tim x,y biet
a,3x-2y=x.2y=2x:y
b,/4x-3/+/3xy-5/=0
Những câu hỏi liên quan
Tim so nguyen x,y biet
a) (x+5) mu 2 + (2y - 8 ) mu 2 = 0
b)(x + 3).(2y - 1 ) = 5
a: \(\left(x+5\right)^2>=0\forall x\)
\(\left(2y-8\right)^2>=0\forall y\)
Do đó: \(\left(x+5\right)^2+\left(2y-8\right)^2>=0\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+5=0\\2y-8=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-5\\y=4\end{matrix}\right.\)
b: \(\left(x+3\right)\left(2y-1\right)=5\)
=>\(\left(x+3\right)\left(2y-1\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)
=>\(\left(x+3;2y-1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(-2;3\right);\left(2;1\right);\left(-4;-2\right);\left(-8;0\right)\right\}\)
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Thực hiện phép chia:
a. (-2x^5+3x^2-4x^3):2x^2
b .(x^3-2x^2y+3xy^2):(-1/2x)
c. (3x^2y^2+6x^2y^3-12xy^2):3xy
d. (4x^3-3x^2y+5xy^2):0,5x
e. (18x^3y^5-9x^2y^2+6xy^2):3xy^2
f. (x^4+2x^2y^2+y^4):(x^2+y^2)
sau bạn đăng tách ra cho mn cùng giúp nhé
a, \(\left(-2x^5+3x^2-4x^3\right):2x^2=-x^3+\frac{3}{2}-2x\)
b, \(\left(x^3-2x^2y+3xy^2\right):\left(-\frac{1}{2}x\right)=-\frac{x^2}{2}+xy-\frac{3y^2}{2}\)
c, \(\left(3x^2y^2+6x^3y^3-12xy^2\right):3xy=xy+2x^2y^2-4y\)
d, \(\left(4x^3-3x^2y+5xy^2\right):\frac{1}{2}x=2x^2-\frac{3xy}{2}+\frac{5y^2}{2}\)
e, \(\left(18x^3y^5-9x^2y^2+6xy^2\right):3xy^2=6x^2y^3-3x+2\)
f, \(\left(x^4+2x^2y^2+y^4\right):\left(x^2+y^2\right)=\left(x^2+y^2\right)^2:\left(x^2+y^2\right)=x^2+y^2\)
a.4x^2y-3xy^2+xy+xy-x^2y+5xy^2
b.x^2+2y^2+3xy+x^2-3y^2+4xy
c.2x^y-3xy+4xy^2-5x^2y+2xy^2
d.(2x^3+3x^2-4x+1)-(3x+4x^3-5)
Tìm x
a) x(x^3+3x-4x)-(4x+3x^2)=20
b) (2\3xy-x^2+3xy^3)(2x^2-3xy^2+x^2y)
c) (2xy+3xy^2-x^2y)(xy+x^2y+y^2)
1) PTTNT
a) x^2 - 4x^2y + 4xy
b)x^2 + 3x + x - 3y
2) Tim GTLN
-2x^2 + 3x - 5
3) tim x,y thuoc z
3xy + 6x - y = 7
Bài 2:
\(A=-2x^2+3x-5\)
\(=-2\left(x^2+\frac{3x}{2}-\frac{5}{2}\right)\)
\(=-2\left(x^2-\frac{3x}{2}+\frac{9}{16}\right)-\frac{31}{8}\)
\(=-2\left(x-\frac{3}{4}\right)^2-\frac{31}{8}\le-\frac{31}{8}\)
Dấu = khi \(-2\left(x-\frac{3}{4}\right)^2=0\Leftrightarrow x-\frac{3}{4}=0\Leftrightarrow x=\frac{3}{4}\)
Vậy \(Max_A=-\frac{31}{8}\Leftrightarrow x=\frac{3}{4}\)
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Bài 1:
a)x2-4x2y+4xy
=x(x-4xy+y)
b)đề sai
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Bài 3:
3yx + 6x - y = 7
<=> x(3y+6) - (3y+6) = 27
<=> (3y+6)(x+1) = 27
Ta có bảng sau:
| x+1 | 1 | -1 | 3 | -3 | 9 | -9 | 27 | -27 | |
| 3y+6 | 27 | -27 | 9 | -9 | 3 | -3 | 1 | -1 | |
| x | 0 | -2 | 2 | -4 | 8 | -10 | 26 | -28 | |
| y | 7 | -11 | 1 | -5 | -1 | -3 | \(-\frac{5}{3}\) | \(-\frac{7}{3}\) |
Vậy...
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Tìm x,y,z biết: a) x^2+y^2-4x+4y+8=0 b) 5x^2-4xy+y^2=0 c) x^2+2y^2+z^2-2xy-2y-4z+5=0 d) 3x^2+3y^2+3xy-3x+3y+3=0 e) 2x^2+y^2+2z^2-2xy-2xz+2yz-2z-2z-2x+2=0
a) x2+y2-4x+4y+8=0
⇔ (x-2)2+(y+2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)
b)5x2-4xy+y2=0
⇔ x2+(2x-y)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
c)x2+2y2+z2-2xy-2y-4z+5=0
⇔ (x-y)2+(y-1)2+(z-2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)
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b: Ta có: \(5x^2-4xy+y^2=0\)
\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)
\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
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d)3x2+3y2+3xy-3x+3y+3=0
⇔ 6x2+6y2+6xy-6x+6y+6=0
⇔ 3(x+y)2+3(x-1)2+3(y+1)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
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giải phương trình :
a) x^2+2y^2-2xy+4x-3y-26=0
b)x^2+3y^2+2xy-2x-4y-3=0
c)2x^2+y^2+3xy+3x+2y+2=0
d)3x^2-y^2-2xy-2x-2y+8=0
Xem chi tiết
a) x^2+2y^2-2xy+4x-3y-26=0
b)x^2+3y^2+2xy-2x-4y-3=0
c)2x^2+y^2+3xy+3x+2y+2=0
d)3x^2-y^2-2xy-2x-2y+8=0
giải phương trình :
a) x^2+2y^2-2xy+4x-3y-26=0
b)x^2+3y^2+2xy-2x-4y-3=0
c)2x^2+y^2+3xy+3x+2y+2
d)3x^2-y^2-2xy-2x-2y+8=0
Xem chi tiết
a) x^2+2y^2-2xy+4x-3y-26=0
b)x^2+3y^2+2xy-2x-4y-3=0
c)2x^2+y^2+3xy+3x+2y+2
d)3x^2-y^2-2xy-2x-2y+8=0
tinh:
a)x^2.(4x-7x^3)
b)(x-2).(x^2+2x+4)
c)(x4-x^3-3x^2+x+2):x^2-1
d)(3xy-4x^3y^4+6x^2y^3):2
ptdt thanh nhan tu
x^3-4x
b)2x^2y+2xy^2-x-y
tim x:
x^2-5=0
ai lam dc cau nao lam ho minh nhe,mai mih can roi thank
Bài 2:
a: \(=x\left(x^2-4\right)=x\left(x-2\right)\left(x+2\right)\)
b: \(=2xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(2xy-1\right)\)
Bài 3:
=>x^2=5
hay \(x=\pm\sqrt{5}\)
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