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5 tháng 8 2020
Bài làm:
a) \(\left(x^4-2x^2y+y^2\right)\div\left(y-x^2\right)\)
\(=\left(x^2-y\right)^2\div\left(y-x^2\right)\)
\(=\left(y-x^2\right)^2\div\left(y-x^2\right)\)
\(=y-x^2\)
b) \(\left(x^2-2xy^2+y^4\right)\div\left(x-y^2\right)\)
\(=\left(x-y^2\right)^2\div\left(x-y^2\right)\)
\(=x-y^2\)
AH
Akai Haruma
Giáo viên
28 tháng 6 2023
Lời giải:
$2x^2+y^2+2xy-8x-6y+30$
$=(x^2+y^2+2xy)+x^2-8x-6y+30$
$=(x+y)^2-6(x+y)+(x^2-2x)+30$
$=(x+y)^2-6(x+y)+9+(x^2-2x+1)+20$
$=(x+y-3)^2+(x-1)^2+20\geq 20$
Vậy GTNN của biểu thức là $20$ khi $x+y-3=x-1=0$
$\Leftrightarrow x=1; y=2$
Q
20 tháng 6 2017
a) \(-\left(x+2\right)\cdot\left(x^2-1x+4\right)\)
\(=-\left(x+2\right)\cdot\left(x^2-x+4\right)\)
\(=-\left(x^3-x^2+4x+2x^2-2x+8\right)\)
\(=-\left(x^3+x^2+2x+8\right)\)
\(=-x^3-x^2-2x-8\)
b) \(-\left(x+2y\right)\cdot\left(x^2-2xy+y^2\right)\)
\(=-\left(x^3-2x^2y+xy^2+2x^2y-4xy^2+2y^3\right)\)
\(=-\left(x^3-3xy^2+2y^3\right)\)
\(=-x^3+3xy^2-2y^3\)
c) \(-\left(5-a\right)\cdot\left(25+5a+a^2\right)\)
\(=-\left(125-a^3\right)\)
\(=-125+a^3\)
d) \(-\left(x-2y\right)\cdot\left(x^2+2xy+4y^2\right)\)
\(=-\left(x^3-8y^3\right)\)
\(=-x^3+8y^3\)
HH
20 tháng 6 2018
Giải:
5) \(-x^2+x-\dfrac{1}{2}\)
\(=-x^2+x-\dfrac{1}{4}+\dfrac{3}{4}\)
\(=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\le\dfrac{3}{4}\)
\(\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)
Vậy ...
6) \(-\dfrac{1}{4}x^2+x-2\)
\(=-\dfrac{1}{4}x^2+x-1-1\)
\(=-\left(\dfrac{1}{4}x^2-x+1\right)-1\)
\(=-\left(\dfrac{1}{2}x-1\right)^2-1\le-1\)
\(\Leftrightarrow\dfrac{1}{2}x-1=0\Leftrightarrow x=2\)
Vậy ...
7) \(-\dfrac{1}{9}x^2-\dfrac{1}{3}x+1\)
\(=-\dfrac{1}{9}x^2-\dfrac{1}{3}x-\dfrac{1}{4}+\dfrac{5}{4}\)
\(=-\left(\dfrac{1}{9}x^2+\dfrac{1}{3}x+\dfrac{1}{4}\right)+\dfrac{5}{4}\)
\(=-\left(\dfrac{1}{3}x+\dfrac{1}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{1}{3}x+\dfrac{1}{2}=0\Leftrightarrow x=-\dfrac{3}{2}\)
Vậy ...
8) \(-2x^2+2xy-2y^2+2x+2y-8\)
\(=-x^2+2xy-y^2+2x-x^2+2y-y^2-1-1-6\)
\(=-\left(x^2-2xy+y^2\right)-\left(x^2-2x+1\right)-\left(y^2-2y+1\right)-6\)
\(=-\left(x-y\right)^2-\left(x-1\right)^2-\left(y-1\right)^2-6\le-6\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y-1=0\end{matrix}\right.\Leftrightarrow x=y=1\)
Vậy ...
5 tháng 11 2017
\(\text{1) }\dfrac{x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^2-1}\\ =\dfrac{\left(x^7+x^6\right)+\left(x^5+x^4\right)+\left(x^3+x^2\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(x^6+x^4+x^2+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6+x^4+x^2+1}{x-1}\)
\(\text{3) }\dfrac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\\ =\dfrac{\left(x^2-2xy+y^2\right)+\left(2xz-2yz\right)+z^2}{\left(x^2-2xy+y^2\right)-z^2}\\ =\dfrac{\left(x-y\right)^2+2\left(x-y\right)z+z^2}{\left(x-y\right)^2-z^2}\\ =\dfrac{\left(x-y+z\right)^2}{\left(x-y+z\right)\left(x-y-z\right)}\\ =\dfrac{x-y+z}{x-y-z}\)
ND
1
ND
Nguyễn Đức Trí
VIP
6 tháng 8 2023
\(A=\left(x^2-4y^2\right)\left(x^2-2xy+4y^2\right)\left(x^2+2xy+4y^2\right)\)
\(A=\left(x-2y\right)\left(x+2y\right)\left(x^2-2xy+4y^2\right)\left(x^2+2xy+4y^2\right)\)
\(A=\left(x-2y\right)\left(x^2+2xy+4y^2\right)\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
\(A=\left[x^3-\left(2y\right)^3\right]\left[x^3+\left(2y\right)^3\right]\)
\(A=\left[x^3-8y^3\right]\left[x^3+8y^3\right]\)
\(A=x^6-64y^6\)
\(x^2y^4+2xy^2+1=\left(xy^2\right)^2+2.xy^2.1+1^2=\left(xy^2+1\right)^2\)
Áp dụng hằng đẳng thức thứ nhất: \(\left(a+b\right)^2=a^2+2ab+b^2\)
\(2xy^2+x^2y^4+1\)
\(=\left(xy^2\right)^2+2xy^2+1\)
\(=\left(xy^2+1\right)^2\)