5-0x3 + \(\frac{9}{3}\)

= 5-3

=2

Nhầm 5 - 0 x 3 + 9 :3

5-0+3

8

5-0x3+9:3 =5 - 0 +3 = 5 + 3 = 8

5-0x3+9:3= 5-0+3

=8

5-0x3+9:3=5-0+3

               = 5+3

               =8

5-0x3+9:3

=5-0+3

=5+3

=8 ok 

= 8 NHOA  DoAnhTu !

K VÀ KB NHA !

=8 ah bn

=8 nha bạn

5-0*3+9/3

=5-0+3

=8

tick nha

5+0x3+9:3=8 

ban a 

tich minh nha ! minh tra loi nhanh nhat day !

Xiếc áo thuật đây  , muốn xem thêm thì tick 

=2

k mik nha

8 nhé bạn

bn sai hoàn toàn nha 

1. (x + 5)² - (x + 5)(x - 2) = 0

<=> (x + 5 - x + 2)(x + 5) = 0

<=> 7(x + 5) = 0

<=> x + 5 = 0

<=> x = -5

2. x³ + 7x² + 6x = 0

<=> x³ + x² + 6x² + 6x = 0

<=> x²(x + 1) + 6x(x + 1) = 0

<=> (x² + 6x)(x + 1) = 0

<=> x(x + 6)(x + 1) = 0

<=> \(\left[{}\begin{matrix}x=0\\x+6=0\\x+1=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=0\\x=-6\\x=-1\end{matrix}\right.\)

3. (x + 1)² - (2x + 3)² = 0

<=> (x + 1 + 2x + 3)(x + 1 - 2x - 3) = 0

<=> (3x + 4)(-2 - x) = 0

<=> \(\left[{}\begin{matrix}3x+4=0\\-2-x=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{-4}{3}\\x=-2\end{matrix}\right.\)

a)
\(=\left(x+2y\right)\left(x^2-xy+y^2\right)-3xy\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x^2-xy+y^2-3xy\right)\)
\(=\left(x+2y\right)\left(x^2-2xy+y^2\right)\)
\(=\left(x+2\right)\left(x-2\right)^2\)
b)
\(3x\left(2x-1\right)\left(2x+1\right)=0\)
3x=0 =>x=0
hoặc 2x-1=0 => 2x=1=>x=1/2
hoặc 2x+1=0=>2x=-1=>x=-1/2

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

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

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

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

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

2.\(x^2+3x+2=0\)

\(\Leftrightarrow x^2+x+2x+2=0\)

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

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

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

3.\(2x^2+5x+3=0\)

\(\Leftrightarrow2x^2+2x+3x+3=0\)

\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{2}\end{matrix}\right.\)

4.\(x^3+x^2-12x=0\)

\(\Leftrightarrow x\left(x^2+x-12\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\\x=3\end{matrix}\right.\)

a: \(\Leftrightarrow\left(x+2\right)\left(2x-3\right)-\left(x+2\right)\left(x-2\right)=0\)

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

=>(x+2)(x-1)=0

=>x=-2 hoặc x=1

b: =>(x+1)(x+2)=0

=>x=-1 hoặc x=-2

c: =>(2x+3)(x+1)=0

=>x=-1 hoặc x=-3/2

d: =>x(x+4)(x-3)=0

hay \(x\in\left\{0;-4;3\right\}\)

0X3=0

0 x 3 = 0

(Số nào nhân với 0 cũng bằng 0)

3 x 0 = 0

Mình nhanh,đúng nhất.

K mình nha

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