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DT
2
6 tháng 10 2016
x2+y2+z2+2x-4y+6z+14=0
(x+1)2+(y-2)2+(z+3)2=0
=>x+1=0=>x=-1
y-2=0=>y=2
z+3=0=>z=-3
=>x+y+z=............
VD
31 tháng 12 2016
x^2+y^2+z^2+2x-4y+6z+14=0
x^2+y^2+z^2+2x-4y+6z+1+4+9 = 0
(x+1)^2+(y-2)^2+(z+3)^2 =0
=> x+1=0 -> x = -1
=> y-2=0 -> y=2
=> z+3=0->z=-3
vậy x+y+z = -2
PT
0
VT
1
3 tháng 10 2016
\(x^2+y^2+z^2+2x-4y+6z=-14\)
\(x^2+y^2+z^2+2x-4y+6z+14=0\)
\(x^2+2x+1+y^2-4y+4+z^2+6z+9=0\)
\(\left(x+1\right)^2+\left(y-2\right)^2+\left(z+3\right)^2=0\)
\(\left(x+1\right)^2=0\) x+1 = 0 x = -1 | \(\left(y-2\right)^2=0\) y - 2 = 0 y = 2 | \(\left(z+3\right)^2=0\) z + 3 = 0 z = -3 |
vậy x + y + z = -1 + 2 + (-3) = -2
LM
4 tháng 8 2018
b, x2 +y2+z2 +2x-4y-6z+14=0
<=> (x2+2x+1)+(y2-4y+4)+(z2-6z+9)=0
<=> (x+1)2+(y-2)2+(z-3)2=0
=>(x+1)2=(y-2)2=(z-3)2=0
=>x+1=y-2=z-3=0
=> x=-1; y=2; z=3
c, 2x2+y2-6x-4y+2xy+5=0
<=> (x2+y2+4+2xy-4x-4y)+(x2-2x+1)=0
<=> (x+y-2)2+(x-1)2=0
=> (x+y-2)2=(x-1)2=0
=>x+y-2=x-1=0
=>x=1; y=1
21 tháng 10 2017
Câu b:
Ta có: \(x^2 + 4y^2 + z^2 - 2x - 6z + 8y + 15\)
\(= (x^2 - 2x +1) + (4y^2 - 8y + 4) + (z^2 - 6z +9) +1\)
\(= (x-1)^2 + (2y-2)^2 + (z-3)^2 + 1\)
Mà \((x-1)^2 \geq 0; (2y-2)^2 \geq 0; (z-3)^2\geq 0\)
\(\implies\) \((x-1)^2+(2y-2)^2 +(z-3)^2\geq 0\)
\(\implies\)\((x-1)^2+(2y-2)^2 +(z-3)^2+1> 0\)
KN
2
29 tháng 7 2020
Bài làm:
Ta có: \(x^2+4y^2+z^2-2x-6z+8y+15\)
\(=\left(x^2-2x+1\right)+\left(4y^2+8y+4\right)+\left(z^2-6z+9\right)+1\)
\(=\left(x-1\right)^2+4\left(y+1\right)^2+\left(z-3\right)^2+1\ge1>0\left(\forall x,y,z\right)\)
30 tháng 7 2020
x2 + 4y2 + z2 - 2x - 6z + 8y + 15
= ( x2 - 2x + 1 ) + ( 4y2 + 8y + 4 ) + ( z2 - 6z + 9 ) + 1
= ( x - 1 )2 + ( 2y + 2 )2 + ( z - 3 )2 + 1 ≥ 1 > 0 ∀ x,y,z ( đpcm )
11 tháng 6 2017
a) \(\left(x\right)^2+2\left(x\right)\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2+1\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
Nên \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)
b) \(\left(x^2-2x+1\right)+\left(4y^2+8y+1\right)+\left(z^2-6z+9\right)+4\)
\(=\left(x-1\right)^2+\left(2y+1\right)^2+\left(z-3\right)^2+4\)
Vì \(\left(x-1\right)^2+\left(2y+1\right)^2+\left(z-3\right)^2\ge0\forall x,y,z\)
Nên \(\left(x-1\right)^2+\left(2y+1\right)^2+\left(z-3\right)^2+4>0\forall x,y,z\)
CM
4 tháng 12 2018
a) \(\left(3x-5\right)\left(2x+3\right)-\left(2x-3\right)\left(3x+7\right)-2x\left(x-4\right)\)
\(=\left(6x^2-x-15\right)-\left(6x^2+5x-21\right)-\left(2x^2-8x\right)\)
\(=6x^2-x-15-6x^2-5x+21-2x^2+8x\)
\(=-2x^2+2x+6\)
\(=-2\left(x^2-x-3\right)\)
b) \(\left(x^2+2\right)^2-\left(x+2\right)\left(x-2\right)\left(x^2+4\right)\)
\(=\left(x^2+2\right)^2-\left(x^2-4\right)\left(x^2+4\right)\)
\(=\left(x^2+2\right)^2-\left(x^4-16\right)\)
\(=\left(x^4+4x^2+4\right)-\left(x^4-16\right)\)
\(=x^4+4x^2+4-x^4+16\)
\(=4x^2+20\)
\(=4\left(x^2+5\right)\)
c) \(\left(2x-y\right)^2-2\left(x+3y\right)^2-\left(1+3x\right)\left(3x-1\right)\)
\(=\left(4x^2-4xy+y^2\right)-2\left(x^2+6xy+9y^2\right)-\left(9x^2-1\right)\)
\(=4x^2-4xy+y^2-2x^2-16xy-18y^2-9x^2+1\)
\(=-7x^2-20xy-17y^2+1\)
d) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)\)
\(=\left(x^6-3x^4+3x^2-1\right)-\left(x^6-1\right)\)
\(=x^6-3x^4+3x^2-1-x^6+1\)
\(=-3x^4+3x^2\)
\(=-3x^2\left(x^2-1\right)\)
\(=-3x^2\left(x-1\right)\left(x+1\right)\)
e) \(\left(2x-1\right)^2-2\left(4x^2-1\right)+\left(2x+1\right)^2\)
\(=\left(2x-1\right)^2-2\left(2x-1\right)\left(2x+1\right)+\left(2x+1\right)^2\)
\(=\left[\left(2x-1\right)-\left(2x+1\right)\right]^2\)
\(=\left(2x-1-2x-1\right)^2\)
\(=\left(-2\right)^2=4\)
g) \(\left(x-y+z\right)^2+\left(y-z\right)^2-2\left(x-y+z\right)\left(z-y\right)\)
\(=\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2\)
\(=\left(x-y+z+y+z\right)^2\)
\(=\left(x+2z\right)^2\)
h) \(\left(2x+3\right)^2+\left(2x+5\right)^2-\left(4x+6\right)\left(2x+5\right)\)
\(=\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\)
\(=\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\)
\(=\left(2x+3-2x-5\right)^2\)
\(=\left(-2\right)^2=4\)
i) \(5x^2-\dfrac{10x^3+15x^2-5x}{-5x}-3\left(x+1\right)\)
\(=5x^2-\dfrac{-5x\left(-2x^2-3x+1\right)}{-5x}-3\left(x+1\right)\)
\(=5x^2-\left(-2x^2-3x+1\right)-3\left(x+1\right)\)
\(=5x^2+2x^2+3x-1-3x-3\)
\(=7x^2-4\)
S
14 tháng 7 2019
\(x^2+y^2+z^2=4x-2y+6z-14\Leftrightarrow x^2-4x+y^2+2y+z^2-6z+14=0\Leftrightarrow\left(x^2-4x+4\right)+\left(y^2+2y+1\right)+\left(z^2-6z+9\right)=0\Leftrightarrow\left(x-2\right)^2+\left(y+1\right)^2+\left(z-3\right)^2=0matkhac:\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\\\left(y+1\right)^2\ge0\\\left(z-3\right)^2\ge0\end{matrix}\right.\Rightarrow\left(x-2\right)^2+\left(y+1\right)^2+\left(z-3\right)^2\ge0mà:\left(x-2\right)^2+\left(y+1\right)^2+\left(z-3\right)^2=0\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(y+1\right)^2=0\\\left(z-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+1=0\\z-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\\z=3\end{matrix}\right..Vậy:x=2;y=-1;z=3\)
\(^{x^2+y^2+z^2+2x-4y+6z=-14}\)
\(=x^2+2x+1+y^2-4y+4+z^2+6z+9=-14+14=0\)\(=\left(x+1\right)^2+\left(y-2\right)^2+\left(z+3\right)^2=0\)\(\Rightarrow\left(x+1\right)^2=0;\left(y-2\right)^2=0;\left(z+3\right)^2=0\)\(\Rightarrow x+1=0;y-2=0;z+3=0\)\(\Rightarrow x=-1;y=2;z=-3\Rightarrow x+y+z=-2\)
-2
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