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]

\(\Rightarrow y_{max}=y\left(1\right)=\dfrac{4}{\sqrt{2}}=2\sqrt{2}\)

\(y_{min}=y\left(3\right)=\dfrac{6}{\sqrt{10}}=\dfrac{3\sqrt{10}}{5}\)

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 \(sinx=t\Rightarrow t\in\left[-1;1\right]\)

\(y=f\left(t\right)=t^3-3t^2+2\Rightarrow f'\left(t\right)=3t^2-6t=0\Rightarrow\left[{}\begin{matrix}t=0\\t=2\notin\left[-1;1\right]\end{matrix}\right.\)

\(f\left(-1\right)=-2\) ; \(f\left(1\right)=0\) ; \(f\left(0\right)=2\)

\(\Rightarrow y_{min}=-2\) ; \(y_{max}=2\)

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 $x=\frac{2}{\sqrt{3}}$

\(P=-3x^2-4x\sqrt{y}+16x--y+12\sqrt{y}+1999\)

\(=-2\left(x^2+2x\sqrt{y}+y\right)+12\left(x+\sqrt{y}\right)-18-x^2+4x-4+2021\)

\(=-2\left(x+\sqrt{y}\right)^2+12\left(x+\sqrt{y}\right)-18-\left(x-2\right)^2+2021\)

\(=-2\left(x+\sqrt{y}-3\right)^2-\left(x-2\right)^2+2021\)\(\le2021\) với mọi x và y không âm

Dấu = xảy ra <=> x=2 và y=1

Vậy maxP=2021 

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 \(\Leftrightarrow\)\(x=2\)

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

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

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