a) \(B=\left(x^2+2x+1\right)+\left(y^2-2.2.y+2^2\right)=\left(x+1\right)^2+\left(y-2\right)^2\)

thay x=99 và y=102 vào B ta có:

\(B=\left(99+1\right)^2+\left(102-2\right)^2=100^2-100^2=0\)

b) 

b) \(2x^2+16x+32-2y^2=2\left(x^2+8x+16-y^2\right)=2\left(\left(x+4\right)^2-y^2\right)=2\left(x+4-y\right)\left(x+4+y\right)\)

\(x^2-3x+2x-6=x\left(x-3\right)+2\left(x-3\right)=\left(x-3\right)\left(x+2\right)=0\)

nếu x-3=0=>x=3

nếu x+2=0=>x=-2

nhớ k cả 3 nhé thanks bn nhìu

1.

\(A=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(2x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{2x-9-\left(x^2-9\right)+\left(2x^2-8\right)}{\left(x-2\right)\left(x-3\right)}\)

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

\(=\dfrac{x+4}{x-3}\)

b.

\(A=2\Rightarrow\dfrac{x+4}{x-3}=2\Rightarrow x+4=2\left(x-3\right)\)

\(\Rightarrow x=10\) (thỏa mãn)

2.

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

a)\(x^2+7x+6\)

\(=x^2+6x+x+6\)

\(=x\left(x+6\right)+\left(x+6\right)\)

\(=\left(x+1\right)\left(x+6\right)\)

b)\(x^4+2016x^2+2015x+2016\)

\(=x^4+2016x^2+\left(2016x-x\right)+2016\)

\(=\left(x^4-x\right)+\left(2016x^2+2016x+2016\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2016\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2016\right)\)

Bài 3:

Từ \(a^2+b^2+c^2+3=2\left(a+b+c\right)\)

\(\Rightarrow a^2+b^2+c^2+3-2a-2b-2c=0\)

\(\Rightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)

\(\Rightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\) (1)

Ta thấy:\(\begin{cases}\left(a-1\right)^2\ge0\\\left(b-1\right)^2\ge0\\\left(c-1\right)^2\ge0\end{cases}\)

\(\Rightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\) (2)

Từ (1) và (2) \(\Rightarrow\begin{cases}\left(a-1\right)^2=0\\\left(b-1\right)^2=0\\\left(c-1\right)^2=0\end{cases}\)

\(\Rightarrow\begin{cases}a-1=0\\b-1=0\\c-1=0\end{cases}\)\(\Rightarrow\begin{cases}a=1\\b=1\\c=1\end{cases}\)

\(\Rightarrow a=b=c=1\Rightarrow H=1\cdot1\cdot1+1^{2014}+1^{2015}+1^{2016}=1+1+1+1=4\)

Bài 2: 

\(A=\left(x+y\right)^3-3xy\left(x+y\right)+3xy=1^3-3xy+3xy=1\)

Bài 3:

\(M=x^6-x^4-x^4+x^2+x^3-x\)

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

\(=8x^3-8x+8\)

\(=8\cdot8+8=72\)

Bài 1:

a)x²-10x+9

=x²-x-9x+9

=x(x-1)-9(x-1)

=(x-9)(x-1)

b)x²-2x-15

=x²+3x-5x-15

=x(x+3)-5(x+3)

=(x-5)(x+3)

c)3x²-7x+2

=3x²-x-6x+2

=x(3x-1)-2(3x-1)

=(x-2)(3x-1)x^3-12+x^2

d)x³-12+x²

=x³+3x²+6x-2x²-6x-12

=x(x²+3x+6)-2(x²+3x+6)

=(x-2)(x²+3x+6)

bài 3:

a)-1/2

b)1/2

Phân tích đa thức thành nhân tử

a) 2( x + 1 ) - 3y( x + 1 ) = ( x + 1 )( 2 - 3y )

b) x² - 5x + 4 = x² - x - 4x + 4 = x( x - 1 ) - 4( x - 1 ) = ( x - 1 )( x - 4 )

Tìm x

a) x( x - 3 ) + 7x - 21 = 0

<=> x( x - 3 ) + 7( x - 3 ) = 0

<=> ( x - 3 )( x + 7 ) = 0

<=> x - 3 = 0 hoặc x + 7 = 0

<=> x = 3 hoặc x = -7

b) ( x - 2 )² + x( 3 - x ) = 6

<=> x² - 4x + 4 + 3x - x² = 6

<=> -x + 4 = 6

<=> -x = 2

<=> x = -2

\(A=\frac{x-2}{x}\)và \(B=\frac{x}{x-2}-\frac{2x}{x^2-4}\)( x ≠ 0 ; x ≠ ±3 )

a) Tại x = 23 ( tmđk ) => \(A=\frac{23-2}{23}=\frac{21}{23}\)

b) P = A.B

\(=\frac{x-2}{x}\times\left(\frac{x}{x-2}-\frac{2x}{x^2-4}\right)\)

\(=\frac{x-2}{x}\times\left(\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{2x}{\left(x-2\right)\left(x+2\right)}\right)\)

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

\(=\frac{1}{x}\times\frac{x^2}{x+2}=\frac{x}{x+2}\)

Để P = 4 => \(\frac{x}{x+2}=4\)

=> 4( x + 2 ) = x

=> 4x + 8 - x = 0

=> 3x + 8 = 0

=> x = -8/3 ( tmđk )