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a: =1+1=2
b: \(=\sin x\left(1-\cos^2x\right)=sinx\cdot sin^2x=sin^3x\)
c: \(=cos^2x\left(1+tg^2x\right)=cos^2x\cdot\dfrac{1}{cos^2x}=1\)
d: \(=\dfrac{cos58^0}{cos58^0}=1\)
a) sin230 - cos670 = sin230 - sin230 =0
b)sin100 + sin400 - cos500 - cos800 = sin100 + sin400 - sin400 - sin100 = (sin100 - sin100) +(sin400 - sin400) = 0
\(A=\sin^6\alpha+cos^6\alpha+3\sin^2\alpha\cos^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right).\)vì\(\sin^2\alpha+\cos^2\alpha=1\)
\(=\left(\sin^2\alpha+\cos^2\alpha\right)^3=1^3=1\)
\(B=2\left(\cos^2\alpha+\sin^2\alpha\right)=2.1=2\)
\(C=\frac{-4\cos\alpha\sin\alpha}{\sin\alpha\cos\alpha}=-4\)
a) x -\(\sqrt{2x-9}=0\) ĐKXĐ: x\(\ge\frac{9}{2}\)
<=> x=\(\sqrt{2x-9}\)
<=> x2=2x-9 (vì x>0)
<=> x2-2x+1=8
<=>(x-1)2=8
<=>\(\left[{}\begin{matrix}x-1=2\sqrt{2}\\x-1=-2\sqrt{2}\end{matrix}\right.\)
<=>x=\(2\sqrt{2}+1\)(vì x>0) (thỏa mãn)
\(a.\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}-\dfrac{3}{3-\sqrt{6}}=\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\dfrac{\sqrt{3}.\sqrt{3}}{\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}=\sqrt{6}-\dfrac{\sqrt{3}}{\sqrt{3}-\sqrt{2}}=\dfrac{3\sqrt{2}-3\sqrt{3}}{\sqrt{3}-\sqrt{2}}=\dfrac{-3\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}=-3\) \(b.\left(2\sqrt{2}-\sqrt{3}\right)^2-2\sqrt{3}\left(\sqrt{3}-2\sqrt{2}\right)=\left(2\sqrt{2}-\sqrt{3}\right)\left(2\sqrt{2}+\sqrt{3}\right)=8-3=5\) \(c.\left(\dfrac{1}{3-\sqrt{5}}-\dfrac{1}{3+\sqrt{5}}\right):\dfrac{5-\sqrt{5}}{\sqrt{5}-1}=\dfrac{3+\sqrt{5}-3+\sqrt{5}}{9-5}:\sqrt{5}=\dfrac{2\sqrt{5}}{4}.\dfrac{1}{\sqrt{5}}=\dfrac{\sqrt{5}}{2}.\dfrac{1}{\sqrt{5}}=\dfrac{1}{2}\) \(d.\left(3-\dfrac{a-2\sqrt{a}}{\sqrt{a}-2}\right)\left(3+\dfrac{\sqrt{ab}-3\sqrt{a}}{\sqrt{b}-3}\right)=\left(3-\sqrt{a}\right)\left(3+\sqrt{a}\right)=9-a\)
a.1+sin^2x+cos^2x=1+(sin^2+cos^2)=1+1=2