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\(\left(x^4-1\right)^2+\left(x^2+1\right)^2=\left(x^2-1\right)^2.\left(x^2+1\right)^2+\left(x^2+1\right)^2\)
\(=\left(x^2+1\right)^2\left[\left(x^2-1\right)^2+1\right]=\left(x^2+1\right)^2\left(x^4-2x^2+2\right)\)
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\(\left(\dfrac{1}{2}x^2-\dfrac{1}{3}y\right)\left(\dfrac{1}{4}x^4+\dfrac{1}{6}x^2y+\dfrac{1}{9}y^2\right)\\ =\left(\dfrac{1}{2}x^2\right)^3-\left(\dfrac{1}{3}y\right)^3\\ =\dfrac{1}{8}x^6-\dfrac{1}{27}y^3.\)
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a,
$xy^2+x^2y+(-2xy^2)=xy^2-2xy^2+x^2y=-xy^2+x^2y$
b,
$12x^2y^3z^4+(-7x^2y^3z^4)=12x^2y^3z^4-7x^2y^3z^4=5x^2y^3z^4$
c,
$-6xy^3-(-6xy^3)+6x^3=-6xy^3+6xy^3+6x^3=0+6x^3=6x^3$
d,
$\frac{-x^2}{2}+\frac{7}{2}x^2+x=(\frac{7}{2}-\frac{1}{2})x^2+x$
$=3x^2+x$
e,
$2x^3+3x^3-\frac{1}{3}x^3=(2+3-\frac{1}{3})x^3=\frac{14}{3}x^3$
f,
$5xy^2+\frac{1}{2}xy^2+\frac{1}{4}xy^2=(5+\frac{1}{2}+\frac{1}{4})xy^2$
$=\frac{23}{4}xy^2$
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a: A+B
=x^2y+xyz+7y^2-25xy-xyz+x^2y-7y^2+xy
=-24xy+2x^y
A-B=x^2y+xyz+7y^2-25xy+xzy-x^2y+7y^2-xy
=2xyz+14y^2-26xy
b: Bậc của A là 3
bậc của B là 3
c: Khi x=-3;y=-1/2;z=0 thì:
A=9*(-1/2)+0+7*(-1/2)^2-25*(-3)*(-1/2)
=-9/2+7/4-75/2
=-42+7/4=-161/4
B=(-3)*(-1)*(-1/2)*0+(-3)^2*(-1/2)-7*1/4+(-3)*(-1/2)
=-9/2-7/4+3/2
=-3-7/4=-19/4
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\(=\left(x^2+2x+1\right)+\left(y^2-8y+16\right)=\left(x+1\right)^2+\left(y-4\right)^2\ge0\forall x,y\)
dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=4\end{matrix}\right.\)
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\(a,=4x^2-4x+1-4x^2+4-x^2-x+6=-x^2-5x+11\\ b,=8x^3+27-8x^3+72x=72x+27\)
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a) \(=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2}{x^2+x+1}-\dfrac{1}{x-1}=\dfrac{x^2+2+2\left(x-1\right)-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
b) \(=\dfrac{1}{x+2}+\dfrac{3}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x+2\right)\left(x-2\right)+3\left(x+2\right)+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x+6}{\left(x+2\right)^2}\)
c) \(=\dfrac{x^2+xy+y^2-3xy+\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{x^2-2xy+y^2+x^2-2xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{2\left(x^2-2xy+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{2\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{2\left(x-y\right)}{x^2+xy+y^2}\)
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Bài 1:
a) \(2x\left(x^2-5x+6\right)=2x^3-10x^2+12x\)
b) \(\left(7x^5+14x^2y^3-28x^3y^2\right):7x^2=x^3+2y^3-4xy^2\)
Bài 2:
\(x^2+y^2+2x-8y+17=\left(x^2+2x+1\right)+\left(y^2-8y+16\right)=\left(x+1\right)^2+\left(y-4\right)^2\ge0\forall x,y\)
Ta có:
\(A=-1^2+2^2-3^2+4^2-...-\left(n-1\right)^2+n^2\) (đã sửa đề)
\(A=\left(2^2-1^2\right)+\left(4^2-3^2\right)+...+\left[n^2-\left(n-1\right)^2\right]\)
\(A=\left(2-1\right)\left(2+1\right)+\left(4-3\right)\left(4+3\right)+...+\left(n-n+1\right)\left(n+n-1\right)\)
\(A=1+2+3+4+...+\left(n-1\right)+n\)
\(A=\frac{\left(n+1\right)\left[\left(n-1\right)\div1+1\right]}{2}=\frac{n\left(n+1\right)}{2}\)
xin lỗi, nhưng bạn có thể giải đề này hộ mình được ko?
sao bạn phải sửa đề vậy?