giai pt luong giac! hay thi zo! bi' wa thi hoi!

[newtons007]

[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
Chắc suất Đại học top - Giữ chỗ ngay!!

ĐĂNG BÀI NGAY để cùng trao đổi với các thành viên siêu nhiệt tình & dễ thương trên diễn đàn.

a)2cos^3(x)+cos2x+sinx = 0
b)cos(4x/3)= cos^2(x)
kem theo 2 cau tim GTLN :
C)Y= kan(2sin^2(x)-1)/(2sin^2(x)
d)sin^6(x).cos^8(x)
co gang nha ae ! mai tui giai ko piet dung ko ? ae giup do~

 

[tuyn]

a)2cos^3(x)+cos2x+sinx = 0

[TEX]PT \Leftrightarrow 2cos^3x+1-2sin^2x+sinx=0 \Leftrightarrow 2cosx(1-sin^2x)+(1-sinx)(2sinx-1)=0 \Leftrightarrow (1-sinx)[2cosx(1+sinx)+2sinx-1]=0 \Leftrightarrow \left[\begin{sinx=1}\\{2(sinx+cosx)+sin2x = 0(t=sinx+cosx)}[/TEX]

b)cos(4x/3)= cos^2(x)

[TEX]PT \Leftrightarrow 2cos(4x/3)=1+cos2x \Leftrightarrow 4cos^2(2x/3)-2=1+4cos^3(2x/3)-3cos(2x/3)(t=cos(2x/3)[/TEX]

kem theo 2 cau tim GTLN :
C)Y= kan(2sin^2(x)-1)/(2sin^2(x)

Không hiểu

d)y=sin^6(x).cos^8(x)

[TEX]y \geq 0 \Rightarrow Miny=0 khi sin2x=0[/TEX]
[TEX]sin^6x.cos^8x=sin^6x.(1-sin^2x)^4=sin^2x.sin^2x.sin^2x.(1-sin^2x)(1-sin^2x)(1-sin^2x)(1-sin^2x)=\frac{1}{64.81}4sin^2x.4sin^2x.4sin^2x.(3-3sin^2x)(3-3sin^2x)(3-3sin^2x)(3-3sin^2x) \leq \frac{1}{64.81}.(\frac{4sin^2x+4sin^2x+4sin^2x+(3-3sin^2x)+(3-3sin^2x)+(3-3sin^2x)+(3-3sin^2x)}{7})^7=\frac{1}{64.81}.(\frac{12}{7})^7 \Rightarrow Maxy=\frac{1}{64.81}.(\frac{12}{7})^7 Khi 4sin^2x=3-3sin^2x[/TEX]

 

[newtons007]

y= (sin^2(x))^3.(cos^2(x))^4 ta co 1= sin^2(x) +cos^2(x)= \frac{sin^2(x)}{3} + \frac{sin^2(x)}{3}+ \frac{sin^2(x)}{3} + \frac{cos^2(x)}{4} + \frac{cos^2(x)}{4}+ \frac{cos^2(x)}{4}+ \frac{cos^2(x)}{4}\geq 7\sqrt[7]{\frac{ (sin^2(x))^3.(cos^2(x))^4}{3^3.4^4}}\Leftrightarrow y= (sin^2(x))^3.(cos^2(x))^4\leq\frac{3^3.4^4}{7^7}
vay maxy=\frac{3^3.4^4}{7^7} khi \frac{sin^2(x)}{3}= \frac{cos^2(x)}{4} \Leftrightarrow 2sinx=\sqrt[2]{3}cosx
sua lai cau c) y=\frac{\sqrt[2]{2sin^x -1}}{2sin^2x}

 

[tuyn]

[TEX]y= (sin^2(x))^3.(cos^2(x))^4[/TEX] ta co [TEX]1= sin^2(x) +cos^2(x)= \frac{sin^2(x)}{3} + \frac{sin^2(x)}{3}+ \frac{sin^2(x)}{3} + \frac{cos^2(x)}{4} + \frac{cos^2(x)}{4}+ \frac{cos^2(x)}{4}+ \frac{cos^2(x)}{4}\geq 7\sqrt[7]{\frac{ (sin^2(x))^3.(cos^2(x))^4}{3^3.4^4}}\Leftrightarrow y= (sin^2(x))^3.(cos^2(x))^4\leq\frac{3^3.4^4}{7^7}[/TEX]
vay [TEX]maxy=\frac{3^3.4^4}{7^7}[/TEX] khi [TEX] \frac{sin^2(x)}{3}= \frac{cos^2(x)}{4} \Leftrightarrow 2sinx=\sqrt[2]{3}cosx[/TEX]
sua lai cau [TEX]c) y=\frac{\sqrt[2]{2sin^x -1}}{2sin^2x}[/TEX]

Công thức Toán phải để trong TEX chứ bạn
:-SS:-SS:-SS:-SS:-SS:-SS:-SS:-SS