\(M=sin^4x+cos^4x.\\ =1-\dfrac{1}{2}sin^22x.\\ sin^22x\in\left[0;1\right].\\ \Rightarrow MaxM=1-\dfrac{1}{2}.0=1.\\ "="\Leftrightarrow sin^22x=0.\\ \Leftrightarrow2x=\dfrac{\pi}{2}+k2\pi.\\ \Leftrightarrow x=\dfrac{\pi}{4}+k\pi,k\in Z.\)

Vậy \(MaxM=1\Leftrightarrow x=\dfrac{\pi}{4}+k\pi,k\in Z.\)

Bài 2:

132 em được phát số quyển vở là:

\(132\times6=792\) (quyển vở).

Bài 1:

6kg gạo tẻ hết số tiền là:

\(6\times4500=27000\) (đổng).

2kg gạo nếp hết số tiền là:

\(2\times6800=13600\) (đổng).

Số gạo cần mua hết tất cả số tiền là:

\(27000+13600=40600\) (đổng).

\(1,cos\left(2x+\dfrac{\pi}{6}\right)=0.\\ \Leftrightarrow2x+\dfrac{\pi}{6}=k\pi.\\ \Leftrightarrow x=\dfrac{-\pi}{12}+k\dfrac{\pi}{2},k\in Z.\)

\(2,sin\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)=1.\\ \Leftrightarrow\dfrac{\pi}{4}-\dfrac{x}{2}=\dfrac{\pi}{2}+k2\pi.\\ \Leftrightarrow-\dfrac{x}{2}=\dfrac{\pi}{4}+k2\pi.\\ \Leftrightarrow x=\dfrac{-\pi}{2}-k\pi,k\in Z.\)

\(3,sin\left(3x-1\right)=\dfrac{1}{2}.\\ \Leftrightarrow sin\left(3x-1\right)=sin\dfrac{\pi}{6}.\\ \Leftrightarrow\left[{}\begin{matrix}3x-1=\dfrac{\pi}{6}+k2\pi.\\3x-1=\pi-\dfrac{\pi}{6}+k2\pi.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{\pi}{6}+1+k2\pi.\\3x=\dfrac{5}{6}\pi+1+k2\pi.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{18}+\dfrac{1}{3}+k\dfrac{2}{3}\pi.\\x=\dfrac{5}{18}\pi+\dfrac{1}{3}+k\dfrac{2}{3}\pi.\end{matrix}\right.\) \(\left(k\in Z\right).\)

\(4,sin4x=sin\left(x+60^o\right).\\ \Leftrightarrow sin4x=sin\left(x+\dfrac{\pi}{3}\right).\Leftrightarrow\left[{}\begin{matrix}4x=x+\dfrac{\pi}{3}+k2\pi.\\4x=\pi-x-\dfrac{\pi}{3}+k2\pi.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{9}+k\dfrac{2}{3}\pi.\\x=\dfrac{2}{15}\pi+k\dfrac{2}{5}\pi.\end{matrix}\right.\) \(\left(k\in Z\right).\)

\(5,cos\left(x-\dfrac{5\text{​​}\pi}{4}\right)=cos\left(2x+\text{​​}\dfrac{\pi}{4}\right).\\ \Leftrightarrow\left[{}\begin{matrix}x-\dfrac{5\text{​​}\pi}{4}=2x+\text{​​}\dfrac{\pi}{4}+k2\pi.\\x-\dfrac{5\text{​​}\pi}{4}=-2x-\text{​​}\dfrac{\pi}{4}+k2\pi.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\pi-k2\pi.\\x=\dfrac{\pi}{3}+k\dfrac{2\pi}{3}\end{matrix}\right.\) \(\left(k\in Z\right).\)

\(6,sin2x=cos\left(x+\text{​​}\dfrac{\text{​​}\pi}{3}\right).\\ sin2x=sin\left(\dfrac{\pi}{6}-x\right).\\ \Leftrightarrow\left[{}\begin{matrix}2x=\text{​​}\text{​​}\dfrac{\pi}{6}-x+k2\pi.\\2x=\pi-\dfrac{\pi}{6}+x+k2\pi.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{18}+k\dfrac{2}{3}\pi.\\x=\dfrac{5}{6}\pi+k2\pi.\end{matrix}\right.\)\(\left(k\in Z\right).\)

\(a,x^2+5x+\dfrac{25}{4}=x^2+5x+\left(\dfrac{5}{2}\right)^2=\left(x+\dfrac{5}{2}\right)^2.\\ b,16x^2-8x+1=\left(4x-1\right)^2.\\ c,4x^2+12xy+9y^2=\left(2x+3y\right)^2.\)

\(N=sin^4x-cos^4x.\\ =\left(sin^2x+cos^2x\right)\left(sin^2x-cos^2x\right).\\ =-\left(cos^2x-sin^2x\right).\\ =-2cos^2x+1.\\ cos^2x\in\left[0;1\right].\\ \Rightarrow MaxN=-2.0+1=1.\\ "="\Leftrightarrow cos^2x=0.\\ \Leftrightarrow x=k\pi,k\in Z.\)

\(Q=sin^6x+cos^6x=1-\dfrac{3}{4}sin^22x.\)

Ta có:

 \(sin^22x\in\left[0;1\right].\\ \Rightarrow MaxQ=1-\dfrac{3}{4}.0=1.\)

\("="\Leftrightarrow sin^22x=0.\\ \Leftrightarrow2x=\dfrac{\pi}{2}+k\text{​​}\pi.\\ \Leftrightarrow x=\dfrac{\pi}{4}+k\dfrac{\pi}{2},k\in Z.\)

:>>Vâng ạ

E thấy mấy bài tự luận trên lớp lúc hỏi max thì cũng tính luôn x nên e làm vậy ạ.

\(sin^2x\in\left[0;1\right].\\ \Rightarrow MaxM=5-2.0=5.\\"="\Leftrightarrow sin^2x=0.\\ \Leftrightarrow x=\dfrac{\pi}{2}+k\pi,k\in Z.\)