C

Đẳng thức nào đúng:

 A.(x-3)^2=x^2+6x+9                              B.(x-3)^2=9-x^2

C.(x-3)^2=9-6x+x^2                               D.(x-3)^2=x^2-3x+9

Câu C

Em ơi em đăng câu hỏi thì dùng công thức toán học để viết chữ viết thế này ai hiểu đúng được em?

1: =>x^2+4x-21=0

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

=>x=3 hoặc x=-7

2: =>(2x-5-4)(2x-5+4)=0

=>(2x-9)(2x-1)=0

=>x=9/2 hoặc x=1/2

3: =>x^3-9x^2+27x-27-x^3+27+9(x^2+2x+1)=15

=>-9x^2+27x+9x^2+18x+9=15

=>18x=15-9-27=-21

=>x=-7/6

6: =>4x^2+4x+1-4x^2-16x-16=9

=>-12x-15=9

=>-12x=24

=>x=-2

7: =>x^2+6x+9-x^2-4x+32=1

=>2x+41=1

=>2x=-40

=>x=-20

mọi người ơi giúp với tui đang cần gấp .Ai làm nhanh nhất thì tui sẽ h cho

`a,3(x-2)^2+9(x-1)=3(x^2+x-3)`

`<=>3(x^2-4x+4)+9x-9=3x^2+3x-9`

`<=>3x^2-12x+12+9x-9=3x^2+3x-9`

`<=>3x^2-3x+3=3x^2+3x-9`

`<=>6x=12`

`<=>x=12`

`b,(x+3)^2-(x-3)=6x+18`

`<=>(x+3-x+3)(x+3+x-3)+6x+18`

`<=>6.2x=6(x+3)`

`<=>2x=x+3`

`<=>x=3`

`c,(2x+7)^2=9(x+2)^2`

`<=>(2x+7)^2=(3x+6)^2`

`<=>(3x+6-2x-7)(3x+6+2x+7)=0`

`<=>(x-1)(5x+13)=0`

`<=>` $\left[ \begin{array}{l}x-1=0\\5x+13=0\end{array} \right.$

`<=>` $\left[ \begin{array}{l}x=1\\5x=-13\end{array} \right.$

`<=>` $\left[ \begin{array}{l}x=1\\x=-\dfrac{13}{5}\end{array} \right.$

a) Ta có: \(3\left(x-2\right)^2+9\left(x-1\right)=3\left(x^2+x-3\right)\)

\(\Leftrightarrow3\left(x^2-4x+4\right)+9x-9=3x^2+3x-9\)

\(\Leftrightarrow3x^2-12x+12+9x-9-3x^2-3x+9=0\)

\(\Leftrightarrow-6x+12=0\)

\(\Leftrightarrow-6x=-12\)

hay x=2

Vậy: x=2

Lời giải:

a. $x(3x+1)+(x-1)^2-(2x+1)(2x-1)=0$

$\Leftrightarrow (3x^2+x)+(x^2-2x+1)-(4x^2-1)=0$

$\Leftrightarrow 3x^2+x+x^2-2x+1-4x^2+1=0$

$\Leftrightarrow (3x^2+x^2-4x^2)+(x-2x)+(1+1)=0$

$\Leftrightarrow -x+2=0$

$\Leftrightarrow x=2$

b.

$(x+1)^3+(2-x)^3-9(x-3)(x+3)=0$

$\Leftrightarrow [(x+1)+(2-x)][(x+1)^2-(x+1)(2-x)+(2-x)^2]-9(x-3)(x+3)=0$

$\Leftrightarrow 3[x^2+2x+1-(x-x^2+2)+(x^2-4x+4)]-9(x-3)(x+3)=0$

$\Leftrightarrow 3(3x^2-3x+3)-9(x^2-9)=0$

$\Leftrightarrow 9(x^2-x+1)-9(x^2-9)=0$

$\Leftrightarrow 9(x^2-x+1-x^2+9)=0$
$\Leftrightarrow 9(-x+10)=0$

$\Leftrightarrow -x+10=0\Leftrightarrow x=10$

c.

$(x-1)^3-(x+3)(x^2-3x+9)+3x^2=25$

$\Leftrightarrow (x^3-3x^2+3x-1)-(x^3+3^3)+3x^2=25$

$\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2=25$
$\Leftrightarrow (x^3-x^3)+(-3x^2+3x^2)+3x-28=25$

$\Leftrightarrow 3x-28=25$

$\Leftrightarrow x=\frac{53}{3}$

d.

$(x+2)^3-(x+1)(x^2-x+1)-6(x-1)^2=23$
$\Leftrightarrow (x^3+6x^2+12x+8)-(x^3+1)-6(x^2-2x+1)=23$

$\Leftrightarrow x^3+6x^2+12x+8-x^3-1-6x^2+12x-6=23$

$\Leftrightarrow (x^3-x^3)+(6x^2-6x^2)+(12x+12x)+(8-1-6)=23$
$\Leftrightarrow 24x+1=23$

$\Leftrgihtarrow 24x=22$

$\Leftrightarrow x=\frac{11}{12}$

e.

$(x+3)(x^2-3x+9)-x(x-2)(x+2)+11=0$

$\Leftrightarrow x^3+3^3-x(x^2-4)+11=0$

$\Leftrightarrow x^3+27-x^3+4x+11=0$

$\Leftrightarrow (x^3-x^3)+4x+(27+11)=0$

$\Leftrightarrow 4x+38=0$

$\Leftrightarrow x=\frac{-19}{2}$

f.

$x(x-3)-x+3=0$

$\Leftrightarrow x(x-3)-(x-3)=0$

$\Leftrightarrow (x-3)(x-1)=0$

$\Leftrightarrow x-3=0$ hoặc $x-1=0$

$\Leftrightarrow x=3$ hoặc $x=1$

a: \(\dfrac{5x+y^2}{x^2y}-\dfrac{5y-x^2}{xy^2}\)

\(=\dfrac{5xy+y^3-x\left(5y-x^2\right)}{x^2y^2}\)

\(=\dfrac{5xy+y^3-5xy+x^3}{x^2y^2}=\dfrac{x^3+y^3}{x^2y^2}\)

b: \(\dfrac{x+9}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x\left(x+3\right)}\)

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

`9/2xx7/3-4/3xx9/2`

`=9/2xx(7/3-4/3)`

`=9/2xx3/3`

`=9/2xx1`

`=9/2`