=[(x-1)(x+6)][(x+2)(x+3)]

=[x^2+5x-6][x^2+5x+6]

Đặt: x^2+5x=a

Ta có:

=(a-6)(a+6)

=a^2-36

Min A=-36

<=>a=0

<=>x=0 hoặc x=-5

cho anh 1 like

a: \(x^2-5x+6\)

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

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

=(x-2)(x-3)

b: \(4x^2-9\)

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

=(2x-3)(2x+3)

c: \(x^3+2x^2-x-2\)

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

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

=(x+2)(x-1)(x+1)

Bài 1:

1 (x+3)²=x²+6x+9

2

a, 2x²(3x-5x³)+10x⁵-5x³=6x³-10x⁵+10x⁵-5x³=x³

b, (x+3)(x²-3x+9)+(x-9)(x+3)=(x³+27)+(x²-6x-27)=x³+x²-6x

Bài 2:

a, x²-25x=0

\(\Leftrightarrow x\left(x-25\right)=0\)

\(\Leftrightarrow\begin{cases}x=0\\x-25=0\end{cases}\)

\(\Leftrightarrow\begin{cases}x=0\\x=25\end{cases}\)

b, (4x-1)²-9=0

\(\Leftrightarrow\left(4x-1-3\right)\left(4x-1+3\right)=0\)

\(\Leftrightarrow\left(4x-4\right)\left(4x+2\right)=0\)

\(\Leftrightarrow4\left(x-1\right)2\left(2x+1\right)=0\)

\(\Leftrightarrow8\left(x-1\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\begin{cases}x-1=0\\2x+1=0\end{cases}\)

\(\Leftrightarrow\begin{cases}x=1\\x=\frac{-1}{2}\end{cases}\)

Bài 3:

a, 3x²-18x+27=3(x²-6x+9)=3(x-3)²

b, xy-y²-x+y=y(x-y)-(x-y)=(y-1)(x-y)

c, x²-5x-6=x²-6x+x-6=x(x-6)+(x-6)=(x+1)(x-6)

Bài 4:

a, ( 12x³y³-3x²y³+4x²y⁴):6x²y³=(12x³y³:6x²y³)-(3x²y³:6x²y³)+(4x²y⁴:6x²y³)

=2x-1/2 + 2/3y

b, bạn ơi mình không biết cách vẽ đường kẻ để chia ý , nếu bạn biết thì chỉ cho mình rồi mình làm cho

Bài 5 :

b, A = x(2x-3)

A= 2x²-3x

A= 2(x²-3/2x)

A= 2(x²-2x3/4+9/16-9/16)

A=2[(x-3/4)²-9/16]

A=2(x-3/4)²-9/8

A=2(x-3/4)²+(-9/8)

Vì (x-3/4)² \(\ge\)0 \(\forall x\)

-> 2(x-3/4)² \(\ge0\forall x\)

-> 2(x-3/4)²+(-9/8)\(\ge-\frac{9}{8}\forall x\)

Vậy MinA= -9/8

Bài 1:

1. Khai triển hằng đẳng thức

(x+3)² = x²+6x+9

2. Thực hiện phép tính

a) 2x²(3x-5x³)+10x⁵-5x³

=6x³-10x⁵+10x⁵-5x³

=x³

b)(x+3)(x²-3x+9)+(x-9)(x+3)

=(x³+27)+(x²+3x-9x-27)

=x³+27+x²+3x-9x-27

=x³+x²-6x

Bài 2:

a) x²-25x=0

\(\Leftrightarrow\)x(x-25)=0

\(\Leftrightarrow\) \(\left[\begin{matrix}x=0\\x-25=0\end{matrix}\right.\)

\(\Leftrightarrow\left[\begin{matrix}x=0\\x=25\end{matrix}\right.\)

Vậy x=0 hoặc x=25

b)(4x-1)² - 9=0

\(\Leftrightarrow\)(4x-1+3)(4x-1-3)=0

\(\Leftrightarrow\)(4x+2)(4x-4)=0

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

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

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

Vậy x=1 hoặc x=\(\frac{-1}{2}\)

Bài 3:

a) 3x²-18x+27

=3(x²-6x+9)

=3(x-3)²

b) xy-y²-x+y

=(xy-y²)-(x-y)

=y(x-y)-(x-y)

=(x-y)(y-1)

c) x²-5x-6

=x²-6x+x-6

=(x²-6x)+(x-6)

=x(x-6)+(x-6

=(x-6)(x+1)

Bài 4:

a) (12x³y³-3x²y³+4x²y⁴) : 6x²y³

=x²y³(12x-3+4y): 6x²y³

=(12x-3+4y) : 6

= (12x : 6)-(3 : 6)+(4y : 6)

=2x-\(\frac{1}{2}\)+\(\frac{2y}{3}\)

b) (6x³-19x²+23x-12) : (2x-3)

=(3x²-5x+4)(2x-3) : (2x-3)

=3x²-5x+4

d,(x+y)\(^{3^{ }}\)-x\(^3\)-y\(^3\)

= (x+y)\(^3\)-(x+y)(x\(^2\)-xy+y\(^2\))

=(x+y)(x\(^2\)+2xy+y\(^2\)-x\(^2\)+xy-y\(^2\))

=3xy(x+y)

e,x\(^3\)+27+(x+3)(x-9)

=(x+3)(x\(^2\)-3x+9)+(x+3)(x-9)

=(x+3)(x\(^2\)-3x+9+x-9)

=x(x+3)(x-2)

f,x\(^2\)-5x+6

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

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

=(x-2)(x-3)

Dễ như này chắc gì nó đã đăng câu hỏi :))

Khi qua thi học kì xong, mệt => không onl :))

Chiều về có rảnh làm thử cho :>>