tìm x biết 2x^2-8x+y^2+2y+9=0
Những câu hỏi liên quan
tìm x, y biết
a) \(x^2+2y^2+2xy-2x+2=0\)
b) \(2x^2-8x+y^2+2y+9=0\)
\(x^2+2y^2+2xy-2x+2=0.\)
\(\Leftrightarrow\left(x^2+y^2+1+2xy-2x-2y\right)+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow\left(x+y-1\right)^2+\left(y+1\right)^2=0\)
Mà \(\left(x+y-1\right)^2\ge0,\left(y+1\right)^2\ge0\)
Suy ra \(\hept{\begin{cases}\left(x+y-1\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x+y=1\\y=-1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=-1\end{cases}.}\)
\(2x^2-8x+y^2+2y+9=0\)
\(\Leftrightarrow\left(2x^2-8x+8\right)+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow2\left(x^2-4x+4\right)+\left(y+1\right)^2=0\)
\(\Leftrightarrow2\left(x-2\right)^2+\left(y+1\right)^2=0\)
Mà \(2\left(x-2\right)^2\ge0,\left(y+1\right)^2\ge0\)
Suy ra \(\hept{\begin{cases}2\left(x-2\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=-1\end{cases}}}\)
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b1) tìm x biết:
a)z^2-2x-=0
b)x^2+8x-9=0
c)x^2-14x+13=0
d)x^2-5x-6=0
b2)Tìm x;y;z biết:
a)2x=3y,9y=5z và x-2y+z=10
b) 3x=4y=7z và x-2y+5z=20
chiu roi
ban oi
tk nhe@@@@@@@@@@
ai tk minh minh tk lai!!
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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
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a.x mũ 2 + 2x + y mũ 2 - 6y + 19=0
b. x mũ 2 + y mũ 2 + 2x +4y+ 5=0
5x mũ 2+ 5y mũ 2+ 8xy + 2y-2x + 2=0
d.2x mũ 2 - 8x + y mũ 2 + 2y+ 9=0
Tìm x, y biết:a, 2x2 - 8x + y2 + 2y + 9 0b, x2 + y2 + x - xy + 1/2 0c, x2 - 4xy + 92 + 9 2xy + 6x - x2d, 5x2 + 5y2 + 8xy + 2y - 2x + 2 0e, 4x2 + 13y2 +12xy + 4x - 2y + 5 0f, x2 + 2y2 - 2xy + 2x + 2 - 4y 0g, 8x3 + (x + 8)2 8(x + 2)(x2 - 2x + 4)h, (3x +1)(9x2 + 1 - 3x) - (3 - x)2 (3x - 2)3i, 2x2 + 5y2 - 4xy - 4x - 14y + 29 0j, 4x2 + 9y2 - 12x - 32y - 2xy + 44 0
Đọc tiếp
Tìm x, y biết:
a, 2x2 - 8x + y2 + 2y + 9 = 0
b, x2 + y2 + x - xy + 1/2 = 0
c, x2 - 4xy + 92 + 9 = 2xy + 6x - x2
d, 5x2 + 5y2 + 8xy + 2y - 2x + 2 = 0
e, 4x2 + 13y2 +12xy + 4x - 2y + 5 =0
f, x2 + 2y2 - 2xy + 2x + 2 - 4y = 0
g, 8x3 + (x + 8)2 = 8(x + 2)(x2 - 2x + 4)
h, (3x +1)(9x2 + 1 - 3x) - (3 - x)2 = (3x - 2)3
i, 2x2 + 5y2 - 4xy - 4x - 14y + 29 = 0
j, 4x2 + 9y2 - 12x - 32y - 2xy + 44 = 0
1, Tìm x, y biết:a, 2x2 - 8x + y2 + 2y + 9 0b, x2 + y2 + x - xy + 1/2 0c, x2 - 4xy + 92 + 9 2xy + 6x - x2d, 5x2 + 5y2 + 8xy + 2y - 2x + 2 0e, 4x2 + 13y2 +12xy + 4x - 2y + 5 0f, x2 + 2y2 - 2xy + 2x + 2 - 4y 0g, 8x3 + (x + 8)2 8(x + 2)(x2 - 2x + 4)h, (3x +1)(9x2 + 1 - 3x) - (3 - x)2 (3x - 2)3i, 2x2 + 5y2 - 4xy - 4x - 14y + 29 0j, 4x2 + 9y2 - 12x - 32y - 2xy + 44 0v
Đọc tiếp
1, Tìm x, y biết:
a, 2x2 - 8x + y2 + 2y + 9 = 0
b, x2 + y2 + x - xy + 1/2 = 0
c, x2 - 4xy + 92 + 9 = 2xy + 6x - x2
d, 5x2 + 5y2 + 8xy + 2y - 2x + 2 = 0
e, 4x2 + 13y2 +12xy + 4x - 2y + 5 =0
f, x2 + 2y2 - 2xy + 2x + 2 - 4y = 0
g, 8x3 + (x + 8)2 = 8(x + 2)(x2 - 2x + 4)
h, (3x +1)(9x2 + 1 - 3x) - (3 - x)2 = (3x - 2)3
i, 2x2 + 5y2 - 4xy - 4x - 14y + 29 = 0
j, 4x2 + 9y2 - 12x - 32y - 2xy + 44 = 0v
a,\(2x^2-8x+y^2+2y+9=0\)
\(\Rightarrow2\left(x^2-4x+4\right)+\left(y^2+2y+1\right)=0\)
\(\Rightarrow2\left(x-2\right)^2+\left(y+1\right)^2=0\)
Mà \(2\left(x-2\right)^2\ge0\forall x\); \(\left(y+1\right)^2\ge0\forall y\)
\(\Rightarrow2\left(x-2\right)^2+\left(y+1\right)^2\ge0\forall x;y\)
Dấu "=" xảy ra<=> \(\hept{\begin{cases}2\left(x-2\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=-1\end{cases}}}\)
Vậy x=2;y=-1
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Tìm x, biết:
a) (x+2)2 + (x-3)2 = 2x (x+7)
b) (x+1)3 + (x-1)3 = 2x3
c) x3 - 3x2 + 3x - 126 = 0
d) x2 + y2 - 2x + 4y + 5 = 0
e) 2x2 + 8x +y2 - 2y + 9 = 0
a)
pt <=> \(x^2+4x+4+x^2-6x+9=2x^2+14x\)
<=> \(2x^2-2x+13=2x^2+14x\)
<=> \(16x=13\)
<=> \(x=\frac{13}{16}\)
b)
pt <=> \(x^3+3x^2+3x+1+x^3-3x^2+3x-1=2x^3\)
<=> \(2x^3+6x=2x^3\)
<=> \(6x=0\)
<=> \(x=0\)
c)
pt <=> \(\left(x^3-3x^2+3x-1\right)-125=0\)
<=> \(\left(x-1\right)^3=125\)
<=> \(x-1=5\)
<=> \(x=6\)
d)
pt <=> \(\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)
<=> \(\left(x-1\right)^2+\left(y+2\right)^2=0\) (1)
CÓ: \(\left(x-1\right)^2;\left(y+2\right)^2\ge0\forall x;y\)
=> \(\left(x-1\right)^2+\left(y+2\right)^2\ge0\) (2)
TỪ (1) VÀ (2) => DÁU "=" XẢY RA <=> \(\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
e)
pt <=> \(2x^2+8x+8+y^2-2y+1=0\)
<=> \(2\left(x+2\right)^2+\left(y-1\right)^2=0\)
TA LUÔN CÓ: \(2\left(x+2\right)^2+\left(y-1\right)^2\ge0\forall x;y\)
=> DẤU "=" XẢY RA <=> \(\hept{\begin{cases}2\left(x+2\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=-2\\y=1\end{cases}}\)
a) ( x + 2 )2 + ( x - 3 )2 = 2x( x + 7 )
<=> x2 + 4x + 4 + x2 - 6x + 9 = 2x2 + 14x
<=> x2 + 4x + x2 - 6x - 2x2 - 14x = -4 - 9
<=> -16x = -13
<=> x = 13/16
b) ( x + 1 )3 + ( x - 1 )3 = 2x3
<=> x3 + 3x2 + 3x + 1 + x3 - 3x2 + 3x - 1 = 2x3
<=> x3 + 3x2 + 3x + x3 - 3x2 + 3x - 2x3 = -1 + 1
<=> 6x = 0
<=> x = 0
c) x3 - 3x2 + 3x - 126 = 0
<=> ( x3 - 3x2 + 3x - 1 ) - 125 = 0
<=> ( x - 1 )3 = 125
<=> ( x - 1 )3 = 53
<=> x - 1 = 5
<=> x = 6
d) x2 + y2 - 2x + 4y + 5 = 0
<=> ( x2 - 2x + 1 ) + ( y2 + 4y + 4 ) = 0
<=> ( x - 1 )2 + ( y + 2 )2 = 0 (*)
\(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(y+2\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\)
Đẳng thức xảy ra ( tức (*) ) <=> \(\hept{\begin{cases}x-1=0\\y+2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
e) 2x2 + 8x + y2 - 2y + 9 = 0
<=> 2( x2 + 4x + 4 ) + ( y2 - 2y + 1 ) = 0
<=> 2( x + 2 )2 + ( y - 1 )2 = 0 (*)
\(\hept{\begin{cases}2\left(x+2\right)^2\ge0\forall x\\\left(y-1\right)^2\ge0\forall y\end{cases}}\Rightarrow2\left(x+2\right)^2+\left(y-1\right)^2\ge0\forall x,y\)
Đẳng thức xảy ra ( tức xảy ra (*) ) <=> \(\hept{\begin{cases}x+2=0\\y-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-2\\y=1\end{cases}}\)
a. \(\left(x+2\right)^2+\left(x-3\right)^2=2x\left(x+7\right)\)
\(\Leftrightarrow x^2+4x+4+x^2-6x+9=2x^2+14x\)
\(\Leftrightarrow2x^2-2x+13=2x^2+14x\)
\(\Leftrightarrow16x=13\)
\(\Leftrightarrow x=\frac{13}{16}\)
b. \(\left(x+1\right)^3+\left(x-1\right)^3=2x^3\)
\(\Leftrightarrow x^3+3x^2+3x+1+x^3-3x^2+3x-1=2x^3\)
\(\Leftrightarrow2x^3+6x=2x^3\)
\(\Leftrightarrow6x=0\)
\(\Leftrightarrow x=0\)
c. \(x^3-3x^2+3x-126=0\)
\(\Leftrightarrow\left(x-6\right)\left(x^3+3x+21\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x^3+3x+21=0\left(ktm\right)\end{cases}\Leftrightarrow}x=6\)
d. \(x^2+y^2-2x+4y+5=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+2\right)^2=0\)
\(\hept{\begin{cases}\left(x-1\right)^2\ge0\\\left(y+2\right)^2\ge0\end{cases}}=0\Leftrightarrow\orbr{\begin{cases}x-1=0\\y+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\y=-2\end{cases}}}\)
e. \(2x^2+8x+y^2-2y+9=0\)
\(\Leftrightarrow2\left(x+2\right)^2+\left(y-1\right)^2=0\)
\(\hept{\begin{cases}2\left(x+2\right)^2\ge0\\\left(y-1\right)^2\ge0\end{cases}=0\Leftrightarrow\orbr{\begin{cases}x+2=0\\y-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-2\\y=1\end{cases}}}\)
tính
a)6x^3y-8x^2y^2+4xy
b)x^2-4x+xy-4y
c)x^2-y^2-6x+9
tìm x
a)x^2-2x-3=0
a)biết 8x^3+12x^2y+6xy^2+y^3=27
tính x(2x+y)+xy+1/2y
b) cho x^2+y^2+z(x-y+1)=0
tính x+y
c) cho a+b+c=9,a^2+b^2+c^2=53
tính ab, ac, bc
Đề: Biết \(8x^3+12x^2y+6xy^2+y^3=27\) . Tính \(A=x\left(2x+y\right)+xy+\frac{1}{2}y^2\)
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Ta có:
\(8x^3+12x^2y+6xy^2+y^3=27\)
\(\Leftrightarrow\) \(\left(2x+y\right)^3=27\)
\(\Leftrightarrow\) \(2x+y=3\)
Do đó:
\(A=3x+xy+\frac{1}{2}y^2\)
\(=3x+\frac{1}{2}y\left(2x+y\right)\)
\(=3x+\frac{3}{2}y\)
\(=\frac{3}{2}\left(2x+y\right)\)
\(A=\frac{9}{2}\)
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