a) \(\Leftrightarrow\left(x+5\right)^2=4^2\)
    \(\Leftrightarrow x+5=4\)
    \(\Leftrightarrow x=-1\)
Vậy...
b) \(\Leftrightarrow2\left(x-4\right)+x\left(x-4\right)=0\)
    \(\Leftrightarrow\left(x-4\right)\left(2+x\right)=0\)
    \(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\2+x=0\end{matrix}\right.\)
    \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
Vậy...

Có dãy số 1+2+.....+x có tất cả x số hạng.

Theo công thức tính tổng dãy số ta có: \(\frac{\left(1+x\right)\cdot x}{2}=2415\)

\(\Rightarrow\)(1+x).x=2415.2=4830

\(\Rightarrow\)(1+x).x=70.69 [Vì (1+x).x là 2 số tự nhiên liên tiếp nhé bn]

\(\Rightarrow\)x=69

Chú ý dấu "." là dấu nhân nha bn

a: =>x+3/7=2+9/21=52/21

hay x=52/21-9/21=43/21

b: =>1/12-x=-11/9

hay x=1/12+11/9=47/36

e 2 - x - 3/7 = -3/7

2-x=0

x=2

f x=-11/9 - 13/36 + 5/18

-x= -47/36

x=47/36

\(a,=\dfrac{x+3+4x-3}{xy}=\dfrac{5x}{xy}=\dfrac{5}{y}\\ b,=\dfrac{x-2+4x-3}{x-1}=\dfrac{5\left(x-1\right)}{x-1}=5\\ c,=\dfrac{x^2+4-6x+5}{x-3}=\dfrac{\left(x-3\right)^2}{x-3}=x-3\\ d,=\dfrac{4-x^2+2x^2-2x+5-4x}{x-3}=\dfrac{\left(x-3\right)^2}{x-3}=x-3\)

kết quả thôi nha không nó dài lắm:
e) =\(\dfrac{3}{2x}\)
f) =\(\dfrac{2x}{\left(x+2\right)\left(x-2\right)}\)
g) =\(\dfrac{2}{x+1}\)
h) =\(\dfrac{4x+5}{x+2}\)

b: \(=\dfrac{x+3+x-1+x+4}{x+2}=\dfrac{3x+6}{x+2}=3\)

\(a,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{3x^2}\\ b,=\dfrac{4x-8+2x+4-5x+6}{\left(x+2\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\\ d,=\dfrac{1}{\left(x+3\right)^2}-\dfrac{1}{\left(x-3\right)^2}+\dfrac{x}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{x^2-6x+9-x^2-6x-9+x^3-9x}{\left(x+3\right)^2\left(x-3\right)^2}=\dfrac{x^3-21x}{\left(x+3\right)^2\left(x-3\right)^2}\)

a)   (x+2)² + (x + 1)(x - 1) - x(2x + 5)

    =   x² + 4x +4 + x² - 1 - 2x² - 5x

    = 3 - x

b)    (35x⁶y⁷ - 25x³y⁵): 5x³y⁵ - \(\dfrac{1}{3}x^3y^2\)

    = 7x³y² - 5 - \(\dfrac{1}{3}x^3y^2\)

     = \(\dfrac{20}{3}\)x³y² -5

c)   \(\dfrac{x-3}{x+3}+\dfrac{3}{3-x}+\dfrac{6x}{x^2-9}\)

= \(\dfrac{x-3}{x+3}-\dfrac{3}{x-3}+\dfrac{6x}{x^2-9}\)

= \(\dfrac{\left(x-3\right)^2-3\left(x+3\right)+6x}{x^2-9}\)

= \(\dfrac{x^2-6x+9-3x-9+6x}{x^2-9}\)

= \(\dfrac{x^2-3x}{x^2-9}\)

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

=\(\dfrac{x}{x+3}\)