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a)sin a-sin a.cos^2 a=sin a(1-cos^2 a)=sin a(sin^2 a)=sin^3 a
b)sin^4a+cos^4a+2sin^2acos^2a=(sin^2a+cos^2a)^2=1^2=1
\(A=\sqrt{1-4a+4a^2}-2a=\sqrt{\left(1-2a\right)^2}-2a=\left|1-2a\right|-2a\)
Nếu \(a\le\frac{1}{2}\)thì: \(A=1-2a-2a=1-4a\)
Nếu \(a>\frac{1}{2}\)thì: \(A=2a-1-2a=-1\)
ta có:\(\sqrt{\left(1-2a\right)^2}-2a=|1-2a|-2a\)
th1:neu 1-2a <0 <=>1<2a<=>1/2<a:
l1-2al=2a-1
=>2a-1-2a=-1
th2:neu 1-2a>=0=>1>=2a=>1/2>a ta co:
l1-2al=1-2a
=>1-2a-2a=1-4a
\(sin^2a+cos^2a-sin^4a-2cos^2a+sin^2a\)
\(=2sin^2a-cos^2a-sin^4a\)
\(=2sin^2a-cos^2a-\left(\frac{1-cos2a}{2}\right)^2\)
khai triển ra rồi quy đồng lên
\(=\frac{8sin^2a-4cos^2a-1+2cos2a-cos^22a}{4}\)
Mà \(2cos2a=2\left(cos^2a-1\right)=4cos^2-2\)
\(\Rightarrow\frac{8sin^2a-cos^22a-3}{4}\)
Mà \(-cos^22a=sin^22a-1=4sin^2cos^2-1\)
\(\Rightarrow\frac{8sin^2a+4sin^2a.cos^2a-4}{4}\)
\(=\frac{4sin^2a.\left(2-cos^2a\right)-4}{4}\)
\(=sin^2a\left(1+sin^2a\right)-1\)
\(=sin^4a-cos^2a\)
\(\frac{2}{2a-1}.\sqrt{5x^4\left(1-4a+4a^2\right)}\)
\(=\frac{2}{2a-1}.\sqrt{5x^4\left(2a-1\right)^2}\)
\(=\frac{2}{2a-1}.x^2.\left(2a-1\right).\sqrt{5}\)
\(=2\sqrt{5}x^2\)
\(B=\frac{1}{2a-1}.\sqrt{5a^4\left(2a-1\right)^2}=\sqrt{5}a^2.\frac{\left|2a-1\right|}{2a-1}\)
Nếu \(a>\frac{1}{2}\) thì \(B=\sqrt{5}a^2\)
Nếu \(a< \frac{1}{2}\) thì \(B=-\sqrt{5}a^2\)
\(B=\frac{1}{2a-1}\sqrt{5a^4\left(1-4a+4a^2\right)}\)
\(B=\frac{2\left|a\right|}{2a-1}\sqrt{5\left[1-2.2a+\left(2a\right)^2\right]}\)
\(B=\frac{2a}{2a-1}\sqrt{5\left(1-2a\right)^2}\)
\(B=\frac{2a\left|1-2a\right|}{2a-1}\sqrt{5}\)
\(=\frac{2a\left(2a-1\right)}{2a-1}\sqrt{5}=2a\sqrt{5}\)
\(ĐKXĐ:a\ne\frac{1}{2}\)
\(B=\frac{1}{2a-1}.\sqrt{5a^4.\left(1-4a+4a^2\right)}\)
\(=\frac{1}{2a-1}.\sqrt{5a^4.\left(1-2a\right)^2}\)
\(=\frac{1}{2a-1}.\sqrt{5}.\sqrt{a^4}.\sqrt{\left(1-2a\right)^2}\)
\(=\frac{1}{2a-1}.\sqrt{5}.a^2.\left|1-2a\right|=\frac{\sqrt{5}.a^2.\left|1-2a\right|}{2a-1}\)
+) Nếu \(a< \frac{1}{2}\)\(\Rightarrow\left|1-2a\right|=1-2a=-\left(2a-1\right)\)
\(\Rightarrow B=\frac{-\sqrt{5}.a^2.\left(2a-1\right)}{2a-1}=-\sqrt{5}.a^2\)
+) Nếu \(a>\frac{1}{2}\)\(\Rightarrow\left|1-2a\right|=-\left(1-2a\right)=-1+2a=2a-1\)
\(\Rightarrow B=\frac{\sqrt{5}.a^2.\left(2a-1\right)}{2a-1}=\sqrt{5}.a^2\)