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已知函数fx等于2cosx的平方 f(x)=cos2x+2sinx=1-2(six)^2+2sinx=-2(sinx-1/2)^2+

火烧 2022-06-06 07:26:47 1088
f x =co 2x+2 i x=1-2 ix ^2+2 i x=-2 i x-1/2 ^2+ f x =co 2x+2 i x=1-2 ix ^2+2 i x=-2 i x-1/2 ^2+f x =

f(x)=cos2x+2sinx=1-2(six)^2+2sinx=-2(sinx-1/2)^2+  

f(x)=cos2x+2sinx=1-2(six)^2+2sinx=-2(sinx-1/2)^2+

f(x)=cos2x+2sinx
=1-2(sinx)^2 +2sinx
=-2(sinx)^2 +2sinx+1
=-2[(sin)^2-sinx]+1
=-2[(sin)^2-sinx+1/4-1/4]+1
=-2[(sin)^2-sinx+1/4]++1/2+1
=-2(sinx-1/2)^2+3/2

(3)-x+(2x-2)-(3x+5) (4)3a^2 +a^2-(2a^2-2a)+(3a-a^2)

解(3)
原式=-x+2x-2-3x-5
=(-x+2x-3x)+(-2-5)
=-2x-7
解(4)
原式=3a²+a²-2a²+2a+3a-a²
=(3a²+a²-2a²-a²)+(2a+3a)
=a²+5a

计算:(1)(-2ab2)2?(3a2b-2ab-1); (2)4(a-b)2-(2...

(1)原式=12a4b5-8a3b5-4a2b4;
(2)原式=4a2-8ab+4b2+2ab-4a2+b2-2ab=5b2-8ab;
(3)原式=x2-(y-1)2=x2-y2+2y-1;
(4)原式=4a2-2ab2+

1 4

b4.

(2+1)(2^1+1)(2^2+1)(2^3+1)(2^4+1)...(2^2n+1) 怎么算

原式=3×1×(2^1+1)(2^2+1)(2^3+1)…(2^2n+1)
=3×(2^1-1)(2^1+1)(2^2+1)(2^3+1)…(2^2n+1)
=3×(2^2-1)(2^2+1)(2^3+1)…(2^2n+1)
=3×(2^3-1)(2^3+1)…(2^2n+1)
……………………………………
=3×(2^2n-1)(2^2n+1)
=3×[2^(n+1)-1]
望采纳

(2a+b)^2-(2a-b)(a+b)-2(a-2b)(a+2b),其中a=1/2,b=-2

(2a+b)^2-(2a-b)(a+b)-2(a-2b)(a+2b)
原式=4a²+4ab+b²-(2a²+2ab-ab-b²)-2(a²-4b²)
=4a²+4ab+b²-2a²-ab+b²-2a²+8b²
=3ab+10b²
代入 =3*1/2*(-2)+10*4
=-3+40
=37

(2+1)(2∧2+1)(2∧4+1)(2∧8+1)(2∧16+1)(2∧32+1)-2∧64求值

-1。你先去括号,就会发现规律,即 2的63次方加到2的0次方,再减去2的64次方。最后用一下等比数列求和公式。

已知函数fx等于2cosx的平方 f(x)=cos2x+2sinx=1-2(six)^2+2sinx=-2(sinx-1/2)^2+

(1)计算:(-3)2-2-3+30;(2)化简:(x3)2÷(-x)2+(-2x)2?(-x2

(1)原式=9-

1 8

+1=9

7 8


(2)原式=x6÷x2+4x2?(-x2)=x4-4x4=-3x4.

计算:(1)-1 4 +〔1-(1-0.5×2)〕÷|2-(-3) 2 |;(2)(-1) 10 ×2+(-2) 3 ÷4+(-2 2 );(3

(1)原式=-1+(1-1+1)÷7=-1+

1 7 =-

6 7


(2)原式=1×2-8÷4-4=2-2-4=-4;
(3)原式=-a+b-3b+6a=5a-2b;
(4)原式=

1 2

x-4y-

3 2

x+3y=-x-y.

高数 为什么2(sinx/2)^2<2(x/2)^2=x^2/2

因为sin(x)<x

1. 2(x+1)^2=18 2. 9(x-2)^2=(2x+1)^2 3. x(x-2)=3

1. 2(x+1)^2=18
(x+1)^=9
(x+1)^=3^
x+1=3
x=2
2. 9(x-2)^2=(2x+1)^2
(3(x-2))^2=(2x+1)^2
3(x-2)=2x+1
x=7
3. x(x-2)=3
x^2-2x+1=4
(x-1)^2=2^2
x-1=2
x=3

  
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