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.

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

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

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

$\Leftrgihtarrow 24x=22$

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

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

a: \(\left(\frac{x}{x^2-25}-\frac{x-5}{x^2+5x}\right):\frac{2x-5}{x^2+5x}+\frac{x}{5-x}\)

\(=\left(\frac{x}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{x\left(x+5\right)}\right):\frac{2x-5}{x\left(x+5\right)}-\frac{x}{x-5}\)

\(=\frac{x^2-\left(x-5\right)^2}{x\left(x-5\right)\left(x+5\right)}\cdot\frac{x\left(x+5\right)}{2x-5}-\frac{x}{x-5}\)

\(=\frac{\left(x-x+5\right)\left(x+x-5\right)}{\left(x-5\right)\left(2x-5\right)}-\frac{x}{x-5}=\frac{5}{x-5}-\frac{x}{x-5}=\frac{5-x}{x-5}\)

=-1

b: \(\left(\frac{9}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)\)

\(=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)

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

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

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

đéo bt

mong bạn nói văn minh

Ta có:

a) ( 45 – 5 x 9 ) x 1 x 2 x 3 x 4 x 5 x 6 x 7

= (45 – 45) x 1 x 2 x 3 x 4 x 5 x 6 x 7

= 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7

= 0

b) (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x (72 – 8 x 8 – 8)

= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x (72 – 64 – 8)

= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x 0

= 0

c) (36 – 4 x 9) : (3 x 5 x 7 x 9 x 11)

= (36 – 36) : (3 x 5 x 7 x 9 x 11)

= 0 : (3 x 5 x 7 x 9 x 11)

= 0

d) (27 – 3 x 9) : 9 x 1 x 3 x 5 x 7

= (27 – 27) : 9 x 1 x 3 x 5 x 7

= 0 : 9 x 1 x 3 x 5 x 7

=0

a) ( 45 – 5 x 9 ) x 1 x 2 x 3 x 4 x 5 x 6 x 7

= 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7

b) (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x (72 – 8 x 8 – 8)

= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x 0

c) (36 – 4 x 9) : (3 x 5 x 7 x 9 x 11)

= 0 : (3 x 5 x 7 x 9 x 11)

d) (27 – 3 x 9) : 9 x 1 x 3 x 5 x 7

= 0 : 9 x 1 x 3 x 5 x 7                                                                                                                  Nếu đúng thì k cho mình nhé bạn!

rảnh à

100

Nhập biểu thức ở dấu Σ cho dễ nhìn em ạ

dấu đấy là dấu j ạ

Đề như này hả bạn? :V

\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-9x^2+27x-27-\left(x^3-27\right)+9\left(x+1\right)^2=15\)

\(\Leftrightarrow-9x^2+27x+9\left(x^2+2x+1\right)=15\)

\(\Leftrightarrow45x=6\)\(\Leftrightarrow x=\dfrac{2}{15}\)

Vậy...

5: \(\Leftrightarrow9\left(x^2-5x-4\right)=36\left(x+1\right)+8\left(x^2-10x\right)\)

\(\Leftrightarrow9x^2-45x-36-36x-36-8x^2+80x=0\)

\(\Leftrightarrow x^2-x-72=0\)

=>(x-9)(x+8)=0

=>x=9 hoặc x=-8

6: \(\Leftrightarrow x^2-9=9x-x^2-9+x\)

\(\Leftrightarrow2x^2-10x=0\)

=>2x(x-5)=0

=>x=0 hoặc x=5

5, <=> 9x^2 - 45x - 36 = 36x + 36 + 8x^2 - 80x 

<=> x^2 - x - 72 = 0 <=> x = 9 ; x = -8 

6, <=> x^2 - 9 = 9x - x^2 - 9 + x = 10x - x^2 - 9 

<=> 2x^2 - 10x = 0 <=> x = 0 ; x = 5 

7, <=> (x-1)^2 = (3x+3)^2 

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

<=> (-2x-4)(4x+2) = 0 <=> x = -2;x=-1/2

8, = (x^2-10x-15)(x^2-10x+25)