(y^2-2y+1)-9=0
Những câu hỏi liên quan
Tìm các cặp số (x;y) nguyên
x^2-2y+x-2=-7
(x+2)^2+y^2-2y+1=0
x^2-6x+9+(2y-4)^4=0
Tìm x,y thuộc Z thỏa mãn:
a,|2x+4|+|y-6|=0
b,|x-5|+|2y-2|=0
c,(x-2)(2y+1)=8
d,(8x)(4y+1)=20
e,(x+1)(xy-1)=3
g,(x+y-2)(2y+1)=9
a , |2x+4|+|y-6|=0
=> 2 x + 4 = 0 => x = 0
=> y - 6 = 0 => y = 6
Vậy x = 0 và y = 6
Đúng 0
Bình luận (0)
Bài 2 :Giaỉ các phương trình sau
a) y(y2 -1)y2-5y+60
b)y(y-dfrac{1}{2})(2y+5)0
c)4y2+14y
d)y2-2y80
e)(2y-1)2-(y+3)20
f)2y2-11y0
g)(2y-3)(y+1)+y(y-2)3(y+2)2
h)(y2-2y+1)-90
i)y2+5y+60
k) y2+7y+2o
l)y2-y-120
m)x2+2x+70
n)y3-y2-21y+450
p)2y3-5y2+8y-30
q) (y+3)2 +(y+5)20
Đọc tiếp
Bài 2 :Giaỉ các phương trình sau
a) y(y2 -1)=y2-5y+6=0
b)y(y-\(\dfrac{1}{2}\))(2y+5)=0
c)4y2+1=4y
d)y2-2y=80
e)(2y-1)2-(y+3)2=0
f)2y2-11y=0
g)(2y-3)(y+1)+y(y-2)=3(y+2)2
h)(y2-2y+1)-9=0
i)y2+5y+6=0
k) y2+7y+2=o
l)y2-y-12=0
m)x2+2x+7=0
n)y3-y2-21y+45=0
p)2y3-5y2+8y-3=0
q) (y+3)2 +(y+5)2=0
c.
\(4y^2+1=4y\)
\(\Leftrightarrow4y^2-4y+1=0\)
\(\Leftrightarrow4y^2-2y-2y+1=0\)
\(\Leftrightarrow2y\left(2y-1\right)-\left(2y-1\right)=0\)
\(\Leftrightarrow\left(2y-1\right)^2=0\)
\(\Leftrightarrow y=0\)
d.
\(y^2-2y=80\)
\(\Leftrightarrow y^2-2y-80=0\)
\(\Leftrightarrow y^2-10y+8y-80=0\)
\(\Leftrightarrow y\left(y-10\right)+8\left(y-10\right)=0\)
\(\Leftrightarrow\left(y+8\right)\left(y-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y+8=0\\y-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=-8\\y=10\end{matrix}\right.\)
Đúng 0
Bình luận (1)
1)\(\begin{cases}x^4+2xy+6y-7x^2-2x^2y+9=0\\2x^2y-x^3=10\end{cases}\)
2)\(\begin{cases}2x^3-x^2y+x^2+y^2-2xy-y=0\\xy+x-2=0\end{cases}\)
Bài 1 Tìm cặp số (x;y) thỏa mãn biểu thức sau
2x^2+y^2-2xy-10x+6y+13=0
x^2+7y^2-4xy-2x-2y+4=0
11x^2+y^2-6xy-14x+2y+9=0
Tìm x,y biết :
1,(x-3)(y-1)=7
2,xy+3x-7y=21
3,xy+3x-2y=11
4,(x+1)(y-1)=-2
5,|x|=2x-6
6,|2y-4|<2
7,x(x+2)<0
8,x(x-y)=5
9,x(x-2)<0
10,(x+2)(3-x)>0
11,(x-2y)(y-1)=5
Giải phương trình:
1. \(\left\{{}\begin{matrix}5x-2y=-9\\4x+3y=2\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}2x+y-4=0\\x+2y-5=0\end{matrix}\right.\)
3. \(\left\{{}\begin{matrix}2x+3y-7=0\\x+2y-4=0\end{matrix}\right.\)
4. \(\left\{{}\begin{matrix}5x+6y=17\\9x-y=7\end{matrix}\right.\)
1)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}15x-6y=-27\\8x+6y=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2y=5x+9\\23x=-23\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(-1;2\right)\)
2)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}2x+y=4\\2x+4y=10\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-3y=-6\\x=5-2y\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(1;2\right)\)
3)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=14\\3x+6y=12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y=4-x\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(2;1\right)\)
4)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}5x+6y=17\\54x-6y=42\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}59x=59\\y=9x-7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(1;2\right)\)
Đúng 4
Bình luận (0)
Bất phương trình nào sau đây là bất phương trình bậc nhất hai ẩn?
a) \(2x^2+3y>0\)
b) 2x + \(3y^2\le0\)
c) 2x + 3y > 0
d) \(2x^2-y^2+3x-2y< 0\)
e) 3y < 1
f) x - 2y \(\le1\)
g) x \(\le0\)
h) y > 0
i) 4(x-1) + 5(y-3) > 2x - 9
Bất phương trình bậc nhất 2 ẩn :
\(2x+3y>0\Rightarrow Câu\) \(C\)
\(x-2y\le1\Rightarrow Câu\) \(f\)
\(4\left(x-1\right)+5\left(y-3\right)>2x-9\)
\(\Leftrightarrow4x-4+5y-15-2x+9>0\)
\(\Leftrightarrow2x+5y-10>0\) \(\Rightarrow Câu\) \(i\)
Đúng 0
Bình luận (0)
CMR:
a) 4x^2-6x+9>0 với mọi số thực x
b) x^2+2y^2-2xy+y+1>0 với mọi số thực x,y
a. Ta có : \(4x^2-6x+9=4x^2-6x+\dfrac{9}{4}+\dfrac{27}{4}\)
\(=\left[\left(2x\right)^2-6x+\left(\dfrac{3}{2}\right)^2\right]+\dfrac{27}{4}\)
\(=\left(2x-\dfrac{3}{2}\right)^2+\dfrac{27}{4}\)
Vì \(\left(2x-\dfrac{3}{2}\right)^2\ge0\forall x\)
nên \(\left(2x-\dfrac{3}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}>0\forall x\)
b.Ta có : \(x^2+2y^2-2xy+y+1=\left(x^2+y^2-2xy\right)+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}\)
\(=\left(x-y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Vì \(\left(x-y\right)^2\ge0\forall x;y\)
\(\left(y+\dfrac{1}{2}\right)^2\ge0\forall y\)
nên \(\left(x-y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}>0\forall x;y\)
Đúng 0
Bình luận (0)