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\(x^2+14y^2+t^2+2xy+6yt-12y+9=0\)
\(\Leftrightarrow\)\(\left(x^2+2xy+y^2\right)+\left(t^2+6yt+9y^2\right)+\left(4y^2-12y+9\right)=0\)
\(\Leftrightarrow\)\(\left(x+y\right)^2+\left(t+3y\right)^2+\left(2y-3\right)^2=0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}x+y=0\\t+3y=0\\2y-3=0\end{cases}}\) \(\Leftrightarrow\)\(\hept{\begin{cases}x=-1,5\\t=-4,5\\y=1,5\end{cases}}\)
\(x^2+2xy+y^2+9y^2+6yt+t^2+4y^2-12y+9=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(3y+t\right)^2+\left(2y-3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\3y+t=0\\2y-3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3}{2}\\t=\frac{-9}{2}\\x=\frac{-3}{2}\end{matrix}\right.\)
pt <=> (x2 + 2xy + y2) + (t2 + 6yt + 9y2) + (4y2 - 12y + 9) = 0
<=> (x + y)2 + (t + 3y)2 + (2y - 3)2 = 0
<=> \(\left\{{}\begin{matrix}\left(x+y\right)^2=0\\\left(t+3y\right)^2=0\\\left(2y-3\right)^2=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=-y=-\dfrac{3}{2}\\t=-3y=-\dfrac{9}{2}\\y=\dfrac{3}{2}\end{matrix}\right.\)
Vậy ...
a)\(x^2+4y^2+6x-12y+18=0\)
\(\Leftrightarrow\left(x^2+2\cdot x\cdot3+9\right)+\left[\left(2y\right)^2-2\cdot2y\cdot3+9\right]=0\)
\(\Leftrightarrow\left(x+3\right)^2+\left(2y-3\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+3=0\\2y-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=\dfrac{3}{2}\end{matrix}\right.\)
b)\(2x^2+2y^2+2xy-10x-8y+41=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2-2\cdot x\cdot5+25\right)+\left(y^2-2.y.4+16\right)=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x-5\right)^2+\left(y-4\right)^2\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=0\\x-5=0\\y-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=5\\y=4\end{matrix}\right.\)(vô lý)
1.
a. x2 - 2x + 1 = 0
x2 - 2x*1 + 12 = 0
(x-1)2 = 0
............( tới đây tui bí rùi tự suy nghĩ rùi lm tiếp ik)
1, Tìm x biết:
a, x2 - 2x +1 = 0
(x-1)2 = 0
x-1 = 0
x = 1. Vậy ...
b, ( 5x + 1)2 - (5x - 3) ( 5x + 3) = 30
25x2 +10x + 1 - (25x2 -9) = 30
25x2 +10x + 1 - 25x2 +9 = 30
10x + 10 =30
10(x+1) = 30
x+1 =3
x = 2. vậy ...
c, ( x - 1) ( x2 + x + 1) - x ( x +2 ) ( x - 2) = 5
(x3 - 1) - x(x2 -4) = 5
x3 - 1 - x3 + 4x = 5
4x - 1 = 5
4x = 6
x = \(\dfrac{3}{2}\) .vậy ...
d, ( x - 2)3 - ( x - 3) ( x2 + 3x + 9 ) + 6 ( x + 1)2 = 15
x3 - 6x2 + 12x - 8 - (x3 - 27) + 6 (x2 + 2x +1) =15
x3 - 6x2 + 12x - 8 - x3 + 27 + 6x2 + 12x +6 =15
24x + 25 = 15
24x = -10
x = \(\dfrac{-5}{12}\) vậy ...
1. a) Ta có: 2x2 - x + 1 = x(2x + 1) - 2x + 1 = x(2x + 1) - (2x + 1) + 2 = (x - 1)(2x + 1) + 2
Do (x - 1)(2x + 1) \(⋮\)2x + 1
=> 2 \(⋮\)2x + 1
=> 2x + 1 \(\in\)Ư(2) = {1; -1; 2; -2}
Do : 2x + 1 là số lẻ => 2x + 1 \(\in\){1; -1}
+) 2x + 1 = 1 => 2x = 0 => x = 0
+) 2x + 1 = -1 => 2x = -2 => x = -1
b) 2x + y + 2xy - 3 = 0
=> 2x(1 + y) + (1 + y) = 4
=> (2x + 1)(1 + y) = 4
=> 2x + 1;1 + y \(\in\)Ư(4) = {1; -1;2 ;-2; 4; -4}
Do: 2x + 1 là số lẻ => 2x + 1 \(\in\){1; -1}
=> 1 + y \(\in\){4; -4}
Lập bảng :
| 2x + 1 | 1 | -1 |
| 1 + y | 4 | -4 |
| x | 0 | -1 |
| y | 3 | -5 |
Vậy ....
c) x2 + 2xy = 0
=> x(x + 2y) = 0
=> \(\hept{\begin{cases}x=0\\x+2y=0\end{cases}}\)
=> \(\hept{\begin{cases}x=0\\2y=0\end{cases}}\)
=> \(\hept{\begin{cases}x=0\\y=0\end{cases}}\)
Vậy x = y = 0
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(t^2+6yt+9y^2\right)+\left(4y^2-12y+9\right)=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(t+3y\right)^2+\left(2y-3\right)^2=0\)
Dấu '=' xảy ra khi y=3/2; x=-3/2; t=-3y=-9/2