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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 – 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) 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+2y2+3y2)-1
P= 4y2 - 1
câu b) giống sai đề rùi bn!!!!!!!
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.