Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)\(P+Q=\left(x^2y+xy^2-5x^2y^2+x^3\right)+\left(3xy^2-x^2y+x^2y^2\right)\)
=\(x^2y+xy^2-5x^2y^2+x^3+3xy^2-x^2y+x^2y^2\)
=\(x^2y-x^2y+xy^2+3xy^2-5x^2y^2+x^2y^2+x^3\)
=\(4xy^2-4x^2y^2+x^3\)
b)\(M+N=\left(x^3+xy+y^2-x^2y^2-2\right)+\left(x^2y^2+5-y^2\right)\)
=\(x^3+xy+y^2-x^2y^2-2+x^2y^2+5-y^2\)
=\(x^3+xy+y^2-y^2-x^2y^2+x^2y^2-2+5\)
=\(x^3+xy+3\)
Bài dài nên chắc sẽ có sai sót, nếu đúng bạn nha
a) Ta có: P = x2y + xy2 – 5x2y2 + x3 và Q = 3xy2 – x2y + x2y2
=> P + Q = x2y + xy2 – 5x2y2 + x3 + 3xy2 – x2y + x2y2
= x3 – 5x2y2 + x2y2 + x2y – x2y + xy2 + 3xy2
= x3 – 4x2y2 + 4xy2
b) Ta có: M = x3 + xy + y2 – x2y2 – 2 và N = x2y2 + 5 – y2.
=> M + N = x3 + xy + y2 – x2y2 – 2 + x2y2 + 5 – y2
= x3 – x2y2 + x2y2 + y2 – y2 + xy - 2 + 5
= x3 + xy + 3.
a)
P + Q = x2y + xy2 – 5x2y2 + x3 + 3xy2 – x2y + x2y2
= x3 – 5x2y2 + x2y2 + x2y – x2y + xy2 + 3xy2
= x3 – 4x2y2 + 4xy2
b)
M + N = x3 + xy + y2 – x2y2 – 2 + x2y2 + 5 – y2
= x3 – x2y2 + x2y2 + y2 – y2 + xy - 2 + 5
= x3 + xy + 3.
a, \(A=x^3-x^2y+3x^2-xy+y^2-4y+x+2\)
\(=x^3-x^2y+3x^2-\left(xy-y^2+3y\right)-y+x+3-1\)
\(=x^2\left(x-y+3\right)-y\left(x-y+3\right)+\left(x-y+3\right)-1\)
Thay x-y+3=0 vào A
\(A=x^2.0-y.0+0-1=-1\)
b, \(B=x^3-2x^2y+3x^2+xy^2-3xy-2y+2x+4\)
\(=x^3-x^2y-x^2y+3x^2+xy^2-3xy-2y+2x+4\)
\(=x^3-x^2y+3x^2-x^2y+xy^2-3xy+2x-2y+6-2\)
\(=x^2\left(x-y+3\right)-xy\left(x-y+3\right)+2\left(x-y+3\right)-2\)
Thay x-y+3=0 vào B
\(B=x^2.0-xy.0+2.0-2=-2\)
a) (5x2y-5xy2+xy) + (xy-x2y2+5xy2)
= 5x2y-5xy2+xy+xy-x2y2+5xy2
= 5x2y+(5xy2-5xy2)+(xy+xy)-x2y2
= 5x2y+2xy-x2y2
b) (x2+y2+z2) + (x2-y2+z2)
= x2+y2+z2+x2-y2+z2
= (x2+x2)+(y2-y2)+(z2+z2)
= 2x2+2z2
a)( \(5x^2y\)\(-\) \(5xy^2\) \(+\) \(xy\)) + (\(xy\) \(-\) \(x^2y^2\) \(+\) \(5xy^2\))
= \(5x^2y-5xy^2+xy+xy-x^2y^2+5xy^2\)
= \(5x^2y+2xy-x^2y^2\)
b) \(\left(x^2+y^2+z^2\right)+\left(x^2-y^2+z^2\right)\)
= \(x^2+y^2+z^2+x^2-y^2+z^2\)
=\(2x^2+2z^2\)
=\(2\left(x+z\right)^2\)
1) \(A=2xy^2+3xy-xy^2+5xy^2+5xy+1\)
a, \(A=2xy^2+3xy-xy^2+5xy^2+5xy+1\)
= \(6xy^2+8xy+1\)
b, giá trị của biểu thức tại x = 1 và y = 2 là:
\(A=6.1.2^2+8.1.2+1=41\)
2) và 3) bạ vt khó hiểu wa
2) đề bài này là tìm b.a.c á bn, ghi đề chưa rõ lắm nên tui cx pó tay
3)
a/ Có: \(4x+9=0\)
\(\Leftrightarrow4x=-9\Rightarrow x=-\dfrac{9}{4}\)
vậy.............
b/ Có: \(-5x+6=0\)
\(\Leftrightarrow-5x=-6\Rightarrow x=\dfrac{6}{5}\)
Vậy....................
c/ có: \(x^2-4=0\)
\(\Leftrightarrow x^2=4\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy ..................
d/ Có: \(9-x^2=0\)
\(\Leftrightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Vậy.............
e/ Có: \(\left(y+2\right)\left(3-y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y+2=0\\3-y=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=-2\\y=3\end{matrix}\right.\)
Vậy...............
p/s: bài 3 này thuộc dạng cơ bản nên lần sau nhớ suy nghĩ trc khi đăng câu hỏi
\(A=x^3-y^3-21xy\)
\(A=\left(x-y\right).\left(x^2+xy+y^2\right)-21xy\)
\(A=7.\left(x^2+xy+y^2\right)-21xy\)
\(A=7.\left(x^2+xy+y^2+3xy\right)\)
\(A=7.\left(x^2+2xy+y^2+2xy\right)\)
\(A=7.\text{[}\left(x+y\right)^2+2xy\text{]}\)
\(A=7.\left(7^2+2xy\right)\)
\(A=7^3+14xy\)
Ngáo rồi @@
\(\)
\(A=x^3-y^3-21xy\)
\(\Rightarrow A=\left(x-y\right)\left(x^2+xy+y^2\right)-21xy\)
\(\Rightarrow A=7\left(x^2+xy+y^2\right)-21xy\)
\(\Rightarrow A=7\left(x^2+xy+y^2-3xy\right)\)
\(\Rightarrow A=7\left(x^2+y^2-2xy\right)\)
\(\Rightarrow A=7\left(x-y\right)^2\)
\(\Rightarrow A=7.7^2\)
\(\Rightarrow A=7.49\)
\(\Rightarrow A=343\)
\(\left\{{}\begin{matrix}f\left(x\right)=3x^4+5yx^2-3yx+y^4+z^2\\M\left(x\right)=ax^4+bx^2+cx+D\end{matrix}\right.\)
\(f\left(x\right)+M\left(x\right)=\left(3+a\right)x^4+\left(5y+a\right)x^2+\left(-3y+c\right)x+y^4+z^2+D\)\(\Leftrightarrow\left\{{}\begin{matrix}a=-3\\b=-5y\\c=3y\end{matrix}\right.\)\(\Rightarrow M\left(x\right)=-3x^4-5yx^2+3yx+y^4+z^2+D\) với D tùy ý không chứa x
\(\int f\left(x\right)dx=x^3+C\)
\(\sum a\left(b^2-1\right)\left(c^2-1\right)\)
\(a\left(b^2-1\right)\left(c^2-1\right)+b\left(a^2-1\right)\left(c^2-1\right)+c\left(b^2-1\right)\left(a^2-1\right)\)
\(\begin{matrix}\sum a\left(b^2-1\right)\left(c^2-1\right)=\sum\left(ab^2-a\right)\left(c^2-1\right)=\sum\left(ab^2c^2-ab^2-ac^2+a\right)\\\left(ab^2c^2-ab^2-ac^2+a\right)+\\\left(a^2bc^2-ba^2-bc^2+b\right)+\\\left(a^2b^2c-b^2c-a^2c+c\right)\end{matrix}\)
\(a+b+c\Rightarrow a+b=abc-c\) \(\Rightarrow\sum ab\left(a+b\right)=\sum ab\left(abc-c\right)=\sum a^2b^2c-abc\)
\(\left[abc\left(bc+ac+ab\right)\right]-\left[ab\left(a+b\right)+ac\left(a+c\right)+bc\left(b+c\right)\right]+\left[\left(a+b+c\right)\right]\)
\(\sum a^2b^2c-abc=\left(-abc+a^2b^2c\right)+\left(-abc+a^2bc^2\right)+\left(-abc+ab^2c^2\right)=-3abc+abc\left(ab+bc+ac\right)\)
\(\left[abc\left(bc+ac+ab\right)\right]+3abc-abc\left(ab+bc+ac\right)+\left(a+b+c\right)=3abc+abc=4abc=VP\)
Ta có: P = x2y + xy2 – 5x2y2 + x3 và Q = 3xy2 – x2y + x2y2
⇒ P + Q = (x2y + xy2 – 5x2y2 + x3) + (3xy2 – x2y + x2y2)
= x2y + xy2 – 5x2y2 + x3 + 3xy2 – x2y + x2y2
= x3 +(– 5x2y2 + x2y2)+ (x2y – x2y) + (xy2+ 3xy2)
= x3 – 4x2y2 + 0 + 4xy2
= x3 – 4x2y2 + 4xy2