6.4

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

\(=cos2x-\sqrt{3}sin2x+2\)

\(=2\left(\dfrac{1}{2}cos2x-\dfrac{\sqrt{3}}{2}sin2x\right)+2\)

\(=2cos\left(2x-\dfrac{\pi}{3}\right)+2\)

Do \(-1\le cos\left(2x-\dfrac{\pi}{3}\right)\le1\)

\(\Rightarrow0\le y\le4\)

\(y_{min}=0\) khi \(cos\left(2x-\dfrac{\pi}{3}\right)=-1\)

\(y_{max}=4\) khi \(cos\left(2x-\dfrac{\pi}{3}\right)=1\)

6.5

Ủa nhìn bài 7 thì đây là chương trình lớp 11 (pt lượng giác) chứ đâu phải lớp 10?

Vậy giải theo kiểu lớp 11 nghe:

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

\(\Leftrightarrow2y+y.cosx=2+cosx+3sinx\)

\(\Leftrightarrow3sinx+\left(1-y\right).cosx=2y-2\)

Theo điều kiện có nghiệm của pt lượng giác bậc nhất:

\(9+\left(1-y\right)^2\ge\left(2y-2\right)^2\)

\(\Leftrightarrow\left(y-1\right)^2\le3\)

\(\Rightarrow1-\sqrt{3}\le y\le1+\sqrt{3}\)

7.

\(\Leftrightarrow\left(m+1\right)\left[\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)\right]+cos4x=0\)

\(\Leftrightarrow\left(m+1\right)\left(1-3sin^2x.cos^2x\right)+cos4x=0\)

\(\Leftrightarrow\left(m+1\right)\left(1-\dfrac{3}{4}sin^22x\right)+cos4x=0\)

\(\Leftrightarrow\left(m+1\right)\left(1-\dfrac{3}{8}\left(1-cos4x\right)\right)+cos4x=0\)

\(\Leftrightarrow\left(m+1\right)\left(\dfrac{5}{8}+\dfrac{3}{8}cos4x\right)+cos4x=0\)

\(\Leftrightarrow5\left(m+1\right)+\left(3m+11\right)cos4x=0\)

\(\Leftrightarrow\left(3m+11\right)cos4x=-5\left(m+1\right)\)

- Với \(m=-\dfrac{11}{3}\) pt vô nghiệm

- Với \(m\ne-\dfrac{11}{3}\)

\(\Rightarrow cos4x=\dfrac{-5\left(5m+1\right)}{3m+11}\)

Do \(-1\le cos4x\le1\) nên pt có nghiệm khi:

\(-1\le\dfrac{-5\left(m+1\right)}{3m+11}\le1\)

Tới đây chắc bạn tự làm tiếp được đúng ko? Tách ra làm 2 BPT rồi sau đó giao nghiệm thôi

2.

\(\Leftrightarrow cos2x-cos8x-sin3x+cos5x-2sin5x.cos5x=0\)

\(\Leftrightarrow2sin5x.sin3x-sin3x+cos5x-2sin5x.cos5x=0\)

\(\Leftrightarrow sin3x\left(2sin5x-1\right)-cos5x\left(2sin5x-1\right)=0\)

\(\Leftrightarrow\left(sin3x-cos5x\right)\left(2sin5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos5x=sin3x=cos\left(\dfrac{\pi}{2}-3x\right)\\sin5x=\dfrac{1}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{16}+\dfrac{k\pi}{4}\\x=-\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{30}+\dfrac{k2\pi}{5}\\x=\dfrac{\pi}{6}+\dfrac{k2\pi}{5}\end{matrix}\right.\)

3.

\(\Leftrightarrow1+sinx=cosx-cos3x+2sinx.cosx+1-2sin^2x\)

\(\Leftrightarrow sinx=2sin2x.sinx+2sinx.cosx-2sin^2x\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\1=2sin2x+2cosx-2sinx\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow4sinx.cosx+2cosx-2sinx-1=0\)

\(\Leftrightarrow2cosx\left(2sinx+1\right)-\left(2sinx+1\right)=0\)

\(\Leftrightarrow\left(2cosx+1\right)\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\cosx=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\)

4.

\(\Leftrightarrow2sin2x.cosx+sin2x=2cos2x.cosx+cos2x\)

\(\Leftrightarrow sin2x\left(2cosx+1\right)=cos2x\left(2cosx+1\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2cosx+1=0\\sin2x=cos2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=-\dfrac{1}{2}\\tan2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{2\pi}{3}+k2\pi\\x=\dfrac{\pi}{8}+\dfrac{k\pi}{2}\\\end{matrix}\right.\)

\(y'=\dfrac{\left(-2x+2\right)\left(x-3\right)-\left(-x^2+2x+c\right)}{\left(x-3\right)^2}=\dfrac{-x^2+6x-6-c}{\left(x-3\right)^2}\)

\(\Rightarrow\) Cực đại và cực tiểu của hàm là nghiệm của: \(-x^2+6x-6-c=0\) (1)

\(\Delta'=9-\left(6+c\right)>0\Rightarrow c< 3\)

Gọi \(x_1;x_2\) là 2 nghiệm của (1) \(\Rightarrow\left\{{}\begin{matrix}-x_1^2+6x_1-6=c\\-x_2^2+6x_2-6=c\end{matrix}\right.\)

\(\Rightarrow m-M=\dfrac{-x_1^2+2x_1+c}{x_1-3}-\dfrac{-x_2^2+2x_2+c}{x_2-3}=4\)

\(\Leftrightarrow\dfrac{-2x_1^2+8x_1-6}{x_1-3}-\dfrac{-2x_2^2+8x_2-6}{x_2-3}=4\)

\(\Leftrightarrow2\left(1-x_1\right)-2\left(1-x_2\right)=4\)

\(\Leftrightarrow x_2-x_1=2\)

Kết hợp với Viet: \(\left\{{}\begin{matrix}x_2-x_1=2\\x_1+x_2=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1=2\\x_2=4\end{matrix}\right.\)

\(\Rightarrow c=2\)

Có 1 giá trị nguyên

is not different

is not different 

Đặt \(cosx-sinx=t\Rightarrow-\sqrt{2}\le t\le\sqrt{2}\)

\(t^2=1-2sinx.cosx\Rightarrow sinx.cosx=\dfrac{1-t^2}{2}\)

Pt trở thành:

\(t\left(1+\dfrac{1-t^2}{2}\right)+1=0\)

\(\Leftrightarrow t^3-3t-2=0\)

\(\Leftrightarrow\left(t-2\right)\left(t+1\right)^2=0\Rightarrow\left[{}\begin{matrix}t=2\left(loại\right)\\t=-1\end{matrix}\right.\)

\(\Rightarrow cosx-sinx=-1\)

\(\Leftrightarrow\sqrt[]{2}cos\left(x+\dfrac{\pi}{4}\right)=-1\)

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

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{4}\right)=cos\left(\dfrac{3\pi}{4}\right)\)

\(\Leftrightarrow...\)

0,3625

\(\text{0,29 x 8 x 1,25 = ?}\)

\(0.29X8=2.32\)

\(2.32X1.25=\text{2.9}\)

0,29 x 8 x 1,25 

= 0.29 x 10

 = 2,9