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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 7 2018
a, \(4x^2+4x+1=\left(2x\right)^2+2.2x.1+1^2\)
\(=\left(2x+1\right)^2\)
b, \(x^2y^4-1=\left(xy^2\right)^2-1^2=\left(xy^2-1\right)\left(xy^2+1\right)\)
d, \(8x^3-12x^2y+6xy^2-y^3=\left(2x\right)^3-3.\left(2x\right)^2y+3.2x.y^2-y^3=\left(2x-y\right)^3\)
e, \(1-2y+y^2=y^2-2y+1=y^2-2.y.1+1^2=\left(y-1\right)^2\)
f, \(\left(x-y\right)^2-4=\left(x-y\right)^2-2^2=\left(x-y-2\right)\left(x-y+2\right)\)
g,\(16x^2-\left(x-y\right)^2=\left(4x\right)^2-\left(x-y\right)^2=\left(4x-x+y\right)\left(4x+x-y\right)=\left(3x+y\right)\left(5x-y\right)\)
Chúc bạn học tốt.
30 tháng 8 2020
a) 16x2 - 9
= ( 4x )2 - 32
= ( 4x - 3 )( 4x + 3 )
b) 9a2 - 25b4
= ( 3a )2 - ( 5b2 )2
= ( 3a - 5b2 )( 3a + 5b2 )
c) 81 - y4
= 92 - ( y2 )2
= ( 9 - y2 )( 9 + y2 )
= ( 32 - y2 )( 9 + y2 )
= ( 3 - y )( 3 + y )( 9 + y2 )
d) ( 2x + y )2 - 1
= ( 2x + y )2 - 12
= ( 2x + y - 1 )( 2x + y + 1 )
e) ( x + y + z )2 - ( x - y - z )2
= [ x + y + z - ( x - y - z ) ][ x + y + z + ( x - y - z ) ]
= [ x + y + z - x + y + z ][ x + y + z + x - y - z ]
= [ 2y + 2z ].2x
= 2[ y + z ].2x
= 4x[ y + z ]
21 tháng 7 2022
a: \(\Leftrightarrow x^3+8-x^3-3x=5\)
=>3x=3
hay x=1
b: \(\Leftrightarrow x^3-8-x\left(x^2-1\right)=8\)
\(\Leftrightarrow x^3-8-x^3+x=8\)
=>x=16
c: =>x2+2=3
=>x2=1
=>x=1 hoặc x=-1
f: \(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+6y+9\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+3\right)^2=0\)
=>x=1 và y=-3
XN
3
31 tháng 7 2017
áp dụng công thức này mà lm câu a,b,e nhá:
\(A=ax^2+bx+c=a\left(x+\dfrac{b}{2a}\right)^2+\dfrac{4ac-b^2}{4a}\\ \left[{}\begin{matrix}A\ge\dfrac{4ac-b^2}{4a}\left(với\text{ }\text{ }\text{ }a\ge0\right)\\A\le\dfrac{4ac-b^2}{4a}\left(với\text{ }a< 0\right)\end{matrix}\right.\)
31 tháng 7 2017
\(C=x^2+2xy+y^2+4y^2=\left(x+y\right)^2+4y^2\ge0\)
đẳng thức xảy ra khi x=y=0
vậy MIN C=0 tại x=y=0
15 tháng 8 2017
1.
= 4x\(^{^{ }2}\)-4x-9x+9
=4x(x-1)-9(x-1)
=(4x-9)(x-1)
AH
Akai Haruma
Giáo viên
12 tháng 5 2020
i)
$I=x^4+4x^3-x^2-14x+6$
$=(x^4+4x^4+4x^2)-5x^2-14x+6$
$=(x^2+2x)^2-6(x^2+2x)+9+x^2-2x-3$
$=(x^2+2x-3)^2+(x^2-2x+1)-4$
$=(x-1)^2(x+3)^2+(x-1)^2-4$
$=(x-1)^2[(x+3)^2+1]-4\geq -4$
Vậy $I_{\min}=-4$ khi $(x-1)^2[(x+3)^2+1]=0\Leftrightarrow x=1$
k)
$K=x^4+2x^3-10x^2-16x+45$
$=(x^4+2x^3+x^2)-11x^2-16x+45$
$=(x^2+x)^2-12(x^2+x)+x^2-4x+45$
$=(x^2+x)^2-12(x^2+x)+36+(x^2-4x+4)+5$
$=(x^2+x-6)^2+(x-2)^2+5$
$=[(x-2)(x+3)]^2+(x-2)^2+5$
$=(x-2)^2[(x+3)^2+1]+5\geq 5$
Vậy $K_{\min}=5$ khi $(x-2)^2[(x+3)^2+1]=0\Leftrightarrow x=2$
AH
Akai Haruma
Giáo viên
12 tháng 5 2020
g)
$G=x^4+4x^3+10x^2+12x+11$
$=(x^4+4x^3+4x^2)+6x^2+12x+11$
$=(x^2+2x)^2+6(x^2+2x)+11$
Đặt $x^2+2x=t$. Khi đó $t=x^2+2x=(x+1)^2-1\geq -1\Rightarrow t+1\geq 0$
$\Rightarrow G=t^2+6t+11=(t+1)^2+4(t+1)+7\geq 7$
Vậy $G_{\min}=7$ khi $t=-1\Leftrightarrow (x+1)^2=0\Leftrightarrow x=-1$
h)
$H=x^4-6x^3+x^2+24x+18$
$=(x^4-6x^3+9x^2)-8x^2+24x+18$
$=(x^2-3x)^2-8(x^2-3x)+18$
$=(x^2-3x)^2-8(x^2-3x)+16+2$
$=(x^2-3x-4)^2+2\geq 2$
Vậy $H_{\min}=2$ khi $x^2-3x-4=0\Leftrightarrow x=4$ hoặc $x=-1$
28 tháng 9 2017
a)x3-7x+6
=x3+0x2-7x+6
=x3-x2+x2-x-6x+6
=(x3-x2)+(x2-x)-(6x-6)
=x2(x-1)+x(x-1)-6(x-1)
=(x-1)(x2+x-6)
=(x-1)(x2-2x+3x-6)
=(x-1)[x(x-2)+3(x-2)]
=(x-1)(x+3)(x-2)
\(\left(x-y\right)^2-4=\left(x-y-2\right)\left(x-y+2\right)\)
\(9-\left(x-y\right)^2=\left(3-x+y\right)\left(3+x-y\right)\)
\(\left(x^2+4\right)^2-16x^2=\left(x^2-4x+4\right)\left(x^2+4x+4\right)=\left(x-2\right)^2\left(x+2\right)^2\)
\((X-y)^2-4=(x-y-2)(x-y+2)\)\((X^2+4)^2-16x^2=(x^2+4)^2-(4x)^2=(x^2+4-4x)(x^2+4+4x)\)
\(9-(x-y)^2=(3-x+y)(3+x-y)\)