Cho 3 số thực dương a,b,c .Chứng minh rằng :
\(1+\dfrac{3}{ab+bc+ca}\ge\dfrac{6}{a+b+c}\)
Chương 1: HÀM SỐ LƯỢNG GIÁC. PHƯƠNG TRÌNH LƯỢNG GIÁC
a/sin3x+cos2x1+2sin xcos2xb/sin^3x+cos^3x2left(sin^5x+cos^5xright)c/dfrac{tan x}{sin x}-dfrac{sin x}{cos x}dfrac{sqrt{2}}{2}d/dfrac{cos xleft(cos x+2sin xright)+3sin xleft(sin x+sqrt{2}right)}{sin2x-1}1e/sin^2x+sin^23x-2cos^22x0f/dfrac{tan x-sin x}{sin^3x}dfrac{1}{cos x}g/sin2xleft(cos x+tan2xright)4cos^2xh/sin^2x+sin^23xcos^2x+cos^23xk/4sin2xdfrac{cos^2x-sin^2x}{cos^6x+sin^6x}mọi người giải giúp em với em đang cần gấp ạ
Đọc tiếp
a/\(\sin3x+\cos2x=1+2\sin x\cos2x\)
b/\(\sin^3x+\cos^3x=2\left(\sin^5x+\cos^5x\right)\)
c/\(\dfrac{\tan x}{\sin x}-\dfrac{\sin x}{\cos x}=\dfrac{\sqrt{2}}{2}\)
d/\(\dfrac{\cos x\left(\cos x+2\sin x\right)+3\sin x\left(\sin x+\sqrt{2}\right)}{\sin2x-1}=1\)
e/\(\sin^2x+\sin^23x-2\cos^22x=0\)
f/\(\dfrac{\tan x-\sin x}{\sin^3x}=\dfrac{1}{\cos x}\)
g/\(\sin2x\left(\cos x+\tan2x\right)=4\cos^2x\)
h/\(\sin^2x+\sin^23x=\cos^2x+\cos^23x\)
k/\(4\sin2x=\dfrac{\cos^2x-\sin^2x}{\cos^6x+\sin^6x}\)
mọi người giải giúp em với em đang cần gấp ạ
Cho ba số thực dương a, b, c thoả mãn a+b+c=2 Chứng minh rằng:
\(\dfrac{ab}{\sqrt{2c+ab}}+\dfrac{bc}{\sqrt{2a+bc}}+\dfrac{ca}{\sqrt{2b+ca}}\le1\)
\(VT=\sqrt{\dfrac{a^2b^2}{c\left(a+b+c\right)+ab}}+\sqrt{\dfrac{b^2c^2}{a\left(a+b+c\right)+bc}}+\sqrt{\dfrac{a^2c^2}{b\left(a+b+c\right)+ac}}\\ VT=\sqrt{\dfrac{a^2b^2}{ac+ab+bc+c^2}}+\sqrt{\dfrac{b^2c^2}{a^2+ac+ab+bc}}+\sqrt{\dfrac{a^2c^2}{ab+bc+b^2+ac}}\\ VT=\sqrt{\dfrac{a^2b^2}{\left(c+a\right)\left(b+c\right)}}+\sqrt{\dfrac{a^2c^2}{\left(b+c\right)\left(a+b\right)}}+\sqrt{\dfrac{b^2c^2}{\left(a+b\right)\left(a+c\right)}}\)
Áp dụng BĐT Cauchy-Schwarz:
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{b^2c^2}{\left(a+b\right)\left(a+c\right)}}\le\dfrac{\dfrac{bc}{a+b}+\dfrac{bc}{a+c}}{2}\\\sqrt{\dfrac{a^2c^2}{\left(a+b\right)\left(b+c\right)}}\le\dfrac{\dfrac{ca}{a+b}+\dfrac{ca}{b+c}}{2}\\\sqrt{\dfrac{a^2b^2}{\left(b+c\right)\left(a+c\right)}}\le\dfrac{\dfrac{ab}{b+c}+\dfrac{ab}{a+c}}{2}\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{\left(\dfrac{bc}{a+b}+\dfrac{ca}{a+b}\right)+\left(\dfrac{ca}{b+c}+\dfrac{ab}{b+c}\right)+\left(\dfrac{bc}{a+c}+\dfrac{ab}{a+c}\right)}{2}\\ \Rightarrow VT\le\dfrac{a+b+c}{2}=\dfrac{2}{2}=1\)
Dấu \("="\Leftrightarrow a=b=c=\dfrac{2}{3}\)
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Cho ba số thực x,y,z thoả mãn : x+2y+3z=18
Cmr : \(\dfrac{2y+3z+5}{1+x}+\dfrac{3z+x+5}{1+2y}+\dfrac{x+2y+5}{1+3z}\ge\dfrac{51}{7}\)
\(VT=\dfrac{2y+3z+5}{1+x}+1+\dfrac{3z+x+5}{2y+1}+1+\dfrac{x+2y+5}{1+3z}+1-3\)
\(VT=\dfrac{x+2y+3z+6}{1+x}+\dfrac{x+2y+3z+6}{1+2y}+\dfrac{x+2y+3z+6}{1+3z}-3\)
\(VT=24\left(\dfrac{1}{1+x}+\dfrac{1}{1+2y}+\dfrac{1}{1+3z}\right)-3\ge\dfrac{24.9}{1+x+1+2y+1+3z}-3=\dfrac{216}{21}-3=\dfrac{51}{7}\)
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Cho hai số a, b dương. Tìm giá trị nhỏ nhất của biểu thức:
\(P=\dfrac{a+b}{\sqrt{a\left(4a+5b\right)}+\sqrt{b\left(4b+5a\right)}}\)
Cho ba số thực dương x,y,z thoả mãn :x+2y+3z=18 .Chứng minh rằng :
\(\dfrac{2y+3z+5}{1+x}+\dfrac{3z+x+5}{1+2y}+\dfrac{x+2y+5}{1+3z}\ge\dfrac{51}{7}\)
sin2\(\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right).tan^2x-cos^2\dfrac{x}{2}\) = 0
⇔ \(\dfrac{1-cos\left(x-\dfrac{\pi}{2}\right)}{2}.tan^2x-\dfrac{1+cosx}{2}=0\)
⇔ \(\dfrac{1-sinx}{2}.tan^2x-\dfrac{1+cosx}{2}=0\)
⇔ tan2x - sin.tan2x - 1 - cosx = 0
⇒ sin2x - sin.cos2x - cos2x - cos3x = 0 (nhân cả 2 vế với cos2x)
⇔ cos2x - sin2x + cos3x - sinx.cos2x = 0 (đổi dấu 2 vế)
⇔ (cosx - sinx)(cosx + sinx) + cos2x (cosx - sinx)
⇔ \(\left[{}\begin{matrix}cosx-sinx=0\left(1\right)\\cosx+sinx+cos^2x=0\left(2\right)\end{matrix}\right.\)
(1) ⇔ \(cos\left(x+\dfrac{\pi}{4}\right)=0\)
(2) ⇔ cosx(cosx + 1) + sinx = 0
⇔ \(cosx.2cos^2\dfrac{x}{2}\) + 2\(sin\dfrac{x}{2}.cos\dfrac{x}{2}\) = 0
⇔ \(\left[{}\begin{matrix}cos\dfrac{x}{2}=0\\cos\dfrac{x}{2}.cosx+sin\dfrac{x}{2}=0\left(\Psi\right)\end{matrix}\right.\)
\(\left(\Psi\right)\) ⇔ \(cos\dfrac{x}{2}.\left(2cos^2\dfrac{x}{2}-1\right)+sin\dfrac{x}{2}=0\)
⇔ \(2cos^3\dfrac{x}{2}+sin\dfrac{x}{2}-cos\dfrac{x}{2}\) = 0
⇔ \(2cos^3\dfrac{x}{2}+sin\dfrac{x}{2}\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}\right)-cos\dfrac{x}{2}\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}\right)=0\)
⇔ \(cos^3\dfrac{x}{2}+sin^3\dfrac{x}{2}+sin\dfrac{x}{2}.cos^2\dfrac{x}{2}-sin^2\dfrac{x}{2}.cos\dfrac{x}{2}=0\)
+ Xét \(cos^3\dfrac{x}{2}=0\), nếu thỏa mãn thì kết luận nghiệm
+ Xét \(cos^3\dfrac{x}{2}\ne0\), chia cả 2 vế cho \(cos^3\dfrac{x}{2}\) đưa về phương trình bậc 3 theo \(tan\dfrac{x}{2}\)
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Tìm Min, Max:
a, \(y=\left|Sinx\right|-\sqrt{Cosx}\)
b, \(y=12Sin^4x+Sin^2x+Cos4x+2Cos^2x\)
Tìm Max:
\(y=8-3Sin^23x+6Sin6x\)
\(y=8-3sin^23x+6sin6x\)
\(=8+\dfrac{3}{2}\left(1-2sin^23x\right)+6sin6x-\dfrac{3}{2}\)
\(=\dfrac{3}{2}cos6x+6sin6x+\dfrac{13}{2}\)
\(=\dfrac{3\sqrt{17}}{2}\left(\dfrac{1}{\sqrt{17}}cos6x+\dfrac{4}{\sqrt{17}}sin6x\right)+\dfrac{13}{2}\)
\(=\dfrac{3\sqrt{17}}{2}cos\left(6x-arccos\dfrac{1}{\sqrt{17}}\right)+\dfrac{13}{2}\)
\(\le-\dfrac{3\sqrt{17}}{2}+\dfrac{13}{2}=\dfrac{13-3\sqrt{17}}{2}\)
\(y_{max}=\dfrac{13-3\sqrt{17}}{2}\Leftrightarrow cos\left(6x-arccos\dfrac{1}{\sqrt{17}}\right)=1\Leftrightarrow x=\dfrac{1}{6}arccos\dfrac{1}{\sqrt{17}}+\dfrac{k\pi}{3}\)
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