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+
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+
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。

(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=
x-4y-
3 2x+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