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a)p+(x2 -2y2)=x2 -y2 +3y2 -1
\(\Rightarrow\)p=(x2 -2y2 +3y2 -1)-(x-2y).(x+2y)
\(\Rightarrow\)p=(x2-2y2 +3y2 -1)-(x2 +2xy-2xy)
\(\Rightarrow\)p=x2 -2y2+3y2 -1-x2 -2xy+2xy
\(\Rightarrow\) p=2y2-1
Vậy P=2y2-1
b)Q-(5x2-xyz)=xy+2x2-3xyz+5
\(\Rightarrow\)Q=(xy+2x2-3xyz+5)+(5x2-xyz)
\(\Rightarrow\)Q=7x2-4xyz+xy+5
Vậy Q=7x2-4xyz+xy+5
a) P + (x2 - 2y2) = x2 - y2 + 3y2 - 1
⇔ P = (x2 - y2 + 3y2 - 1) - (x2 - 2y2)
⇔ P = x2 + 2y2 - 1 - x2 + 2y2
⇔ P = 4y2 - 1
b) Q - (5x2 - xyz) = xy + 2x2 - 3xyz + 5
⇔ Q = (xy + 2x2 - 3xyz + 5) + (5x2 - xyz)
⇔ Q = xy + 2x2 - 3xyz + 5 + 5x2 - xyz
⇔ Q = xy + 7x2 - 4xyz + 5
a) P + (x2 – 2y2) = x2 – y2 + 3y2 – 1
P = (x2 – y2 + 3y2 – 1) - (x2 – 2y2)
P = x2 – y2 + 3y2 – 1 - x2 + 2y2
P = x2 – x2 – y2 + 3y2 + 2y2 – 1
P = 4y2 – 1.
Vậy P = 4y2 – 1.
b) Q – (5x2 – xyz) = xy + 2x2 – 3xyz + 5
Q = (xy + 2x2 – 3xyz + 5) + (5x2 – xyz)
Q = xy + 2x2 – 3xyz + 5 + 5x2 – xyz
Q = 7x2 – 4xyz + xy + 5
Vậy Q = 7x2 – 4xyz + xy + 5.
a) \(B=-\frac{1}{2}x^3y\left(-2xy^2\right)^2\)
\(B=\left(-\frac{1}{2}.-2\right).\left(x^3.x\right)\left(y.y^2\right)^2\)
\(B=1x^4y^5\)
Hệ số: 1
Bậc: 9
Chưa định hình phần b) nó là như nào
2:
a: A(x)=0
=>5x-10-2x-6=0
=>3x-16=0
=>x=16/3
b: B(x)=0
=>5x^2-125=0
=>x^2-25=0
=>x=5 hoặc x=-5
c: C(x)=0
=>2x^2-x-3=0
=>2x^2-3x+2x-3=0
=>(2x-3)(x+1)=0
=>x=3/2 hoặc x=-1
\(a.M+(5x^2-2xy)=6x^2+9xy-y^2
\)
\(M=(6x^2+9xy-y^2)-(5x^2-2xy)\)
\(M=6x^2+9xy-y^2-5x^2+2xy\)
\(M=(6x^2-5x^2)+(9xy+2xy)-y^2\)
\(M=x^2+11xy-y^2\)
Vậy \(M=x^2+11xy-y^2\)
\(b.M+(3x^2y-2xy^3)=2x^2y-4xy^3\)
\(M=(2x^2y-4xy^3)-(3x^2-2xy^3)\)
\(M=
\) \(2x^2-4xy^3-3x^2+2xy^3\)
\(M=(2x^2-3x^2)+(-4xy^3+2xy^3)\)
\(M=-x^2-2xy^3\)
Vậy \(M=-x^2-2xy^3\)
a) M + (5x\(^2\) - 2xy) = 6x\(^2\) + 9xy - y\(^2\)
=> M = (6x\(^2\) + 9xy - y\(^2\)) - (5x\(^2\) - 2xy)
M = 6x\(^2\) + 9xy - y\(^2\) - 5x\(^2\) + 2xy
M = (6x\(^2\) - 5x\(^2\)) + (9xy + 2xy) - y\(^2\)
M = 1x\(^2\) + 11xy - y\(^2\)
a) A+(x2+y2)=5x2+3y2−xy
⇒A=(5x2+3y2−xy)−(x2+y2)
=(5−1)x2+(3−1)y2−xy
=4x2+2y2−xy
b) A−(xy+x2−y2)=x2+y2
⇒A=(x2+y2)+(xy+x2-y2)
=(1+1)x2+(1−1)y2+xy
=2x2+xy
a) A+(x2+y2)=5x2+3y2−xy
⇒A=(5x2+3y2−xy)−(x2+y2)
=(5−1)x2+(3−1)y2−xy
=4x2+2y2−xy
b) A−(xy+x2−y2)=x2+y2
⇒A=(x2+y2)+(xy+x2-y2)
=(1+1)x2+(1−1)y2+xy
=2x2+xy