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Given that the square of x plus 5x minus 991 is equal to 0, find the value of the cubic of x plus 6x minus 986x minus 1011
x ²+ 5x-991=0x ²+ 5x=991x ³+ 6x ²- 986x-1011=x ³+ 5x ²+ x ²+ 5x-991x-1011=x(x ²+ 5x)+(x ²+ 5x)-991x-1011=991x+991-991x-1011=991-1011=-20
The square of x minus 5x minus 2007 is equal to 0. It is a complete process to find the cubic of X-2 (X-2) - (x-1) plus 1
Simplification: (x - 2) ^ 3 / (x - 2) - (x - 1) ^ 2 + 1 = - 2x + 4. Let: - 2x + 4 = a, then: x = 2 - (1 / 2) a, substitute: x ^ 2-5x-2007 = 0, and then simplify to: A1 = - 1 + √ 8053, A2 = - 1 - √ 8053; The test shows that A2 = - 1 - √ 8053 is not acceptable and is rounded off, that is, the value of the square of X-2 (X-2) - (x-1) plus 1 = A1 = - 1 + √ 8053
Find the differential under y = root sign (1-sinx \ 1 + SiNx), Can't anyone do it? It's just a very simple question in differentiation. It's equivalent to derivation. After more than two years, I forgot, or I wouldn't ask,
First, multiply the molecules by √ 1 + SiNx at the same time to obtain:
y=√1-sin ² x/(1+sinx)=|cosx|/(1+sinx)
Then discuss cosx ≥ 0 and COS < 0 according to the situation
1.cosx≥0
y=cosx/(1+sinx);
dy=[-sinx(1+sinx)-cos ² x]/(1+sinx) ²;
=(-sinx-sin ² x-cos ² x)/(1+sinx) ²;
Because sin ² x+cos ² x=1；
So dy = (- sinx-1) (1 + SiNx) ²;=- 1/(1+sinx)
2.cos<0
y=-cosx/(1+sinx);
dy=[sinx(1+sinx)-cos ² x]/(1+sinx) ²= (sinx+sin ² x-cos ² x)/(1+sinx) ²;
Cos ² x-sin ² x=cos2x;
therefore
dy=(sinx-cos2x)/(1+sinx) ²;
Limx approaches the value of SiNx to the power of zero X
lim(x→0)x^sinx
=lim(x→0)e^(sinxlnx)
=lim(x→0)e^(xlnx)
=lim(x→0)e^(lnx/x^-1)
=lim(x→0)e^(-1/x/x^(-2))
=lim(x→0)e^(-x)
=1
Limx → x power of 0 e - x power of E / X
Original formula = LIM (x → 0) (e ^ x-e ^ (- x)) / X
=-lim(x→0)[e^x(e^(-2x)-1)/x
=lim(x→0)[e^x(2x)]/x
=2lim(x→0)e^x=2
What is limx SiNx divided by X to the third power To step
Simultaneous derivation up and down
Rewrite the molecule 1-cosx to 2Sin ² （x/2)
Denominator 3x ² Rewrite to 12 * (x / 2) ²
Replace X / 2 with t
The original formula is equal to 1 / 6
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