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1 tháng 8 2017
\(1.\)
\(x^3z+x^2yz-x^2z^2-xyz^2\)
\(=x^3z-x^2z^2+x^2yz-xyz^2\)
\(=x^2z\left(x-z\right)-xyz\left(x-z\right)\)
\(=\left(x^2z-xyz\right)\left(x-z\right)\)
\(=xz\left(x-y\right)\left(x-z\right)\)
\(2.\)
\(x^2-\left(a+b\right)xy+aby^2\)
\(=x^2-axy-bxy+aby^2\)
\(=x^2-bxy-axy+aby^2\)
\(=x\left(x-by\right)-ay\left(x-by\right)\)
\(=\left(x-ay\right)\left(x-by\right)\)
\(3.\)
\(ab\left(x^2+y^2\right)+xy\left(x^2+y^2\right)\)
\(=abx^2+aby^2+a^2xy+b^2xy\)
\(=abx^2+b^2xy+a^2xy+aby^2\)
\(=bx\left(ax+by\right)+ay\left(ax+by\right)\)
\(=\left(ax+by\right)\left(bx+ay\right)\)
\(4.\)
\(\left(xy+ab\right)^2+\left(ay-bx\right)^2\)
\(=x^2y^2+2abxy+a^2b^2+a^2y^2-2aybx+b^2x^2\)
\(=x^2y^2+a^2b^2+a^2y^2+b^2x^2\)
\(=x^2y^2+b^2x^2+a^2b^2+a^2y^2\)
\(=x^2\left(b^2+y^2\right)+a^2\left(b^2+y^2\right)\)
\(=\left(a^2+x^2\right)\left(b^2+y^2\right)\)
\(5.\)
\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2b-a^2c+b^2c-ab^2+ac^2-bc^2\)
\(=a^2b-ab^2-a^2c-b^2c+ac^2-bc^2\)
\(=ab\left(a-b\right)-c\left(a^2-b^2\right)+c^2\left(a-b\right)\)
\(=ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(ab-ac-bc+c^2\right)\)
\(=\left(a-b\right)\left(ab-bc-ac+c^2\right)\)
\(=\left(a-b\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]\)
\(=\left(a-c\right)\left(b-c\right)\left(a-c\right)\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
\(6.\)
\(16x^2-40xy+2y^2\)
\(=\left(4x\right)^2-2\cdot4\cdot5xy+\left(5y\right)^2\)
\(=\left(4x-5y\right)^2\)
\(7.\)
\(25x^4-10x^2y+y^2\)
\(=\left(5x^2\right)^2-2\cdot5x^2y+y^2\)
\(=\left(5x^2+y\right)^2\)
\(8.\)
\(-16x^4y^6-24x^5y^5-9x^6y^4\)
\(=-\left(4^2x^4y^6+2\cdot4\cdot3x^5y^5+3^2x^6y^4\right)\)
\(=-\left[\left(4x^2y^3\right)^2+2\left(4x^2y^3\right)\left(3x^3y^2\right)+\left(3x^3y^2\right)^2\right]\)
\(=\left(4x^2y^3+3x^3y^2\right)^2\)
\(9.\)
\(16x^2-4y^2-8x+1\)
\(=\left(4x\right)^2-\left(2y\right)^2-8x+1\)
\(=\left(4x\right)^2-8x+1-\left(2y\right)^2\)
\(=\left(4x+1\right)^2-\left(2y\right)^2\)
\(=\left(4x-2y+1\right)\left(4x+2y+1\right)\)
\(10.\)
\(49x^2-25+42xy+9y^2\)
\(=\left(7x\right)^2-5^2+2\cdot7\cdot3xy+\left(3y\right)^2\)
\(=\left(7x\right)^2+2\cdot7\cdot3xy+\left(3y\right)^2-5^2\)
\(=\left(7x+3y\right)^2-5^2\)
\(=\left(7x+5y+5\right)\left(7x+3y-5\right)\)
29 tháng 6 2021
Bài 2 :
\(A=4x^2-2.2x.2+4+1\)
\(=\left(2x-2\right)^2+1\)
Thấy : \(\left(2x-2\right)^2\ge0\)
\(A=\left(2x-2\right)^2+1\ge1\)
Vậy \(MinA=1\Leftrightarrow x=1\)
\(B=\left(5x\right)^2-2.5x.1+1-4\)
\(=\left(5x-1\right)^2-4\)
Thấy : \(\left(5x-1\right)^2\ge0\)
\(\Rightarrow B=\left(5x-1\right)^2-4\ge-4\)
Vậy \(MinB=-4\Leftrightarrow x=\dfrac{1}{5}\)
\(C=\left(7x\right)^2-2.7x.2+4-5\)
\(=\left(7x-2\right)^2-5\)
Thấy : \(\left(7x-2\right)^2\ge0\)
\(\Rightarrow C=\left(7x-2\right)^2-5\ge-5\)
Vậy \(MinC=-5\Leftrightarrow x=\dfrac{2}{7}\)
29 tháng 6 2021
\(1.\)
\(A=-x^2-10x+1=-\left(x^2+10x-1\right)\)
\(=-\left(x^2+2.5x+5^2-5^2-1\right)=-\left[\left(x+5\right)^2-26\right]\)
\(=-\left(x+5\right)^2+26\le26\) dấu "=" xảy ra<=>x=-5
\(B=-4x^2-6x-5=-4\left(x^2+\dfrac{6}{4}x+\dfrac{5}{4}\right)\)
\(=-4\left(x^2+2.\dfrac{3}{4}x+\dfrac{9}{16}+\dfrac{11}{16}\right)\)\(=-4\left[\left(x+\dfrac{3}{2}\right)^2+\dfrac{11}{6}\right]\le-\dfrac{11}{4}\)
\(C=-16x^2+8x-1=-16\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)\)
\(=-16\left(x^2-2.\dfrac{1}{4}x+\dfrac{1}{16}\right)=-16\left(x-\dfrac{1}{4}\right)^2\le0\)
dấu"=" xảy ra<=>x=1/4
AH
Akai Haruma
Giáo viên
21 tháng 8 2023
Lời giải:
a.
$(xy)^2-xy-2=(x^2y^2+xy)-(2xy+2)$
$=xy(xy+1)-2(xy+1)=(xy+1)(xy-2)$
b. Bạn xem lại đoạn $-16x^2$ là dấu - hay + vậy?
AN
25 tháng 10 2017
\(x^8+64\)
\(=x^8+16x^4+64-16x^4\)
\(=\left(x^4\right)^2+2.x^4.8+8^2-16x^4\)
\(=\left(x^4+8\right)^2-\left(4x^2\right)^2\)
\(=\left(x^4+8-4x^2\right)\left(x^4+8+4x^2\right)\)
2 tháng 11 2017
5x2+11x+6
=5x2+5x+6x+6
=(5x2+5x)+(6x+6)
=5x(x+1)+6(x+1)
=(x+1)(5x+6)
TA
2
21 tháng 8 2017
a)\(=3\left(x-5\right)^2\)
b)\(=x\left(y-2x-1\right)\)
c)\(=\left(x+1\right)\left(x-8\right)\)
d)\(=4x\left(x+y+2\right)\left(x+y-2\right)\)
e)\(=\left(y+z\right)\left(x-2\right)\)
g)\(=\left(x+3+y\right)\left(x+3-y\right)\)
coi lại đề đi