Bài 6:

a) \(x\left(x-2\right)+x-2=0\)

\(\Leftrightarrow x\left(x-2\right)+\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

b) \(5x\left(x-3\right)-x+3=0\)

\(\Leftrightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

c) \(3x\left(x-5\right)-\left(x-1\right)\left(2+3x\right)=30\)

\(\Leftrightarrow3x^2-15x-2x-3x^2+2+3x=30\)

\(\Leftrightarrow-14x+2=30\)

\(\Leftrightarrow-14x=28\)

\(\Leftrightarrow x=-2\)

d) \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow x^2+3x+2x+6-x^2-5x+2x+10=0\)

\(\Leftrightarrow2x+16=0\)

\(\Leftrightarrow2x=-16\)

\(\Leftrightarrow x=-8\)

Em cần gấp bây h ạ :<

Bài 1:

a) \(xy+y^2-x-y=y\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(y-1\right)\)

b) \(x^4+x^3+2x^2+x+1\)

\(=x^4+x^2+x^3+x+x^2+1\)

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

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

Để : \(3x^3+2x^2-7x+a⋮3x-1\)<=> \(a-2=0\)

<=> \(a=2\)

Vậy a = 2 

Để \(x^3+3x^2+5x+a⋮x+3\)<=> \(a-66=0\)

<=> \(a=66\)

Vậy a = 66

a: 3x^3+2x^2-7x+a chia hêt cho 3x-1

=>3x^3-x^2+3x^2-x-6x+2+a-2 chia hết cho 3x-1

=>a-2=0

=>a=2

c: =>2x^2-6x+(a+6)x-3a-18+3a+19 chia x-3 dư 4

=>3a+19=4

=>3a=-15

=>a=-5

d: 2x^3-x^2+ax+b chiahêt cho x^2-1

=>2x^3-2x-x^2+1+(a+2)x+b-1 chia hết cho x^2-1

=>a+2=0 và b-1=0

=>a=-2 và b=1

My Nguyễn ơi,bạn truy cập vào đường link này để tìm câu hỏi tương tự của câu a/Bài 1 nhé

https://vn.answers.yahoo.com/question/index?qid=20110206184834AAokV5m&sort=N

Ko biết đợi đứa khác đê

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Dễ nhưng mà dài chết người 

Bài 1:

Ta có: \(x^2-2x+2=x^2-2x+1+1\)

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

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

Ta thấy rằng: \(\left(x-1\right)^2\ge0\) ( Với mọi \(x\in Z\) )

mà 1 > 0

=> \(\left(x-1\right)^2+1\ge0\)

<=> \(x^2-2x+1\ge0\)

Bài 3:

a) 53^2 + 47^2 + 94.53

= 53^2 + 47^2 + 2.47.53

= ( 53 + 47 )^2

= 100^2

= 10000

b) 50^2 - 49^2 + 48^2 - 47^2 + 2^2 - 1^2

= ( 50^2 - 49^2 ) + ( 48^2 - 47^2 ) + ( 2^2 - 1^2 )

= (50+49).(50-49) + (48+47).(48-47) + (2+1).(2-1)

= 50 + 49 + 48 + 47 + 2 + 1

= (49 + 1) + (48 + 2) + 50 + 47

= 50 + 50 + 50 + 47

= 197

Bài 1 : 

x²-2x+2>0 với mọi x

=x²-2.x.1/4+1/16+31/16

=(x-1/4)² + 31/16

Vì (x-1/4)² \(\ge\) 0 nên (x-1/4)² + 31/16 \(\ge\) 0 với mọi x (đfcm)

3: \(\Leftrightarrow a-15=0\)

hay a=15

a: \(\Leftrightarrow3x^3-x^2+3x^2-x-6x+2-a-2⋮3x-1\)

=>-a-2=0

hay a=-2

b: \(-x^2+x-1\)

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

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

\(=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}< 0\forall x\)

c: \(P\left(x\right)=x^2-5x+\dfrac{25}{4}-\dfrac{25}{4}=\left(x-\dfrac{5}{2}\right)^2-\dfrac{25}{4}\ge-\dfrac{25}{4}\forall x\)

Dấu '=' xảy ra khi x=5/2

d: \(f\left(x\right)=x^2-4x+4+5=\left(x-2\right)^2+5\ge5\forall x\)

Dấu '=' xảy ra khi x=2

a) x³ + 2x² + x

= x³ + x² + x² + x

= x² ( x + 1 ) + x ( x + 1 )

= ( x² + x ) ( x + 1 )

a)=x(x²+2x)

b)=x(x²+2xy+y²-9)

d)=x(x²-3x+2)

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

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

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

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