Giải HPT
x4 - y4=240
x3 - 2y3 = 3(x2 - 4y2) - 4(x-8y)
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trở thành một hằng đảng thức. Giá trị trong ô vuông là:
1 tháng 4 2022
⇒(x−1)^2+4(y+1)^2+(z−3)^2≥0
x^2+4y^2+z^2-2x-6z+8y+15
=x^2+4y^2+z^2-2x-6z+8y+1+1+4+9
=(x^2-2x+1)+(4y^2+8y+4)+(z^2-6z+9)+1
=(x-1)^2+4(y+1)^2+(z-3^)2+1
Ta thấy:(x−1)^2≥0
4(y+1)^2≥0
(z−3)^ 2≥0
{(x−1)^24(y+1)^2(z−3)^2≥0
⇒(x−1)^2+4(y+1)^2+(z−3)^2≥0
⇒(x−1)2+4(y+1)2+(z−3)2+1≥0+1=1>0
31 tháng 7 2021
1. x2 - 6x + 9=(x-3)2
2. 25 + 10x + x2=(x+5)2
3. \(\dfrac{1}{4}a^2+2ab^2+4b^4=\left(\dfrac{1}{2}a+2b^2\right)^2\)
4.\(\dfrac{1}{9}-\dfrac{2}{3}y^4+y^8=\left(\dfrac{1}{3}-y^4\right)^2\)
5.x3 + 8y3=(x+8y)(x2-8xy+64y2)
6.8y3 -125=(2y-5)(4y2+10y+25)
7.a6-b3=(a2-b)(a4+a2b+b2)
8 x2 - 10x + 25=(x-2)2
31 tháng 7 2021
1) \(x^2-6x+9=\left(x-3\right)^2\)
2) \(25+10x+x^2=\left(5+x\right)^2\)
3) \(\dfrac{1}{4}a^2+2ab+4b^4=\left(\dfrac{1}{2}a+2b^2\right)^2\)
4) \(\dfrac{1}{9}-\dfrac{2}{3}y^4+y^8=\left(\dfrac{1}{3}-y^4\right)^2\)
5) \(x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
6) \(8y^3-125=\left(2y-5\right)\left(4y^2+10y+25\right)\)
7) \(a^6-b^3=\left(a^2-b\right)\left(a^4+a^2b+b^2\right)\)
8) \(x^2-10x+25=\left(x-5\right)^2\)
9) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
28 tháng 10 2017
a, \(x^4+2x^2+1-x^2\)
= \(\left(x^2+1\right)^2-x^2\)
= \(\left(x^2+x+1\right)\left(x^2-x+1\right)\)
b, \(x^4+x^2+1\)
= \(x^4+2x^2+1-x^2\)
= .. ( như phần a )
c, \(y^4+64\)
= \(\left(y^2+8\right)\left(y^2-8\right)\)
d, \(4xy+3z-12y-xz\)
\(=4y\left(x-3\right)-z\left(x-3\right)\)
\(=\left(x-3\right)\left(4y-z\right)\)
e, \(x^2-4xy+4y^2-z^2+6z-9\)
\(=\left(x-2y\right)^2-\left(z-3\right)^2\)
g, \(x^2-4xy+5x+4y^2-10y\)
\(=\left(x^2-4xy+4y^2\right)+\left(5x-10y\right)\)
\(=\left(x-2y\right)^2+5\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x-2y+5\right)\)
h, \(x^2-7x+6\)
\(=x^2-6x-x+6\)
\(=x\left(x-6\right)-\left(x-6\right)\)
\(=\left(x-6\right)\left(x-1\right)\)
i, \(x^3+5x^2+6x+2\)
\(=x^3+x^2+4x^2+4x+2x+2\)
\(=x^2\left(x+1\right)+4x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+4x+2\right)\)
NH
3
T
19 tháng 10 2023
\(x^2+4y^2+z^2-2x-6z+8y+14=0\\\Leftrightarrow (x^2-2x+1)+(4y^2+8y+4)+(z^2-6z+9)=0\\\Leftrightarrow (x^2-2\cdot x\cdot1+1^2)+[(2y)^2+2\cdot2y\cdot 2+2^2]+(z^2-2\cdot z\cdot3+3^2)=0\\\Leftrightarrow (x-1)^2+(2y+2)^2+(z-3)^2=0\)
Ta thấy: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\forall x\\\left(2y+2\right)^2\ge0\forall y\\\left(z-3\right)^2\ge0\forall z\end{matrix}\right.\)
\(\Rightarrow\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2\ge0\forall x;y;z\)
Mặt khác: \(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2=0\)
nên ta được:
\(\left\{{}\begin{matrix}x-1=0\\2y+2=0\\z-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\\z=3\end{matrix}\right.\)
Vậy: ...
KV
19 tháng 10 2023
\(x^2+4y^2+z^2-2x-6z+8y+14=0\)
\(\left(x^2-2x+1\right)+\left(4y^2+8y+4\right)+\left(z^2-6z+9\right)=0\)
\(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2=0\) (1)
Do \(\left(x-1\right)^2\ge0;\left(2y+2\right)^2\ge0;\left(z-3\right)^2\ge0\)
\(\left(1\right)\Rightarrow\) \(\left(x-1\right)^2=0;\left(2y+2\right)^2=0;\left(z-3\right)^2=0\)
*) \(\left(x-1\right)^2=0\)
\(x-1=0\)
\(x=1\)
*) \(\left(2y+2\right)^2=0\)
\(2y+2=0\)
\(2y=-2\)
\(y=-1\)
*) \(\left(z-3\right)^2=0\)
\(z-3=0\)
\(z=3\)
Vậy x = 1; y = -1; z = 3
NT
1
20 tháng 6 2023
=x^2-6x+9+4y^2-8y+4+2010
=(x-3)^2+(2y-2)^2+2010>=2010
Dấu = xảy ra khi x=3 và y=1
10 tháng 9 2023
a) \(x^2+xy+y^2+1\)
\(=x^2+xy+\dfrac{y^2}{4}-\dfrac{y^2}{4}+y^2+1\)
\(=\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1\)
mà \(\left\{{}\begin{matrix}\left(x+\dfrac{y}{2}\right)^2\ge0,\forall x;y\\\dfrac{3y^2}{4}\ge0,\forall x;y\end{matrix}\right.\)
\(\Rightarrow\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1>0,\forall x;y\)
\(\Rightarrow dpcm\)
10 tháng 9 2023
b) \(...=x^2-2x+1+4\left(y^2+2y+1\right)+z^2-6z+9+1\)
\(=\left(x-1\right)^2+4\left(y^{ }+1\right)^2+\left(z-3\right)^2+1>0,\forall x.y\)
\(\Rightarrow dpcm\)