|2x -6| +( 2y - 8)2 = 0
Những câu hỏi liên quan
Hãy giải các phương trình sau đây :
1, x2 - 4x + 4 = 0
2, 2x - y = 5
3, x + 5y = - 3
4, x2 - 2x - 8 = 0
5, 6x2 - 5x - 6 = 0
6,( x2 - 2x )2 - 6 (x2 - 2x ) + 5 = 0
7, x2 - 20x + 96 = 0
8, 2x - y = 3
9, 3x + 2y = 8
10, 2x2 + 5x - 3 = 0
11, 3x - 6 = 0
1) Ta có: \(x^2-4x+4=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
Vậy: S={2}
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Tim x,y,z biet: a,(3x−2y)6+(y−5z)8+|z−2|0(3x−2y)6+(y−5z)8+|z−2|0 b,3x−2y42x−4z3y−3z23x−2y42x−4z3y−3z2va x+y+z990Gấp gấp gấp!Tim x,y,z biet: a,(3x−2y)6+(y−5z)8+|z−2|0(3x−2y)6+(y−5z)8+|z−2|0 b,3x−2y42x−4z3y−3z23x−2y42x−4z3y−3z2va x+y+z990Gấp gấp gấp!
Đọc tiếp
Tim x,y,z biet: a,(3x−2y)6+(y−5z)8+|z−2|=0(3x−2y)6+(y−5z)8+|z−2|=0
b,3x−2y4=2x−4z3=y−3z23x−2y4=2x−4z3=y−3z2va x+y+z=990
Gấp gấp gấp!
Tim x,y,z biet: a,(3x−2y)6+(y−5z)8+|z−2|=0(3x−2y)6+(y−5z)8+|z−2|=0
b,3x−2y4=2x−4z3=y−3z23x−2y4=2x−4z3=y−3z2va x+y+z=990
Gấp gấp gấp!
vãi 4 năm mà ko mtj thằng nào rep đã thế còn gấp
phân tích đa thức sau thành nhân tử
1,3x^2+x-2
2, 2x^2-3xy-2y^2
3, 2x^2-3xy-2y^2
4, x^2+4xy+2x+3y^2+6
5, x^8+x+1
Tìm x,y biết
1, x^2+2x+5+y^2-4y=0
2,4x^2+y^4-20x-2y=26=0
mik ko bít
I don't now
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Tìm x,y biết
a) (2y-1)^1000-(3+y)^1000=0
b) (x-2/9)^3=(2/3)^6
c) (2x-1)^6=(2x-1)^8
a) \(\left(2y-1\right)^{1000}-\left(3+y\right)^{1000}=0\)
\(\Rightarrow\left(2y-1\right)^{1000}=\left(3+y\right)^{1000}\)
\(\Rightarrow2y-1=3+y\)
\(2y-y=3+1\)
\(y=4\)
b) \(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)
\(\left(x-\frac{2}{9}\right)^3=\left(\left(\frac{2}{3}\right)^2\right)^3\)
\(\Rightarrow x-\frac{2}{9}=\left(\frac{2}{3}\right)^2\)
\(x-\frac{2}{9}=\frac{4}{9}\)
\(x=\frac{2}{3}\)
c) \(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\left(\left(2x-1\right)^3\right)^2=\left(\left(2x-1\right)^4\right)^2\)
\(\Rightarrow\left(2x-1\right)^3=\left(2x-1\right)^4\)
\(8x^3-1=16x^4-1\)
\(16x^4-8x^3=0\)
\(8x^3\left(2x-1\right)=0\)
Nếu \(8x^3=0\) thì \(x^3=0\Rightarrow x=0\)
Nếu \(2x-1=0\)thì \(2x=1\Rightarrow x=\frac{1}{2}\)
Vậy x=0 và x=1/2
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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 các hệ phương trình sau:
a) left{{}begin{matrix}4x^2-4xy-14x-3y^2+y+1005sqrt{xy}+2x+2y6sqrt{y}-8end{matrix}right.
b) left{{}begin{matrix}2x^4+3x^2y+4x^2-2y^2+3y+20sqrt{xleft(y-1right)}+2y+2sqrt{y-1}3x+2sqrt{x}+2end{matrix}right.
c) left{{}begin{matrix}x^6+3x^2-y^3-6y^2-15y-140sqrt{xy+2x-y-2}+6x-2y10end{matrix}right.
d) left{{}begin{matrix}xy+x+yx^2-2y^2xsqrt{2y}-ysqrt{x-1}2x-2yend{matrix}right.
Đọc tiếp
Giải các hệ phương trình sau:
a) \(\left\{{}\begin{matrix}4x^2-4xy-14x-3y^2+y+10=0\\5\sqrt{xy}+2x+2y=6\sqrt{y}-8\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}2x^4+3x^2y+4x^2-2y^2+3y+2=0\\\sqrt{x\left(y-1\right)}+2y+2\sqrt{y-1}=3x+2\sqrt{x}+2\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}x^6+3x^2-y^3-6y^2-15y-14=0\\\sqrt{xy+2x-y-2}+6x-2y=10\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}xy+x+y=x^2-2y^2\\x\sqrt{2y}-y\sqrt{x-1}=2x-2y\end{matrix}\right.\)
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
\(\hept{\begin{cases}2x^2-y^2-7x+2y+6=0\\-7x^3+12x^2y-6xy^2+y^3-2x+2y=0\end{cases}}\)
Ta có: \(-7x^3+12x^2y-6xy^2+y^3-2x+2y=0\)
\(\Leftrightarrow\left(x^2y-x^3\right)-\left(xy^2-x^2y\right)+\left(2x^2y-2x^3\right)+\left(y^3-xy^2\right)-\left(4xy^2-4x^2y\right)+\left(4x^2y-4x^3\right)+\left(2y-2x\right)=0\)\(\Leftrightarrow\left(y-x\right)\left(x^2-xy+2x^2+y^2-4xy+4x^2+2\right)=0\)
\(\Leftrightarrow\left(y-x\right)\left[x^2-x\left(y-2x\right)+\left(y-2x\right)^2+2\right]=0\)
\(\Leftrightarrow\left(y-x\right)\left[\left(x-\frac{y-2x}{2}\right)^2+\frac{3}{4}\left(y-2x\right)^2+2\right]=0\)
Mà \(\left(x-\frac{y-2x}{2}\right)^2+\frac{3}{4}\left(y-2x\right)^2+2>0\left(\forall x,y\right)\)
\(\Rightarrow y-x=0\Leftrightarrow x=y\)
Khi đó \(HPT\Leftrightarrow\hept{\begin{cases}2x^2-y^2-7x+2y+6=0\\x=y\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x^2-x^2-7x+2x+6=0\\x=y\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x^2-5x+6=0\\x=y\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(x-2\right)\left(x-3\right)=0\\x=y\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\in\left\{2;3\right\}\\x=y\end{cases}}\)
Vậy ta có 2 cặp (x;y) thỏa mãn: \(\left(2;2\right);\left(3;3\right)\)
tim x,y thuoc Z biet
|y|.|2x+3|=8
|2x+4|+|y-3|=0
|x-1|+|2y+7|=3
|x+5|+|2y+6| nho hon hoac bang 0