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bài 2
P= (x+1)(x2-x+1)+x-(x-1)(x2+x+1)+2010 với x = -2010
= (x3+1) + x - (x3-1) + 2010
= x3 + 1 + x - x3 + 1 + 2010
= x + 2 + 2010
= 2010 + 2 + 2010
=4022
Q=16x(4x2-5)-(4x+1)(16x2-4x + 1) với x = 1/5
= (4x)3-16.5x - [(4x)3+1]
= (4x)3 - 16.5x - (4x)3 - 1
= -16.5x - 1
= -16.5.1/5 - 1
= -16-1
=-17
a) (x-3)(x2+3x+9)-x(x-4)(x+4)=41
<=> x3 - 33 - x(x2 - 42) = 41
<=> x3 - 27 - x3 + 16x = 41
<=> 16x = 68
<=> x= 4,25
b) (x+2)(x2-2x+4)-x(x2+2)=4
<=> x3 + 23 - x3 - 2x =4
<=> 8 - 2x = 4
<=> 2x = 4
<=> x= 1/2
a/ Ta có : (x2 + x + 1)2 = [x2 + (x + 1)]2 = x4 + 2x2(x + 1) + (x + 1)2 Nên:
A = (x + 1)4 + (x2 + x + 1)2 = (x + 1)4 + x4 + 2x2(x + 1) + (x + 1)2 = [(x + 1)4 + (x + 1)2] + [x4 + 2x2(x + 1)]
= (x + 1)2(x2 + 2x + 2) + x2(x2 + 2x + 2) = (x2 + 2x + 2)(2x2 + 2x + 1).
b/ B = x10 + x5 + 1 Đặt \(|x^5|=t^2\) thì x10 = t4 Ta có B = t4 + t2 + 1 = (t2 + 1)2 - t2 = (t2 - t + 1)(t2 + t + 1)
Vậy : \(B=\left(x^5-\sqrt{|x|^5}+1\right)\left(x^5+\sqrt{|x|^5}+1\right).\)
c/ Nhân đa thức được: C = x2(x4 - 1)(x2 + 2) + 1 = (x6 - x2)(x2 + 2) + 1 = x6 (x2 + 2) - x2 (x2 + 2) + 1
C = x8 + 2x6 - x4 - 2x2 + 1 = x8 + 2x6 - 2x4 + x4 - 2x2 + 1 = (x4)2 + 2x4 (x2 - 1) + (x2 - 1)2
C = (x4 + x2 + 1)2 .
d/ D = 1 + ( a + b + c) + ab + bc + ca) + abc = (1 + a) + (abc + bc) + (b + ab) + (c + ca) = (1 + a) + bc(1 + a) + b(1 + a) + c(1 + a) =
= (1 + a)(1 + bc + b + c) = (1 + a)[(1 + b) + c(1 + b)] = (1 + a)(1 + b)(1 + c).
\(b,\)\(x^{10}+x^5+1\)
\(=x^{10}-x^7+x^7+x^5+x^3-x^3+1\)
\(=x^7\left(x^3-1\right)+x^3\left(x^4+x^2+1\right)-\left(x^3-1\right)\)
\(=x^7\left(x-1\right)\left(x^2+x+1\right)+x^3\left(x^4+2x^2+1-x^2\right)-\left(x-1\right)\left(x^2+x+1\right)\)
\(=x^7\left(x-1\right)\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)\left(x^2-x+1\right)\)\(-\left(x-1\right)\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^8-x^7+x^5-x^4+x^3-x+1\right)\)
\(d,\)\(1+\left(a+b+c\right)+\left(ab+bc+ca\right)+abc\)
\(=1+a+b+c+ab+bc+ca+abc\)
\(=\left(ab+b\right)+\left(abc+bc\right)+\left(ac+c\right)+\left(a+1\right)\)
\(=b\left(a+1\right)+bc\left(a+1\right)+c\left(a+1\right)+\left(a+1\right)\)
\(=\left(a+1\right)\left(b+bc+c+1\right)\)
\(=\left(a+1\right)\left[b\left(c+1\right)+\left(c+1\right)\right]\)
\(=\left(a+1\right)\left(b+1\right)\left(c+1\right)\)
1)a)x+y=60
<=>(x+y)^2=3600
<=>x^2+2xy+y^2=3600(1)
mà xy=35 nên 2xy=2.35=70
(1)<=>x^2+70+y^2=3600
<=>x^2+y^2=3530
<=>(x^2+y^2)^2=12460900
<=>x^4+2x^2.y^2+y^4=12460900(2)
mà xy=35 nên 2x.x.y.y=2450
(2)<=>x^4+y^4=123458450
b)x+y=1
<=>(x+y)^3=1
<=>x^3+3x^2y+3xy^2+y^3=1
<=>x^3+y^3+3xy(x+y)=1
<=>x^3+y^3+3xy=1
=>M=1
x+y=1
<=>x^2+2xy+y^2=1(1)
B=x^3+y^3+3xy(x^2+y^2)+3xy(2xy)
=x^3+y^3+3xy(x^2+2xy+y^2)
=M.1=1(từ(1)
c)
x-y=1
<=>(x-y)^3=1
<=>x^3-3x^2y+3xy^2-y^3=1
<=>x^3-y^3-3xy(x-y)=1
<=>x^3-y^3-3xy=1
=>N=1
a: \(\Leftrightarrow5x^2-20x-41=x^2-10x+25+4x^2+4x+1-x^2+2x+\left(x-1\right)^2\)
\(\Leftrightarrow5x^2-20x-41=4x^2-4x+26+x^2-2x+1\)
\(\Leftrightarrow5x^2-20x-41=5x^2-6x+27\)
=>-14x=68
hay x=-34/7
b: \(\Leftrightarrow x^2-25-x^3+6x^2-12x+8-7x^2+x^3+1=\left(x+3\right)^3-x^3-9x^2\)
\(\Leftrightarrow-12x-16=x^3+9x^2+27x+27-x^3-9x^2=27x+27\)
=>-39x=43
hay x=-43/39
Ta có: \(\frac{x^2-\frac{1}{x^2}}{x^2+\frac{1}{x^2}}=a\Leftrightarrow\frac{x^4+\frac{1}{x^4}-2}{x^4+\frac{1}{x^4}+2}=a^2\) (bình phương 2 vế)
\(\Leftrightarrow1-\frac{4}{x^4+\frac{1}{x^4}+2}=a^2\Leftrightarrow x^4+\frac{1}{x^4}+2=\frac{4}{1-a^2}\Leftrightarrow x^4+\frac{1}{x^4}=\frac{2+2a^2}{1-a^2}\)(1)
Lại có: \(\frac{x^2-\frac{1}{x^2}}{x^2+\frac{1}{x^2}}=a\Leftrightarrow\frac{x^4-\frac{1}{x^4}}{\left(x^2+\frac{1}{x^2}\right)^2}=a\Leftrightarrow\frac{x^4-\frac{1}{x^4}}{x^4+\frac{1}{x^4}+2}=a\)
\(\Leftrightarrow\frac{x^4-\frac{1}{x^4}}{\frac{2+2a^2}{1-a^2}+2}=a\) ( thay (1) vào) \(\Leftrightarrow x^4-\frac{1}{x^4}=\frac{4a}{1-a^2}\) (2)
Từ (1) và (2) \(\Rightarrow M=\frac{x^4-\frac{1}{x^4}}{x^4+\frac{1}{x^4}}=\frac{\frac{4a}{1-a^2}}{\frac{2+2a^2}{1-a^2}}=\frac{2a}{1+a^2}\)