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\(M\left(x\right)+N\left(x\right)\)
\(=5x^3-x^2-4+2x^4-2x^2+2x+1\)
\(=2x^4+5x^3-3x^2+2x-3\)
\(M\left(x\right)-N\left(x\right)\)
\(=5x^3-x^2-4-\left(2x^4-2x^2+2x+1\right)\)
\(=5x^3-x^2-4-2x^4+2x^2-2x-1\)
\(=-2x^4+5x^3+x^2-2x-5\)
\(M\left(x\right)+P\left(x\right)=N\left(x\right)\)
\(\Rightarrow P\left(x\right)=N\left(x\right)-M\left(x\right)\)
\(\Rightarrow P\left(x\right)=2x^4-2x^2+2x+1-\left(5x^3-x^2-4\right)\)
\(\Rightarrow P\left(x\right)=2x^4-2x^2+2x+1-5x^3+x^2+4\)
\(\Rightarrow P\left(x\right)=2x^4-5x^3-x^2+2x+5\)
Ta có:
M = 3xyz - 3x2 + 5xy - 1
N = 5x2 + xyz - 5xy + 3 - y
M + N = 3xyz - 3x2 + 5xy - 1 + 5x2 + xyz - 5xy + 3 - y
= -3x2 + 5x2 + 3xyz + xyz + 5xy - 5xy - y - 1 + 3
= 2x2 + 4xyz - y +2.
M - N = (3xyz - 3x2 + 5xy - 1) - (5x2 + xyz - 5xy + 3 - y)
= 3xyz - 3x2 + 5xy - 1 - 5x2 - xyz + 5xy - 3 + y
= -3x2 - 5x2 + 3xyz - xyz + 5xy + 5xy + y - 1 - 3
= -8x2 + 2xyz + 10xy + y - 4.
N - M = (5x2 + xyz - 5xy + 3 - y) - (3xyz - 3x2 + 5xy - 1)
= 5x2 + xyz - 5xy + 3 - y - 3xyz + 3x2 - 5xy + 1
= 5x2 + 3x2 + xyz - 3xyz - 5xy - 5xy - y + 3 + 1
= 8x2 - 2xyz - 10xy - y + 4.
M = 3xyz - 3x2 + 5xy - 1
N = 5x2 + xyz - 5xy + 3 - y
M + N = 3xyz - 3x2 + 5xy - 1 + 5x2 + xyz - 5xy + 3 - y
= -3x2 + 5x2 + 3xyz + xyz + 5xy - 5xy - y - 1 + 3
= 2x2 + 4xyz - y +2.
M - N = (3xyz - 3x2 + 5xy - 1) - (5x2 + xyz - 5xy + 3 - y)
= 3xyz - 3x2 + 5xy - 1 - 5x2 - xyz + 5xy - 3 + y
= -3x2 - 5x2 + 3xyz - xyz + 5xy + 5xy + y - 1 - 3
= -8x2 + 2xyz + 10xy + y - 4.
N - M = (5x2 + xyz - 5xy + 3 - y) - (3xyz - 3x2 + 5xy - 1)
= 5x2 + xyz - 5xy + 3 - y - 3xyz + 3x2 - 5xy + 1
= 5x2 + 3x2 + xyz - 3xyz - 5xy - 5xy - y + 3 + 1
= 8x2 - 2xyz - 10xy - y + 4.
M + N = \(x^2-2xy+y^2\)+\(y^2+2xy+x^2+1\)
= \(\left(x^2+x^2\right)+\left(-2xy+2xy\right)+\left(y^2+y^2\right)+1\)
= \(2x^2+2y^2+1\)
M - N = \(x^2-2xy+y^2-y^2+2xy+1\)
= \(\left(x^2+x^2\right)+\left(-2xy+2xy\right)+\left(y^2-y^2\right)+1\)
= \(2x^2+1\)
Lời giải:
a)
$M(x)=(x^5+5x^5)-2x^4-4x^3+3x$
$=6x^5-2x^4-4x^3+3x$
$N(x)=-6x^5+(7x^4-5x^4)+(x^3+3x^3)+4x^2-3x-1$
$=-6x^5+2x^4+4x^3+4x^2-3x-1$
b)
$M(-1)=6(-1)^5-2(-1)^4-4(-1)^3+3(-1)=-7$
$N(-2)=-6(-2)^5+2(-2)^4+4(-2)^3+4(-2)^2-3(-2)-1$
$=213$
c)
$M(x)+N(x)=(6x^5-2x^4-4x^3+3x)+(-6x^5+2x^4+4x^3+4x^2-3x-1)$
$=4x^2-1$
$M(x)-N(x)=(6x^5-2x^4-4x^3+3x)-(-6x^5+2x^4+4x^3+4x^2-3x-1)$
$=12x^5-4x^4-8x^3-4x^2+6x+1$
d)
$F(x)=M(x)+N(x)=4x^2-1=0\Leftrightarrow x^2=\frac{1}{4}$
$\Leftrightarrow x=\pm \frac{1}{2}$
Vậy $x=\pm \frac{1}{2}$ là nghiệm của $F(x)$
a, M + N = \(\left(x^2-2xy+y^2\right)+\left(y^2+2xy+x^2+1\right)\)
=\(2x^2+2y^2+1\)
b,M - N =\(\left(x^2-2xy+y^2\right)-\left(y^2+2xy+x^2+1\right)\)
=\(-4xy-1\)
Bài 1
a)M+N=\(x^2y+xy^2-5x^2y^2+x^3+x^3+xy+3xy^2-x^2y+x^2y^2\)
=4xy2-4x2y2+2x3+xy
b)M-N=\(x^2y+xy^2-5x^2y^2+x^3-x^3-xy-3xy^2+x^2y-x^2y^2\)
=\(2x^2y-2xy^2-xy-6x^2y^2\)
\(\left\{{}\begin{matrix}f\left(x\right)=3x^4+5yx^2-3yx+y^4+z^2\\M\left(x\right)=ax^4+bx^2+cx+D\end{matrix}\right.\)
\(f\left(x\right)+M\left(x\right)=\left(3+a\right)x^4+\left(5y+a\right)x^2+\left(-3y+c\right)x+y^4+z^2+D\)\(\Leftrightarrow\left\{{}\begin{matrix}a=-3\\b=-5y\\c=3y\end{matrix}\right.\)\(\Rightarrow M\left(x\right)=-3x^4-5yx^2+3yx+y^4+z^2+D\) với D tùy ý không chứa x
\(\int f\left(x\right)dx=x^3+C\)
\(\sum a\left(b^2-1\right)\left(c^2-1\right)\)
\(a\left(b^2-1\right)\left(c^2-1\right)+b\left(a^2-1\right)\left(c^2-1\right)+c\left(b^2-1\right)\left(a^2-1\right)\)
\(\begin{matrix}\sum a\left(b^2-1\right)\left(c^2-1\right)=\sum\left(ab^2-a\right)\left(c^2-1\right)=\sum\left(ab^2c^2-ab^2-ac^2+a\right)\\\left(ab^2c^2-ab^2-ac^2+a\right)+\\\left(a^2bc^2-ba^2-bc^2+b\right)+\\\left(a^2b^2c-b^2c-a^2c+c\right)\end{matrix}\)
\(a+b+c\Rightarrow a+b=abc-c\) \(\Rightarrow\sum ab\left(a+b\right)=\sum ab\left(abc-c\right)=\sum a^2b^2c-abc\)
\(\left[abc\left(bc+ac+ab\right)\right]-\left[ab\left(a+b\right)+ac\left(a+c\right)+bc\left(b+c\right)\right]+\left[\left(a+b+c\right)\right]\)
\(\sum a^2b^2c-abc=\left(-abc+a^2b^2c\right)+\left(-abc+a^2bc^2\right)+\left(-abc+ab^2c^2\right)=-3abc+abc\left(ab+bc+ac\right)\)
\(\left[abc\left(bc+ac+ab\right)\right]+3abc-abc\left(ab+bc+ac\right)+\left(a+b+c\right)=3abc+abc=4abc=VP\)
Bài 1:
\(A+B=7x^2-3xy+2y^2\)
\(A-B=x^2-7xy+4y^2\)
Bài 2:
a) \(M=6x^2+9xy-y^2-\left(5x^2-2xy\right)\)
\(M=x^2+11xy-y^2\)
b) \(N=\left(3xy-4y^2\right)-\left(x^2-7xy+8y^2\right)\)
\(N=-x^2-12y^2+10xy\)
a, M+N= x^2-2yx+z^2+3yz-z^2+5x^2
= (x^2+5x^2)+(-2yx)+(z^2-z^2)+3yz
=6x^2+-2yx+2z^2+3yz
b, M-N=(x^2-2yx+z^2)-(3yz-z^2+5x^2)
= x^2-2yx+z^2-3yz+z^2-5x^2
=(x^2-5x^2)+(-2yx)+(z^2+z^2)+(-3yz)
=-4x^2+-2yx+2z^2+-3yz
N-M=(3yz-z^2+5x^2)-(x^2-2yx+z^2)
= 3yz-z^2+5x^2-x^2+2yx-z^2
=3yz+(-z^2-z^2)+(5x^2-x^2)+2yx
=3yz+6x^2+2yx