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1/
a)5x – 20y=5(x-4y)
b) 5x.(x – 1) – 3x(x – 1)=2x(x-1)
c) x.(x+y) – 5x – 5y=c) x.(x+y) – 5(x+y)=(x-5)(x+y)
2/
a)x2 + xy + x = x(x+y+1)=77.(77+22+1)=77.100=7700
b) x . ( x – y ) + y . ( y – x )=(x-y)(x-y)=(x-y)2=(53-3)2=2500
3/
a) X + 5x2 = 0
⇒x(x+5)=0
⇒hoặc x=0
x+5=0⇒x=-5
b)x + 1 = ( x + 1 )2
⇒(x + 1)-( x + 1 )2 =0
⇒x(x+1)=0
⇒ hoặc x=0
hoặc x+1=0⇒x=-1
a )
Ta có :
\(x^2+xy+x=x\left(x+y+1\right)\)
Thay \(x=77;y=22\)vào b/t , ta được :
\(77\left(77+22+1\right)=77.100=7700\)
Vậy \(x^2+xy+x=7700\)tại \(x=77;y=22\)
b )
Ta có :
\(x\left(x-y\right)+y\left(y-x\right)\)
\(=x\left(x-y\right)-y\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y\right)\)
\(=\left(x-y\right)^2\)
Thay \(x=53;y=3\)vào b/t , ta được :
\(\left(53-3\right)^2=50^2=2500\)
Vậy \(x\left(x-y\right)+y\left(y-x\right)=2500\) tại \(x=53;y=3\)
a.\(x^2+xy+x=x\left(x+y+1\right)=77\left(77+22+1\right)=77.100=7700\)
b.\(x\left(x-y\right)+y\left(y-x\right)=\left(x-y\right)^2=\left(53-3\right)^2=50^2=2500\)
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a) \(P=x\left(x-y\right)+y\left(x-y\right)=\left(x-y\right)\left(x+y\right)=x^2-y^2=5^2-4^2=9\)
b) \(Q=x\left(x^2-y\right)-x^2\left(x+y\right)+y\left(x^2-x\right)=x^3-xy-x^3-x^2y+x^2y-xy=0\)
\(x^2\left(y-1\right)-4\left(y-1\right)\\ =\left(y-1\right)\left(x^2-4\right)=\left(y-1\right)\left(x-2\right)\left(x+2\right)\)
a.
\(x^2+xy+x=x\left(x+y+1\right)\)
Tại \(x=77;y=22\Rightarrow x\left(x+y+1\right)=77\left(77+22+1\right)=77.100=7700\)
b.
\(x\left(x-y\right)+y\left(y-x\right)=x\left(x-y\right)-y\left(x-y\right)=\left(x-y\right)\left(x-y\right)=\left(x-y\right)^2\)
\(=\left(53-3\right)^2=50^2=2500\)
c.
\(x\left(x-1\right)-y\left(1-x\right)=x\left(x-1\right)+y\left(x-1\right)=\left(x+y\right)\left(x-1\right)\)
\(=\left(2001+1999\right)\left(2001-1\right)=4000.2000=8000000\)