-y-3=-1\\x+y+3=-16\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\frac{17}{2}\\y=-\frac{11}{2}\end{matrix}\right.\left(l\right)\)
TH3: \(\left\{{}\begin{matrix}x-y-3=2\\x+y+3=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=7\end{matrix}\right.\)
TH4: \(\left\{{}\begin{matrix}x-y-3=-2\\x+y+3=-8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=-6\end{matrix}\right.\)
TH5: \(\left\{{}\begin{matrix}x-y-3=16\\x+y+3=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{17}{2}\\y=-\frac{21}{2}\end{matrix}\right.\left(l\right)\)
TH6: \(\left\{{}\begin{matrix}x-y-3=-16\\x+y+3=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\frac{17}{2}\\y=\frac{9}{2}\end{matrix}\right.\left(l\right)\)
TH7: \(\left\{{}\begin{matrix}x-y-3=-8\\x+y+3=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=0\end{matrix}\right.\)
TH8: \(\left\{{}\begin{matrix}x-y-3=8\\x+y+3=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=-6\end{matrix}\right.\)
thử lại
Vậy pt đã cho có nghiệm ...
a, Đặt \(t=cos3x\left(t\in\left[-1;1\right]\right)\)
\(y=9-sin^23x-\sqrt{2}cos3x\)
\(=cos^23x-\sqrt{2}cos3x+8\)
\(\Leftrightarrow y=f\left(t\right)=t^2-\sqrt{2}t+8\)
\(\Rightarrow minf\left(t\right)\le y\le maxf\left(x\right)\)
\(\Rightarrow min\left\{f\left(-1\right);f\left(1\right);f\left(\dfrac{\sqrt{2}}{2}\right)\right\}\le y\le max\left\{f\left(-1\right);f\left(1\right);f\left(\dfrac{\sqrt{2}}{2}\right)\right\}\)
\(\Rightarrow\dfrac{15}{2}\le y\le9+\sqrt{2}\)
\(\Rightarrow y_{max}=9+\sqrt{2}\)
b, Đặt \(t=sin3x\left(t\in\left[-1;1\right]\right)\)
\(y=3sin3x-8cos^23x+4\)
\(=3sin3x+8-8cos^23x-4\)
\(=8sin^23x+3sin3x-4\)
\(\Leftrightarrow y=f\left(t\right)=8t^2+3t-4\)
\(\Rightarrow minf\left(x\right)\le y\le maxf\left(t\right)\)
\(\Rightarrow min\left\{f\left(-1\right);f\left(1\right);f\left(-\dfrac{3}{16}\right)\right\}\le y\le max\left\{f\left(-1\right);f\left(1\right);f\left(-\dfrac{3}{16}\right)\right\}\)
\(\Rightarrow-\dfrac{137}{32}\le y\le7\)
\(\Rightarrow y_{max}=7\)