sin 2 α + sin 2 β = 1 \sin^{2}\alpha+\sin^{2}\beta=1 sin 2 α + sin 2 β = 1 sin α + cos β = 0 \sin\alpha+\cos\beta=0 sin α + cos β = 0 证明：对任意正整数 n n n 2 n + 2 > n 2 2^{n}+2>n^{2} 2 n + 2 > n 2
设实数 a ∈ ( 0 , 1 ) a\in(0,1) a ∈ ( 0 , 1 ) { x n } \{x_{n}\} { x n } x 0 = 1 x_{0}=1 x 0 = 1 n n n x n = 1 − x n − 1 x_{n}=1-x_{n-1} x n = 1 − x n − 1 n n n x n > 1 x_{n}>1 x n > 1
( 2 a 3 + a b 2 + b 3 ) ( a 2 b − a b 2 ) = (2a^{3}+ab^{2}+b^{3})(a^{2}b-ab^{2})= ( 2 a 3 + a b 2 + b 3 ) ( a 2 b − a b 2 ) = 设 λ ≠ 0 , 2 \lambda\neq0,2 λ = 0 , 2 λ 4 − 3 λ 3 − 6 λ 2 + 16 λ λ 2 − 2 λ = \frac{\lambda^{4}-3\lambda^{3}-6\lambda^{2}+16\lambda}{\lambda^{2}-2\lambda}= λ 2 − 2 λ λ 4 − 3 λ 3 − 6 λ 2 + 16 λ =
设函数 f ( x ) = x 2 1 + x 2 f(x)=\frac{x^{2}}{1+x^{2}} f ( x ) = 1 + x 2 x 2 f ( 1 ) + f ( 2 ) + f ( 3 ) + f ( 4 ) + f ( 1 2 ) + f ( 1 3 ) + f ( 1 4 ) = f(1)+f(2)+f(3)+f(4)+f(\frac{1}{2})+f(\frac{1}{3})+f(\frac{1}{4})= f ( 1 ) + f ( 2 ) + f ( 3 ) + f ( 4 ) + f ( 2 1 ) + f ( 3 1 ) + f ( 4 1 ) =
设 f ( x ) f(x) f ( x ) [ 0 , 1 ] [0,1] [ 0 , 1 ] f ( 0 ) = f ( 1 ) f(0)=f(1) f ( 0 ) = f ( 1 ) x 1 , x 2 ∈ [ 0 , 1 ] x_{1},x_{2}\in[0,1] x 1 , x 2 ∈ [ 0 , 1 ] ∣ f ( x 1 ) − f ( x 2 ) ∣ ≤ ∣ x 1 − x 2 ∣ |f(x_{1})-f(x_{2})|\leq|x_{1}-x_{2}| ∣ f ( x 1 ) − f ( x 2 ) ∣ ≤ ∣ x 1 − x 2 ∣ a , b ∈ [ 0 , 1 ] a,b\in[0,1] a , b ∈ [ 0 , 1 ] ∣ f ( a ) − f ( b ) ∣ ≤ 1 2 |f(a)-f(b)|\leq\frac{1}{2} ∣ f ( a ) − f ( b ) ∣ ≤ 2 1
设 x > 0 x>0 x > 0 y = x + 4 x 2 y=x+\frac{4}{x^{2}} y = x + x 2 4
设实数 x , y x,y x , y 3 x 2 + 2 y 2 = 6 3x^{2}+2y^{2}=6 3 x 2 + 2 y 2 = 6 2 x + y 2x+y 2 x + y
函数 f ( x ) = a x 3 + b x 2 − 2 x ( a , b ∈ R , a b ≠ 0 ) f(x)=ax^{3}+bx^{2}-2x(a,b\in R,ab\neq0) f ( x ) = a x 3 + b x 2 − 2 x ( a , b ∈ R , ab = 0 ) x 1 + x 2 < 0 x_{1}+x_{2}<0 x 1 + x 2 < 0 a > 0 , b > 0 a>0,b>0 a > 0 , b > 0 a < 0 , b < 0 a<0,b<0 a < 0 , b < 0 a < 0 , b > 0 a<0,b>0 a < 0 , b > 0 a > 0 , b < 0 a>0,b<0 a > 0 , b < 0
设函数 f ( x ) f(x) f ( x ) a f ( x ) + f ( 1 x ) = a x af(x)+f(\frac{1}{x})=ax a f ( x ) + f ( x 1 ) = a x x ≠ 0 x\neq0 x = 0 a 2 ≠ 1 a^{2}\neq1 a 2 = 1 f ( x ) f(x) f ( x )
已知 cos ( π 12 − θ ) = 1 3 \cos(\frac{\pi}{12}-\theta)=\frac{1}{3} cos ( 12 π − θ ) = 3 1 sin ( 5 π 12 + θ ) \sin(\frac{5\pi}{12}+\theta) sin ( 12 5 π + θ )
已知 ∣ x 1 − 3 ∣ < 1 |x_{1}-3|<1 ∣ x 1 − 3∣ < 1 ∣ x 2 − 3 ∣ < 1 |x_{2}-3|<1 ∣ x 2 − 3∣ < 1 4 < x 1 + x 2 < 8 , ∣ x 1 − x 2 ∣ < 2 4<x_{1}+x_{2}<8,|x_{1}-x_{2}|<2 4 < x 1 + x 2 < 8 , ∣ x 1 − x 2 ∣ < 2 f ( x ) = x 2 − x + 1 f(x)=x^{2}-x+1 f ( x ) = x 2 − x + 1 x 1 ≠ x 2 x_{1}\neq x_{2} x 1 = x 2 ∣ x 1 − x 2 ∣ < ∣ f ( x 1 ) − f ( x 2 ) ∣ < 7 ∣ x 1 − x 2 ∣ |x_{1}-x_{2}|<|f(x_{1})-f(x_{2})|<7|x_{1}-x_{2}| ∣ x 1 − x 2 ∣ < ∣ f ( x 1 ) − f ( x 2 ) ∣ < 7∣ x 1 − x 2 ∣
已知等差数列 { a n } \{a_{n}\} { a n } ( a 1 + a 2 ) + ( a 2 + a 3 ) + ⋯ + ( a n + a n + 1 ) = 2 n ( n + 1 ) ( n ∈ N + ) (a_{1}+a_{2})+(a_{2}+a_{3})+\cdots+(a_{n}+a_{n+1})=2n(n+1)(n\in N_{+}) ( a 1 + a 2 ) + ( a 2 + a 3 ) + ⋯ + ( a n + a n + 1 ) = 2 n ( n + 1 ) ( n ∈ N + ) { a n } \{a_{n}\} { a n } { a n 2 n − 1 } \{\frac{a_{n}}{2^{n-1}}\} { 2 n − 1 a n } n n n S n S_{n} S n
函数 y = − x 2 + ( e x − e − x ) sin x y=-x^{2}+(e^{x}-e^{-x})\sin x y = − x 2 + ( e x − e − x ) sin x [ − 2.8 , 2.8 ] [-2.8,2.8] [ − 2.8 , 2.8 ]
已知曲线 C C C r = 1 r=1 r = 1 x x x l l l { x = 1 + t 2 y = 2 + 3 2 t \begin{cases}x=1+\frac{t}{2}\\y=2+\frac{\sqrt{3}}{2}t\end{cases} { x = 1 + 2 t y = 2 + 2 3 t t t t l l l C C C C C C { x ′ = 3 x y ′ = y \begin{cases}x'=3x\\y'=y\end{cases} { x ′ = 3 x y ′ = y C ′ C' C ′ C ′ C' C ′ M ( x , y ) M(x,y) M ( x , y ) x 3 + 3 y \frac{x}{3}+\sqrt{3}y 3 x + 3 y