tim x,y biet
a, x.(y-3)=25
b, 3x + xy - y= 1
c, 1/x + 1/y = 1/3
CMR
a)(3x-5)(3x+5)=9x^2-25
b)x^3-y^3=(x-y)(x^2+xy+y^2)
c)x^2+y^2=(x+y)^2-2xy
a) \(\left(3x-5\right)\left(3x+5\right)=9x^2-25\Leftrightarrow9x^2+15x-15x-25=9x^2-25\Leftrightarrow9x^2-25=9x^2-25\)(đúng)
b) \(x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)\Leftrightarrow x^3-y^3=x^3+x^2y+xy^2-x^2y-xy^2-y^3\Leftrightarrow x^3-y^3=x^3-y^3\)(đúng)
c) \(x^2+y^2=\left(x+y\right)^2-2xy\Leftrightarrow x^2+y^2=x^2+y^2+2xy-2xy\Leftrightarrow x^2+y^2=x^2+y^2\)(đúng)
a: \(\left(3x-5\right)\left(3x+5\right)\)
\(=9x^2+15x-15x-25\)
\(=9x^2-25\)
b: \(\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x^3+x^2y+xy^2-x^2y-xy^2-y^3\)
\(=x^3-y^3\)
c: \(\left(x+y\right)^2-2xy\)
\(=x^2+2xy+y^2-2xy\)
\(=x^2+y^2\)
CMR
a)(3x-5)(3x+5)=9x^2-25
b)x^3-y^3=(x-y)(x^2+xy+y^2)
c)x^2+y^2=(x+y)^2-2xy
a: Ta có: \(\left(3x-5\right)\left(3x+5\right)\)
\(=9x^2+15x-15x-25\)
\(=9x^2-25\)
b: Ta có: \(\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x^3+x^2y+xy^2-x^2y-xy^2-y^3\)
\(=x^3-y^3\)
c: Ta có: \(\left(x+y\right)^2-2xy\)
\(=x^2+2xy+y^2-2xy\)
\(=x^2+y^2\)
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: \(\left(x+3\right)\left(2y-1\right)=5\)
=>\(\left(x+3\right)\left(2y-1\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)
=>\(\left(x+3;2y-1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(-2;3\right);\left(2;1\right);\left(-4;-2\right);\left(-8;0\right)\right\}\)
Tim giá trị x;y nguyên:
a. xy-3x-y=0
b. x+y+xy=3
c. x+6=y( x-1 )
d. xy=x+y+1992
a,Ta có
xy-3x-y=0
<=>x(y-3)-(y-3)=3
<=>(x-1)(y-3)=3
x-1 | 1 | 3 | -1 | -3 |
y-3 | 3 | 1 | -3 | -1 |
x | 2 | 4 | 0 | -2 |
y | 4 | 6 | 2 | 0 |
b
x+y+xy=3
<=>(x+1)(y+1)=4
Vì x,y thuộc Z nên ta có bảng sau
x-1 | 1 | 4 | -1 | -4 | 2 | -2 |
y-1 | 4 | 1 | -4 | -1 | 2 | -2 |
x | 2 | 5 | 0 | -3 | 3 | -1 |
y | 5 | 2 | -3 | 0 | 3 | -1 |
c,
x+6=y(x-1)
<=>x+6-xy+y=0
<=>x(1-y)-(1-y)+7=0
<=>(x-1)(1-y)=-7
<=>(x-1)(y-1)=7
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Bài 2: Phân tích đa thức thành nhân tử
a) x2−xy+5y−25
b) xy−y2−3x+3y
c) x2(x−3)−4x+12
d) 2a(x+y)−x−y
e) 2x−4+5x2−10x
g) 10ax−5ay−2x+y
h) a2−2a+1−b2
a) x2-xy+5y-25
= x(2-y)+ 5(y-2)
= x(2-y)-5(2-y)
= (x-5)(2-y)
h: \(=\left(a-1-b\right)\left(a-1+b\right)\)
Tim x,y nguyên biết:
x/4=18/x+1
x-4/y-3=4/3 và x-y=5
(x-1)2014+ly-3l=0
(x-3).(2y+1)=7
xy+3x-7y=21
xy+3x-7y=23
Giúp hộ mik vs nhanh lên
1.Tim tat ca cac cap so nguyên sao cho x^3 -x^2y+3x-2y-5=0
2. Cho0<x,y,z =<1 . CMR : x/(1+y+xz) + y/(1+z+xy) +z/(1+x+yz) =< 3/(x+y+z)
bai 1:tim x(chu y dau * la dau nhan)
a)(x+1/4)+(3x-4)+2*(x-3)=1
b)2*(x-3)=3(x+2)-x+1
c)x*(x+3)+x(x-2)=2x*(x-1)
d)(x-1)*3x-2*(x+2)-2x=x(x-1)
a: \(\left(x+\dfrac{1}{4}\right)+\left(3x-4\right)+2\left(x-3\right)=1\)
=>\(x+\dfrac{1}{4}+3x-4+2x-6=1\)
=>\(6x-\dfrac{39}{4}=1\)
=>\(6x=1+\dfrac{39}{4}=\dfrac{43}{4}\)
=>\(x=\dfrac{43}{4}:6=\dfrac{43}{24}\)
b: \(2\left(x-3\right)=3\left(x+2\right)-x+1\)
=>\(2x-6=3x+6-x+1\)
=>2x-6=2x+7
=>-6=7(vô lý)
c: \(x\left(x+3\right)+x\left(x-2\right)=2x\left(x-1\right)\)
=>\(x^2+3x+x^2-2x=2x^2-2x\)
=>3x-2x=-2x
=>3x=0
=>x=0
d: \(\left(x-1\right)\cdot3x-2\left(x+2\right)-2x=x\left(x-1\right)\)
=>\(3x^2-3x-2x-4-2x=x^2-x\)
=>\(3x^2-7x-4-x^2+x=0\)
=>\(2x^2-6x-4=0\)
=>\(x^2-3x-2=0\)
=>\(x=\dfrac{3\pm\sqrt{17}}{2}\)
Tim cac cap x;y thoa man:
a) (x-3).(2.y+1)=5
b) xy+3x-7y=21
c) xy+3x-2y=11
mình mới học lớp 5 thôi !
Thông cảm cho mình nhé Do uyen Linh !