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1
a, 2x2+4x+2-2y2 = 2(x2+2x+1-y2)= 2[(x+1)2-y2 ] = 2(x-y+1)(x+y+1)
b, 2x - 2y - x2 + 2xy - y2= 2(x -y) - (x2 - 2xy + y2) = 2(x-y)-(x-y)2=(x-y)(2-x+y)
c, x2-y2-2y-1=x2-(y2+2y+1)=x2-(y+1)2=(x-y-1)(x+y+1)
d, x2-4x-2xy-4y+y2= x2-2xy+y2-4x-4y=(x-y)
2.
a, x2-3x+2=x2-x-2x+2=x(x-1)-2(x-1)=(x-2)(x-1)
b, x2+5x+6=x2+2x+3x+6=x(x+2)+3(x+2)=(x+3)(x+2)
c, x2+6x-6=
Bài 1 :
x2-2x+2>0 với mọi x
=x2-2.x.1/4+1/16+31/16
=(x-1/4)2 + 31/16
Vì (x-1/4)2 \(\ge\) 0 nên (x-1/4)2 + 31/16 \(\ge\) 0 với mọi x (đfcm)
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
a) 5xy ( x - y ) - 2x + 2y
= 5xy ( x - y ) - 2 ( x - y )
= ( x - y ) ( 5xy - 2 )
b) 6x-2y-x(y-3x)
= 2 ( y - 3x ) - x ( y - 3x )
= ( y - 3x ( ( 2 - x )
c) x2 + 4x - xy-4y
= x ( x + 4 ) - y ( x + 4 )
( x + 4 ) ( x - y )
d) 3xy + 2z - 6y - xz
= ( 3xy - 6y ) + ( 2z - xz )
= 3y ( x - 2 ) + z ( x - 2 )
= ( x - 2 ) ( 3y + z )
a,5xy(x-y)-2x+2y=5xy(x-y)-2(x-y)=(x-y)(5xy-2)
b,6x-2y-x(y-3x)=-2(y-3x)-x(y-3x)=(y-3x)(-2-x)
c,x^2+4x-xy-4y=x(x+4)-y(x+4)=(x+4)(x-y)
d,3xy+2z-6y-xz=(3xy-6y)+(2z-xz)=3y(x-2)+z(2-x)=3y(x-2)-z(x-2)=(x-2)(3y-z)
11)
a,4-9x^2=0
(2-3x)(2+3x)=0
2-3x=0=>x=2/3 hoặc 2+3x=0=>x=-2/3
b,x^2 +x+1/4=0
(x+1/2)^2 =0
x+1/2=0
x=-1/2
c,2x(x-3)+(x-3)=0
(x-3)(2x+1)=0
x-3=0=>x=3 hoặc 2x+1=0=>x=-1/2
d,3x(x-4)-x+4=0
3x(x-4)-(x-4)=0
(x-4)(3x-1)=0
x-4=0=>x=4 hoặc 3x-1=0=>x=1/3
e,x^3-1/9x=0
x(x^2-1/9)=0
x(x+1/3)(x-1/3)=0
x=0 hoặc x+1/3=0=>x=-1/3 hoặc x-1/3=0=>x=1/3
f,(3x-y)^2-(x-y)^2 =0
(3x-y-x+y)(3x-y+x-y)=0
2x(4x-2y)=0
4x(2x-y)=0
x=0hoặc 2x-y=0=>x=y/2
A) \(\left(x-3\right)^2-\left(x+2\right)^2\)
\(=\left(x-3-x-2\right)\left(x-3+x+2\right)\)
\(=-5.\left(2x-1\right)\)
B) \(\left(4x^2+2xy+y^2\right)\left(2x-y\right)-\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
\(=\left(2x\right)^3-y^3-\left[\left(2x\right)^3+y^3\right]\)
\(=8x^3-y^3-8x^3-y^3\)
\(=-2y^3\)
C) \(x^2+6x+8\)
\(=x^2+6x+9-1\)
\(=\left(x+3\right)^2-1\)
\(=\left(x+3-1\right)\left(x+3+1\right)\)
\(=\left(x+2\right)\left(x+4\right)\)
bài 3 A) \(x^2-16=0\)
\(\left(x-4\right)\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
vậy \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
B) \(x^4-2x^3+10x^2-20x=0\)
\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\left(x^3+10x\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^3+10x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x\left(x^2+10\right)=0\\x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
vậy \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Bài 2:
a: \(3x^2-3xy=3x\left(x-y\right)\)
b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)
c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)
d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)
Bài 1:Phân tích các đa thức sau:
\(a,4x^2-6x\\ =2x(2x-3)\\ b,x^3y-2x^2y+5xy\\ = xy(x^2-2x+5)\\ c,2x^2(x+1) +4x(x+1)\\ =2x(x+1)(x+2)\\ d,\frac{2}{5}x.(y-1) -\frac{2}{5}x.(1-y)\\ =\frac{2}{5}x.(y-1)+\frac{2}{5}x.(y-1)\\ =2.\bigg[\frac{2}{5}x.(y-1)\bigg]\)
Bài 2 tính bằng cách hợp lý
\(a, 8,4.84, 5+840.0, 155\\ =8,4.(84,5+100.0,155)\\ =8,4.100\\ =840\\ b, 0,78.1300+50.6, 5-39\\ =(0,78.1300-39)+50. 6,5\\ =0,78.(1300-50)+50. 6,5\\ =0,78.1250+50. 6,5\\ =50.(0,78.25+6,5)\\ =1300\\ c,0, 12.90-110.0, 6+36-25.6\\ =6.(15.0,12-110.0,1+6-25)\\ =6.-28,2\\ =-169.2\)
Bài 3 Phân tích các đa thức sau\(a, (3x+1) ^2-(3x-1) ^2\\ =(3x+1-3x+1)(3x+1+3x-1)\\ =2.6x\\ b, (x+y) ^2-(x-y) ^2\\ =(x+y-x+y)(x+y+x-y)\\ =2y.2x\\ =2.(x-y)\\ c,(x+y)^3-(x-y) ^3\\ =(x+y-x+y)\bigg[(x+y)^2+(x+y)(x-y)+(x-y)^2\bigg]\\ =2y(x^2+2xy+y^2+x^2-xy+xy-y^2+x^2-2xy+y^2)\\ =2y(3x^2+y^2)\)
Bài 1:
\(a,4x^2-6x=2x\left(2x-3\right)\\ b,x^3y-2x^2y+5xy=xy\left(x^2-2x+5\right)\\ c,2x^2\left(x+1\right)+4x\left(x+1\right)=2x\left(x+1\right)\left(x+2\right)\\ d,\frac{2}{5}x\left(y-1\right)-\frac{2}{5}y\left(1-y\right)\\ =\frac{2}{5}\left(y-1\right)\left(x+y\right)\)
Bài 2:
\(a,8,4\cdot84,5+840\cdot0,155\\ =840\left(0,845+0,155\right)\\ =840\cdot1=840\\ b,0,78\cdot1300+50\cdot6,5-39\\ =39\cdot2\cdot13-39+25\cdot2\cdot6,5\\ =39\left(26-1\right)+25\cdot13\\ =39\cdot25+25\cdot13\\ =25\left(39+13\right)\\ =25\cdot52\\ =1300\\ c,0,12\cdot90-110\cdot0,6+36-25\cdot6\\ =6\cdot2\cdot0,9-6\cdot11+6\cdot6-25\cdot6\\ =6\left(1,8-11+6-25\right)\\ =-28,2\cdot6\\ =-169,2\)
Bài 3:
\(a,\left(3x+1\right)^2-\left(3x-1\right)^2\\ =\left(3x+1-3x+1\right)\left(3x+1+3x-1\right)\\ =2\cdot6x\\ =12x\\ b,\left(x+y\right)^2-\left(x-y\right)^2\\ =\left(x+y-x+y\right)\left(x+y+x-y\right)\\ =2y\cdot2x\\ =4xy\\ c,\left(x+y\right)^3-\left(x-y\right)^3\\ =\left(x+y-x+y\right)\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\\ =2y\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\\ =2y\left(3x^2+y^2\right)\)