a: \(y=\sqrt{2}sin\left(x+\dfrac{pi}{4}\right)\)

\(-1< =sin\left(x+\dfrac{pi}{4}\right)< =1\)

=>\(-\sqrt{2}< =y< =\sqrt{2}\)

\(y_{min}=-\sqrt{2}\) khi sin(x+pi/4)=-1

=>x+pi/4=-pi/2+k2pi

=>x=-3/4pi+k2pi

\(y_{max}=\sqrt{2}\) khi sin(x+pi/4)=1

=>x+pi/4=pi/2+k2pi

=>x=pi/4+k2pi

b: \(y=sinx\cdot cos\left(\dfrac{pi}{3}\right)+cosx\cdot sin\left(\dfrac{pi}{3}\right)+3\)

\(=sin\left(x+\dfrac{pi}{3}\right)+3\)

-1<=sin(x+pi/3)<=1

=>-1+3<=sin(x+pi/3)+3<=4

=>2<=y<=4

y min=2 khi sin(x+pi/3)=-1

=>x+pi/3=-pi/2+k2pi

=>x=-5/6pi+k2pi

y max=4 khi sin(x+pi/3)=1

=>x+pi/3=pi/2+k2pi

=>x=pi/6+k2pi

c: \(y=2\cdot\left(sin2x\cdot\dfrac{\sqrt{3}}{2}-cos2x\cdot\dfrac{1}{2}\right)\)

\(=2sin\left(2x-\dfrac{pi}{6}\right)\)

-1<=sin(2x-pi/6)<=1

=>-2<=y<=2

y min=-2 khi sin(2x-pi/6)=-1

=>2x-pi/6=-pi/2+k2pi

=>2x=-1/3pi+k2pi

=>x=-1/6pi+kpi

y max=2 khi sin(2x-pi/6)=1

=>2x-pi/6=pi/2+k2pi

=>2x=2/3pi+k2pi

=>x=1/3pi+kpi

24.

\(cos\left(x-\dfrac{\pi}{2}\right)\le1\Rightarrow y\le3.1+1=4\)

\(y_{max}=4\)

26.

\(y=\sqrt{2}cos\left(2x-\dfrac{\pi}{4}\right)\)

Do \(cos\left(2x-\dfrac{\pi}{4}\right)\le1\Rightarrow y\le\sqrt{2}\)

\(y_{max}=\sqrt{2}\)

b.

\(\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=\dfrac{1}{2}\)

\(\Leftrightarrow cos\left(x-\dfrac{\pi}{6}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{3}+k2\pi\\x-\dfrac{\pi}{6}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)

a) \(y=1-2sinx\)

Ta có: \(-1\le sinx\le1\Rightarrow-2\le2sinx\le2\)

                                   \(\Rightarrow2\ge-2sin2x\ge-2\)

                                   \(\Rightarrow3\ge1-2sinx\ge-1\)

      Vậy \(y_{max}=3,y_{min}=-1\)

a.

\(-1\le sinx\le1\Rightarrow-7\le y\le-3\)

\(y_{min}=-7\) khi \(sinx=-1\)

\(y_{max}=-3\) khi \(sinx=1\)

b.

\(-1\le cos\left(x+\frac{\pi}{3}\right)\le1\Rightarrow1\le y\le5\)

\(y_{min}=1\) khi \(cos\left(x+\frac{\pi}{3}\right)=-1\)

\(y_{max}=5\) khi \(cos\left(x+\frac{\pi}{3}\right)=1\)

c.

\(0\le1-cosx\le2\Rightarrow-5\le y\le3\sqrt{2}-5\)

\(y_{min}=-5\) khi \(cosx=1\)

\(y_{max}=3\sqrt{2}-5\) khi \(cosx=-1\)

d.

ĐKXĐ: \(0\le sinx\Rightarrow0\le sinx\le1\Rightarrow1\le y\le3\)

\(y_{min}=1\) khi \(sinx=0\)

\(y_{max}=3\) khi \(sinx=1\)

Đc dùng hàm ko ¿

Áp dụng bđt \(\sqrt[3]{a_1^3+b_1^3}+\sqrt[3]{b_1^3+b_2^3}+\sqrt[3]{a_3^3+b_3^3}\ge\sqrt[3]{\left(a_1+a_2+a_3\right)^3+\left(b_1+b_2+b_3\right)^3}\)

và bđt \(\left(a+b+c\right)^3\ge27abc\)

Ta thu đc \(M\ge\sqrt[3]{\left(x+y+z\right)^3+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^3}\ge\sqrt[3]{27abc+\frac{27}{abc}}\)

Đặt \(0< t=abc\le\left(\frac{a+b+c}{3}\right)^3\le\frac{1}{8}\)ta thu được

\(P\ge\sqrt[3]{f\left(t\right)}=\sqrt[3]{27t+\frac{27}{t}}\)

Lại có \(f\left(t\right)=27\left(64t+\frac{1}{t}-63t\right)\ge27\left(2\sqrt{64}-\frac{63}{8}\right)\)

 \(\Leftrightarrow f\left(t\right)\ge27\left(16-\frac{63}{8}\right)=\frac{27.65}{8}\)

\(\Rightarrow P\ge\sqrt[3]{\frac{27.65}{8}}=\frac{3}{2}\sqrt[3]{65}\)(Đpcm !)

                            Nguồn : Team toán tỉnh 9B Tiên Lữ !!!!

a: \(A=\dfrac{x^{\dfrac{1}{3}}\cdot y^{\dfrac{1}{2}}+y^{\dfrac{1}{3}}\cdot x^{\dfrac{1}{2}}}{x^{\dfrac{1}{6}}+y^{\dfrac{1}{6}}}=\dfrac{x^{\dfrac{1}{3}}\cdot y^{\dfrac{1}{3}}\left(x^{\dfrac{1}{6}}+y^{\dfrac{1}{6}}\right)}{x^{\dfrac{1}{6}}+y^{\dfrac{1}{6}}}=x^{\dfrac{1}{3}}\cdot y^{\dfrac{1}{3}}=\left(xy\right)^{\dfrac{1}{3}}\)

b: \(B=\dfrac{x^{3+\sqrt{3}}}{y^2}\cdot\dfrac{x^{-\sqrt{3}-1}}{y^{-2}}=\dfrac{x^{3+\sqrt{3}-\sqrt{3}-1}}{y^{2-2}}=x^2\)

a.\(-1\le cosx\le1\Rightarrow-4\le y=3cosx-1\le2\)

b.-1 \(\le sinx\le1\)\(\Rightarrow3\le y=5+2sinx\le7\)  

c.\(\sqrt{3-1}\le\sqrt{3+cos2x}\le\sqrt{3+1}\Rightarrow\sqrt{2}\le y\le2\)

d.\(y=\sqrt{5sinx-1}+2\le\sqrt{5.1-1}+2=4\)

\(y=\sqrt{5sinx-1}+2\ge2\) . " = " \(\Leftrightarrow sinx=\dfrac{1}{5}\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(\dfrac{1}{5}\right)+2k\pi\\x=\pi-arcsin\left(\dfrac{1}{5}\right)+2k\pi\end{matrix}\right.\)  ( k thuộc Z ) 

1.

\(y=\sqrt{5-2\cos ^2x\sin ^2x}=\sqrt{5-\frac{1}{2}(2\cos x\sin x)^2}=\sqrt{5-\frac{1}{2}\sin ^22x}\)

Dễ thấy:

$\sin ^22x\geq 0\Rightarrow y=\sqrt{5-\frac{1}{2}\sin ^22x}\leq \sqrt{5}$

Vậy $y_{\max}=\sqrt{5}$

$\sin ^22x\leq 1\Rightarrow y=\sqrt{5-\frac{1}{2}\sin ^22x}\geq \sqrt{5-\frac{1}{2}}=\frac{3\sqrt{2}}{2}$

Vậy $y_{\min}=\frac{3\sqrt{2}}{2}$

2.

$y=1+\frac{1}{2}\sin 2x\cos 2x=1+\frac{1}{4}.2\sin 2x\cos 2x$

$=1+\frac{1}{4}\sin 4x$

Vì $-1\leq \sin 4x\leq 1$

$\Rightarrow \frac{5}{4}\leq 1+\frac{1}{4}\sin 4x\leq \frac{3}{4}$

$\Leftrightarrow \frac{5}{4}\leq y\leq \frac{3}{4}$
Vậy $y_{\max}=\frac{5}{4}; y_{\min}=\frac{3}{4}$

a.

\(-1\le sin\left(1-x^2\right)\le1\)

\(\Rightarrow y_{min}=-1\) khi \(1-x^2=-\dfrac{\pi}{2}+k2\pi\) \(\Rightarrow x^2=\dfrac{\pi}{2}+1+k2\pi\) (\(k\ge0\))

\(y_{max}=1\) khi \(1-x^2=\dfrac{\pi}{2}+k2\pi\Rightarrow x^2=1-\dfrac{\pi}{2}+k2\pi\) (\(k\ge1\))

b.

Đặt \(\sqrt{2-x^2}=t\Rightarrow t\in\left[0;\sqrt{2}\right]\subset\left[0;\pi\right]\)

\(y=cost\) nghịch biến trên \(\left[0;\pi\right]\Rightarrow\) nghịch biến trên \(\left[0;\sqrt{2}\right]\)

\(\Rightarrow y_{max}=y\left(0\right)=cos0=1\) khi \(x^2=2\Rightarrow x=\pm\sqrt{2}\)

\(y_{min}=y\left(\sqrt{2}\right)=cos\sqrt{2}\) khi  \(x=0\)