Viết biểu thức sau thành tích :
(x - y)\(^3\)- 1 - 3(x - y)(x - y -1)
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BÀI 2 a, x2+x+1=(x2+1/2*2*x+1/4)-1/4+1=(x+1/2)2 +3/4
MÀ (x+1/2)2>=0 với mọi giá trị của x .Dấu"=" xảy ra khi x+1/2=0 =>x=-1/2
=>(x+1/2)2+3/4>=3/4 với mọi giá trị của x .Dấu "=" xảy ra khi x=-1/2
=>x2+x+1 có giá trị nhỏ nhất là 3/4 khi x=-1/2
b,A=y(y+1)(y+2)(y+3)
=>A =[y(y+3)] [(y+1)(y+2)]
=>A=(y2+3y) (y2+3y+2)
Đặt X=y2+3y+1
=>A=(X+1)(X-1)
=>A=X2-1
=>A=(y2+3y+1)2-1
MÀ (y2+3y+1)2>=0 với mọi giá trị của y
=>(y2+3y+1)2-1>=-1
Vậy GTNN của Alà -1
c,B=x3+y3+z3-3xyz
=>B=(x3+y3)+z3-3xyz
=>B=(x+y)3-3xy(x+y)+z3-3xyz
=>B=[(x+y)3+z3]-3xy(x+y+z)
=>B=(x+y+z)(x2+2xy+y2-xz-yz+z2)-3xy(x+y+z)
=>B=(x+y+z)(x2+2xy+y2-xz-yz+z2-3xy)
=>B=(x+y+z)(x2+y2+z2-xy-xz-yz)
a) \(x^2-2=\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\)
b) \(y^3-13=\left(y-\sqrt{13}\right)\left(y^2+\sqrt{13}y+13\right)\)
c) \(2x^2-4=\left(\sqrt{2}x-2\right)\left(\sqrt{2}x+2\right)\)
d) \(\left(x-1\right)^3-\left(y+1\right)^3=\left(x-1-y-1\right)\left[\left(x-1\right)^2+\left(x-1\right)\left(y+1\right)+\left(y+1\right)^2\right]=\left(x-y-2\right)\left(x^2-2x+1+xy-y+x-1+y^2+2y+1\right)=\left(x-y-2\right)\left(x^2+y^2-x+y+xy+1\right)\)
2:
-8x^6-12x^4y-6x^2y^2-y^3
=-(8x^6+12x^4y+6x^2y^2+y^3)
=-(2x^2+y)^3
3:
=(1/3)^2-(2x-y)^2
=(1/3-2x+y)(1/3+2x-y)
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câu 1.
P= 2(x+y)(x-y)+(x-y)^2+(x+y)^2-4y^2
P= (x+y+x-y)^2-(2y)^2
P=(2x-2y)(2x+2y)
P=4(x^2-y^2)
câu 2.
a, x^3-2x^2-4xy^2+x= x(x^2-2x+1)-4xy^2
=x(x-1)^2-4xy^2
=x(x-1-2y)(x-1+2y)
b, (x+1)(x+2)(x+3)(x+4)-24= (x^2+5x+4)(x^2+5x+6)-24
Đặt x^2+5x+4= a
Lúc đó: (x+1)(x+2)(x+3)(x+4)-24= a(a+2)-24
= a^2+2a-24
=a^2+2a+1-25
= (a+1)^2-5^2
= (a+1-5)(a+1+5)
= (a-4)(a+6)
mà ta đặt x^2+5x+4=a => (x+1)(x+2)(x+3)(x+4)-24= (x^2+5x+4-4)(x^2+5x+4+6)
= (x^2+5x)(x^2+5x+10)
câu3. (x+2)^2= 4-x^2
=> (x+2)^2-4+x^2=0
=>. (x+2)^2-(2-x)(2+x)=0
=> (x+2)(x+2-2+x)=0
=> (x+2)2x=0
=> x+2=0 hoặc 2x=0
=> x=-2 hoặc x=0
1)P=2(x^2-y^2)+x^2-2xy+y^2+x^2+2xy+y^2-4y^2=2x^2-2y^2+2x^2+2y^2-4y^2=4x^2-4y^2 . 3) <=> x^2+4x+4-4+x^2=0
<=> 2x^2+4x=0 <=>2x(x+2)=0 <=>2x=0 hay x+2=0 <=>x=0 hay x=-2
= [\(\left(x-y\right)^3\)\(-1\)] - \(\left[3\left(x-y\right)\left(x-y-1\right)\right]\)
= \(\left\{\left(x-y-1\right)\left[\left(x-y\right)^2+2\left(x-y\right)+1\right]\right\}\)- \(\left[3\left(x-y\right)\left(x-y-1\right)\right]\)
= \(^{\left(x-y-1\right)\left[\left(x-y\right)^2+2\left(x-y\right)+1-3\left(x-y\right)\right]}\)
= \(\left(x-y-1\right)\left[\left(x-y\right)^2-2\left(x-y\right)+1\right]\)
= \(\left(x-y-1\right)\left(x-y-1\right)^2\)
= \(^{\left(x-y-1\right)^3}\)
T*ck mình nha. Suy nghĩ bài này cực lắm đó!
= (x-y-1) [(x-y)^2 + (x-y) + 1] - 3(x-y)(x-y-1)
= (x-y-1) [(x-y)^2 + (x-y) + 1 - 3(x-y)]
= (x-y-1) [(x-y)^2 - 2(x-y)+1]
= (x-y-1)(x-y-1)^2
=(x-y-1)^3
\(a,5\left(x-y\right)-3x\left(y-x\right)=5\left(x-y\right)+3x\left(x-y\right)=\left(5+3x\right)\left(x-y\right)\\ b,x^2-4xy+4y^2=\left(x-2y\right)^2\\ c,\left(x+1\right)^2+x\left(5-x\right)=0\\ \Rightarrow x^2+2x+1+5x-x^2=0\\ \Rightarrow7x+1=0\\ \Rightarrow7x=-1\\ \Rightarrow x=-\dfrac{1}{7}\)
a: =(x-y)(5+3x)
c: \(\Leftrightarrow x^2-2x+1+5x-x^2=0\)
hay x=-1/3
\(\left(x-y\right)^3-1-3\left(x-y\right)\left(x-y-1\right)\)
\(=\left(x-y-1\right)\left(x^2+\left(x-y\right)+1\right)-3\left(x-y\right)\left(x-y-1\right)\)
\(=\left(x-y-1\right)\left(x^2\left(x+y\right)+y^2-3\left(x-y\right)\right)\)
( x - y )3 - 1 - 3 ( x - y ) ( x - y - 1 )
= ( x - y - 1 ) [ x2 - ( x - y ) - 1 ) - 3(x - y ) ( x - y - 1 ]
= ( x - y - 1 ) [ x2 ( x - y ) - y 2 - 3 ( x - y ) ]
k mik nha làm ơn đó