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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 9: CMR đẳng thức: a) sin ($sin^{4}$ x – $cos^{4}$x)/ sin x+ cosx = sin x- cos x b) $sin^{6}$ x+$cos^{6}$ x+3.$sin^{2}$x .$cos^{2}$x =1

Toán Lớp 9: CMR đẳng thức:
a) sin ($sin^{4}$ x – $cos^{4}$x)/ sin x+ cosx = sin x- cos x
b) $sin^{6}$ x+$cos^{6}$ x+3.$sin^{2}$x .$cos^{2}$x =1

Comments ( 1 )

  1. Lời giải và giải thích chi tiết:
     Ta có:
    \(\begin{array}{l}
    a,\\
    \dfrac{{{{\sin }^4}x – {{\cos }^4}x}}{{\sin x + \cos x}}\\
     = \dfrac{{{{\left( {{{\sin }^2}x} \right)}^2} – {{\left( {{{\cos }^2}x} \right)}^2}}}{{\sin x + \cos x}}\\
     = \dfrac{{\left( {{{\sin }^2}x – {{\cos }^2}x} \right).\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}}{{\sin x + \cos x}}\\
     = \dfrac{{\left( {{{\sin }^2}x – {{\cos }^2}x} \right).1}}{{\sin x + \cos x}}\\
     = \dfrac{{{{\sin }^2}x – {{\cos }^2}x}}{{\sin x + \cos x}}\\
     = \dfrac{{\left( {\sin x – \cos x} \right)\left( {\sin x + \cos x} \right)}}{{\sin x + \cos x}}\\
     = \sin x – \cos x\\
    b,\\
    {\sin ^6}x + {\cos ^6}x + 3{\sin ^2}x.\cos {x^2}\\
     = {\sin ^6}x + {\cos ^6}x + 3{\sin ^2}x.\cos {x^2}.1\\
     = {\sin ^6}x + {\cos ^6}x + 3{\sin ^2}x.\cos {x^2}.\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\\
     = {\sin ^6}x + {\cos ^6}x + 3{\sin ^4}x.\cos {x^2} + 3{\sin ^2}x.{\cos ^4}x\\
     = {\sin ^6}x + 3{\sin ^4}x.{\cos ^2}x + 3{\sin ^2}x.{\cos ^4}x + {\cos ^6}x\\
     = {\left( {{{\sin }^2}x} \right)^3} + 3.{\left( {{{\sin }^2}x} \right)^2}.{\cos ^2}x + 3.{\sin ^2}x.{\left( {{{\cos }^2}x} \right)^2} + {\left( {{{\cos }^2}x} \right)^3}\\
     = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^3}\\
     = {1^3}\\
     = 1
    \end{array}\)

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222-9+11+12:2*14+14 = ? ( )