\(\frac{sin\frac{a}{2}}{cos\frac{a}{2}}\left(1+\frac{1}{cosa}\right)=\frac{2sin\frac{a}{2}cos\frac{a}{2}}{2cos^2\frac{a}{2}}\left(\frac{cosa+1}{cosa}\right)=\frac{sina}{cosa+1}\left(\frac{cosa+1}{cosa}\right)=\frac{sina}{cosa}=tana\)
\(\frac{sin\frac{a}{2}}{cos\frac{a}{2}}\left(1+\frac{1}{cosa}\right)=\frac{2sin\frac{a}{2}cos\frac{a}{2}}{2cos^2\frac{a}{2}}\left(\frac{cosa+1}{cosa}\right)=\frac{sina}{cosa+1}\left(\frac{cosa+1}{cosa}\right)=\frac{sina}{cosa}=tana\)
1. sin^8(x) - cos^8(x) - 4sin^6(x) + 6sin^4(x) - 4sin^2(x) = 1
2. sin6x+sin4x+sin2x/1+cos2x+cos4x = 2sin2x
3. 1+sin2x /cosx+sinx - 1-tan^2(x/2)/1+tan^2(x/2) = sinx
4. cos4x + 4cos2x + 3 = 8cos^4(x)
5. 1+cosx+cos2x+cos3x/ 2cos^2(x)+cosx-1 = 2cosx
1. sin^8(x) - cos^8(x) - 4sin^6(x) + 6sin^4(x) - 4sin^2(x) = 1
2. sin6x+sin4x+sin2x/1+cos2x+cos4x = 2sin2x
3. 1+sin2x /cosx+sinx - 1-tan^2(x/2)/1+tan^2(x/2) = sinx
4. cos4x + 4cos2x + 3 = 8cos^4(x)
5. 1+cosx+cos2x+cos3x/ 2cos^2(x)+cosx-1 = 2cosx
Bài 1: Cho số thực dương ab + bc + ca =1. Tìm GTLN của
\(P=\dfrac{2a}{\sqrt{1+a^2}}+\dfrac{b}{\sqrt{1+b^2}}+\dfrac{c}{\sqrt{1+c^2}}\)
Bài 2: Cho x,y,z là số thực dương thỏa mãn x+y+z=xyz . CMR:
\(\dfrac{1+\sqrt{1+x^2}}{x}+\dfrac{1+\sqrt{1+y^2}}{y}+\dfrac{1+\sqrt{1+z^2}}{z}\le xyz\)
Giải pt
\(1)4x^2+\sqrt{3x+1}+5=13x\)
\(2)7x^2-13x+8=2x^2.\sqrt[3]{x\left(1+3x-3x^2\right)}\)
\(3)x^3-4x^2-5x+6=\sqrt[3]{7x^2+9x-4}\)
\(4)x^3-5x^2+4x-5=\left(1-2x\right)\sqrt[3]{6x^2-2x+7}\)
\(5)8x^2-13x+7=\left(1+\dfrac{1}{x}\right)\sqrt[3]{3x^2-2}\)
Cho a, b, c là ba số thực dương thõa mãn abc=1. CMR
\(\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\le\frac{1}{2}\)
1. Giải bpt: \(\sqrt{x-2}-2\ge\sqrt{2x-5}-\sqrt{x+1}\)
2. Với \(x\in\left(0;1\right)\) tìm Min \(P=\dfrac{\sqrt{1-x}\left(1+\sqrt{1-x}\right)}{x}+\dfrac{5}{\sqrt{1-x}}\)
Giải hệ pt : \(\left\{{}\begin{matrix}x+y+\dfrac{1}{x}+\dfrac{1}{y}=4\\x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}=4\end{matrix}\right.\)
phương trình nào tương đương với phương trình (3x^2) = 9?
A. (x+1)(x-1) = 0
B. x + 1 =0
C. x - 1 =0
D. x^2 = 2
Cho a, b, c > 0 thỏa mãn ab + bc + ca = 3. Chứng minh rằng: \(\dfrac{1}{1+a^2\left(b+c\right)}+\dfrac{1}{1+b^2\left(a+c\right)}+\dfrac{1}{1+c^2\left(a+b\right)}\le\dfrac{1}{abc}\)
\(\sqrt{x+5-4\sqrt{x+1}}\) + \(\sqrt{x+2-2\sqrt{x+1}}\)=1