đề khó đấy

Câu 1 : sửa x^3 - 8y^3 

chọn C 

Câu 2 : B 

Câu 1.

\(x^2-8y^3=\left(x-2y\right)\left(x^2+2y+4y^2\right)\)

Câu 2.

\(-x^2\left(3-2x\right)=-3x^2+2x^3=2x^3-3x^2\)

Câu 1: C

Câu 2: B

⇒(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

\(x^2+xy+y^2+1.=x^2+2.x.\dfrac{y}{2}+\left(\dfrac{y}{2}\right)^2+\dfrac{3}{4}y^2+1.\\ =\left(x+\dfrac{y}{2}\right)^2+\dfrac{3}{4}y^2+1>0\forall x;y\in R.\\ \Rightarrow x^2+xy+y^2+10\forall x;y\in R.\)

Kkk

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

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) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

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, \(x^3+5x^2+6x+2\)

\(=x^3+x^2+4x^2+4x+2x+2\)

\(=x^2\left(x+1\right)+4x\left(x+1\right)+2\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2+4x+2\right)\)

phần b là 6^4 nhé

nhầm câu c là 6^4

\(x^2+4y^2+z^2-2x-6z+8y+14=0\\\Leftrightarrow (x^2-2x+1)+(4y^2+8y+4)+(z^2-6z+9)=0\\\Leftrightarrow (x^2-2\cdot x\cdot1+1^2)+[(2y)^2+2\cdot2y\cdot 2+2^2]+(z^2-2\cdot z\cdot3+3^2)=0\\\Leftrightarrow (x-1)^2+(2y+2)^2+(z-3)^2=0\)

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: ...

\(x^2+4y^2+z^2-2x-6z+8y+14=0\)

\(\left(x^2-2x+1\right)+\left(4y^2+8y+4\right)+\left(z^2-6z+9\right)=0\)

\(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2=0\) (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

Latex của mình có chút lỗi nên xin phép đánh máy hoàn toàn nhé
x^2+4y^2+z^2-2x-6z+8y+14=0
<=> (x^2-2x+1) +4(y^2+2y+1) +(z^2-6z+9)=0
<=> (x-1)^2+4(y+1)^2+(z-3)^2=0
Do (x-1)^2, (y+1)^2>=0, (z-3)^2>=0
Nên x-1=0, y+1=0, z-3=0
<=> x=1, y=-1, z=3

=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

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 \(\left\{{}\begin{matrix}\left(x+\dfrac{y}{2}\right)^2\ge0,\forall x;y\\\dfrac{3y^2}{4}\ge0,\forall x;y\end{matrix}\right.\)

\(\Rightarrow\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1>0,\forall x;y\)

\(\Rightarrow dpcm\)

b) \(...=x^2-2x+1+4\left(y^2+2y+1\right)+z^2-6z+9+1\)

\(=\left(x-1\right)^2+4\left(y^{ }+1\right)^2+\left(z-3\right)^2+1>0,\forall x.y\)

\(\Rightarrow dpcm\)

b.

$x^2+4y^2+z^2-2x-6z+8y+15=(x^2-2x+1)+(4y^2+8y+4)+(z^2-6z+9)+1$

$=(x-1)^2+(2y+2)^2+(z-3)^2+1\geq 0+0+0+1>0$ với mọi $x,y,z$

Ta có đpcm.