Rút gọn biểu thức: 3(sin^6x+cos^6x)- 2(sin^4x+cos^4x) 28/08/2021 Bởi Amara Rút gọn biểu thức: 3(sin^6x+cos^6x)- 2(sin^4x+cos^4x)
Nếu đề như trên thì: $\begin{array}{l} 3\left( {{{\sin }^6}x + {{\cos }^6}x} \right) – 2\left( {{{\sin }^4}x + {{\cos }^4}x} \right)\\ = 3\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^4}x – {{\sin }^2}x.{{\cos }^2}x + {{\cos }^4}x} \right) – 2\left( {{{\sin }^4}x + {{\cos }^4}x} \right)\\ = 3\left( {{{\sin }^4}x – {{\sin }^2}x{{\cos }^2}x + {{\cos }^4}x} \right) – 2\left( {{{\sin }^4}x + {{\cos }^4}x} \right)\\ = {\sin ^4}x + {\cos ^4}x – 3{\sin ^2}x{\cos ^2}x\\ = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} – 5{\sin ^2}x{\cos ^2}x\\ = 1 – 5{\sin ^2}x{\cos ^2}x \end{array}$ Sửa đề: $3\left( {{{\sin }^4}x + {{\cos }^4}x} \right) – 2\left( {{{\sin }^6}x + {{\cos }^6}x} \right)$ $\begin{array}{l} 3\left( {{{\sin }^4}x + {{\cos }^4}x} \right) – 2\left( {{{\sin }^6}x + {{\cos }^6}x} \right)\\ = 3\left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} – 2{{\sin }^2}x.{{\cos }^2}x} \right] – 2\left[ {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^4}x – {{\sin }^2}x{{\cos }^2}x + {{\cos }^4}x} \right)} \right]\\ = 3\left( {1 – 2{{\sin }^2}x.{{\cos }^2}x} \right) – 2\left( {{{\sin }^4}x – {{\sin }^2}x{{\cos }^2}x + {{\cos }^4}x} \right)\\ = 3\left( {1 – 2{{\sin }^2}x{{\cos }^2}x} \right) – 2\left( {1 – 3{{\sin }^2}x{{\cos }^2}x} \right)\\ = 1 \end{array}$ Bình luận
$3(\sin^6x+\cos^6x)-2(\sin^4x+\cos^4x)$ $=3(\sin^2x+\cos^2x)(\sin^4x+\cos^4x-\sin^2x\cos^2x)-2(\sin^4x+\cos^4x)$ $=3\sin^4x+3\cos^4x-3\sin^2x\cos^2x-2\sin^4x-2\cos^4x$ $=\sin^4x+\cos^4x-3\sin^2x\cos^2x$ $=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x-3\sin^2x\cos^2x$ $=1-5.(\sin x\cos x)^2$ $=1-\dfrac{5}{4}.\sin^22x$ $=1-\dfrac{5}{4}\Big(\dfrac{1}{2}-\dfrac{1}{2}\cos4x\Big)$ $=\dfrac{5}{8}\cos4x+\dfrac{3}{8}$ Bình luận
Nếu đề như trên thì:
$\begin{array}{l} 3\left( {{{\sin }^6}x + {{\cos }^6}x} \right) – 2\left( {{{\sin }^4}x + {{\cos }^4}x} \right)\\ = 3\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^4}x – {{\sin }^2}x.{{\cos }^2}x + {{\cos }^4}x} \right) – 2\left( {{{\sin }^4}x + {{\cos }^4}x} \right)\\ = 3\left( {{{\sin }^4}x – {{\sin }^2}x{{\cos }^2}x + {{\cos }^4}x} \right) – 2\left( {{{\sin }^4}x + {{\cos }^4}x} \right)\\ = {\sin ^4}x + {\cos ^4}x – 3{\sin ^2}x{\cos ^2}x\\ = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} – 5{\sin ^2}x{\cos ^2}x\\ = 1 – 5{\sin ^2}x{\cos ^2}x \end{array}$
Sửa đề: $3\left( {{{\sin }^4}x + {{\cos }^4}x} \right) – 2\left( {{{\sin }^6}x + {{\cos }^6}x} \right)$
$\begin{array}{l} 3\left( {{{\sin }^4}x + {{\cos }^4}x} \right) – 2\left( {{{\sin }^6}x + {{\cos }^6}x} \right)\\ = 3\left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} – 2{{\sin }^2}x.{{\cos }^2}x} \right] – 2\left[ {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^4}x – {{\sin }^2}x{{\cos }^2}x + {{\cos }^4}x} \right)} \right]\\ = 3\left( {1 – 2{{\sin }^2}x.{{\cos }^2}x} \right) – 2\left( {{{\sin }^4}x – {{\sin }^2}x{{\cos }^2}x + {{\cos }^4}x} \right)\\ = 3\left( {1 – 2{{\sin }^2}x{{\cos }^2}x} \right) – 2\left( {1 – 3{{\sin }^2}x{{\cos }^2}x} \right)\\ = 1 \end{array}$
$3(\sin^6x+\cos^6x)-2(\sin^4x+\cos^4x)$
$=3(\sin^2x+\cos^2x)(\sin^4x+\cos^4x-\sin^2x\cos^2x)-2(\sin^4x+\cos^4x)$
$=3\sin^4x+3\cos^4x-3\sin^2x\cos^2x-2\sin^4x-2\cos^4x$
$=\sin^4x+\cos^4x-3\sin^2x\cos^2x$
$=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x-3\sin^2x\cos^2x$
$=1-5.(\sin x\cos x)^2$
$=1-\dfrac{5}{4}.\sin^22x$
$=1-\dfrac{5}{4}\Big(\dfrac{1}{2}-\dfrac{1}{2}\cos4x\Big)$
$=\dfrac{5}{8}\cos4x+\dfrac{3}{8}$