\(x^2+y^2+1\ge xy+x+y\)

\(\Leftrightarrow2\left(x^2+y^2+1\right)\ge2\left(xy+x+y\right)\)

\(\Leftrightarrow x^2-2xy+y^2+y^2-2y+1+x^2-2x+1\ge0\)\(\Leftrightarrow\left(x-y\right)^2-\left(y-1\right)^2-\left(x-1\right)^2\ge0\)

Đúng với mọi x , y

Đẳng thức xảy ra khi \(\left[{}\begin{matrix}\left(x-y\right)^2=0\\\left(y-1\right)^2=0\\\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-y=0\\y-1=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=y\\y=1\\x=1\end{matrix}\right.\Rightarrow x=y=1\)

b, \(A=\dfrac{x-2}{x^3-x^2-x-2}=\dfrac{x-2}{x^3-2x^2+x^2-2x+x-2}\)

\(=\dfrac{x-2}{x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)}\)

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

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

Ta có: \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

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

Dấu " = " xảy ra khi \(\left(x+\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{-1}{2}\)

Vậy \(MAX_A=\dfrac{4}{3}\) khi \(x=\dfrac{-1}{2}\)

a) x² + y² + 1 \(\ge x.y+x+y\) \(\Leftrightarrow\) x²+y² +1-x.y-x-y\(\ge0\)

\(\Leftrightarrow\) 2x² + 2y² + 2 - 2xy - 2x-2y\(\ge0\)

\(\Leftrightarrow\) ( x² + y² - 2 xy ) + ( x² + 1 - 2xy ) + ( y²+1- 2y) \(\ge0\)

\(\Leftrightarrow\) ( x-y )² + ( x-1)² + ( y-1)² \(\ge0\) ( Bất đẳng thức luôn đúng )

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) ( 3 x   -   2 y ) 3 .        b) ( x   -   1 ) ( x   +   3 ) 2 .

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

b: \(27-8y^3=\left(3-2y\right)\left(9+6y+4y^2\right)\)

c: \(y^6+1=\left(y^2+1\right)\left(y^4-y^2+1\right)\)

d: \(64x^3-\dfrac{1}{8}y^3=\left(4x-\dfrac{1}{2}y\right)\left(16x^2+2xy+\dfrac{1}{4}y^2\right)\)

giúp mình câu e,f với

Bn cho vào trg ngoặc đi viết thế này khó hiểu quá

Giải:

Ta có: IA = IB ( I thuộc trung trực của AB )

và SA = SB ( S thuộc trung trực của AB )

Mà CS + BS > BC

=> CS + AS > BI + IC

=> AS + CS > AI + IC

=> đpcm

x 3   –   6 x 2 y   +   12 x y 2   –   8 y 3   =   x 3   –   3 . x 2 . ( 2 y )   +   3 . x . ( 2 y ) 2   –   ( 2 y ) 3     =   ( x   –   2 y ) 3

đáp án cần chọn là: D

Bài 1:

\(1,Sửa:x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\ 2,=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\\ 3,=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\\ 4,=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)

Bài 2:

\(1,=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\\ 2,=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\\ 3,=3\left(x^3-1\right)=3\left(x-1\right)\left(x^2+x+1\right)\)

Bài 3:

\(a,=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y^2\right]=5\left(x-y+1\right)\left(x+y+1\right)\\ b,=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-4y^2\right]\\ =3x\left(x-2y-1\right)\left(x+2y-1\right)\\ c,=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2\\ =\left(a+b\right)\left(a^2b-ab^2+a+b\right)\\ d,=2x\left(x^2-y^2-4x+4\right)=2x\left[\left(x-2\right)^2-y^2\right]\\ =2x\left(x-y-2\right)\left(x+y-2\right)\)

Bài 1;

1) \(x^3-2x-x=x\left(x^2-2x-1\right)\)

2) \(6x^2+12xy+6y^2=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\)

3) \(2y^3+8y^3+8y=10y^3+8y=2y\left(5y^2+4\right)\)

4) \(5x^2-10xy+5y^2=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)

Bài 2:

1) \(x^3-64x=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\)

2) \(8x^2y-18y=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\)

3) \(24x^3-3=3\left(8x^3-1\right)=3\left(2x-1\right)\left(4x^2+2x+1\right)\)

Bài 3:

1) \(5x^2+10x+5-5y^2=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y\right]=5\left(x-y+1\right)\left(x+y+1\right)\)

2) \(3x^3-6x^2+3x-12xy^2=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-\left(2y\right)^2\right]=3x\left(x-2y-1\right)\left(x+2y-1\right)\)

3) \(a^3b-ab^3+a^2+2ab+b^2=ab\left(a^2-b^2\right)+\left(a+b\right)^2=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2=\left(a+b\right)\left(a^2b-ab^2+a+b\right)\)

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