a)\(x^2-4^2+6x-x^2=0\)

\(16+6x=0\)

\(x=\frac{8}{3}\)

b)x=3

\(\left(x+9\right)+\left(x-2\right)+\left(x+7\right)+\left(x-4\right)+\left(x+5\right)+\left(x-6\right)+\left(x+3\right)+\left(x-8\right)+\left(x+1\right)=95\)

\(x+9+x-2+x+7+x-4+x+5+x-6+x+3+x-8+x+1=95\)

\(\left(x\times9\right)+\left(9-2+7-4+5-6+3-8+1\right)=95\)

\(x\times9+5=95\)

\(x\times9=95-5\)

\(x\times9=90\)

\(x=90:9\)

\(x=10\)

(x + 9) + (x - 2) + (x + 7) + (x - 4) + (x + 5) + (x - 6) + (x + 3) + (x - 8) + (x + 1) = 95

x + 9 + x - 2 + x + 7 + x - 4 + x + 5 + x - 6 + x + 3 + x - 8 + x + 1

(x + x + x + x + x + x + x + x + x) + (9 - 8 + 7 - 6 + 5 - 4 + 3 - 2 + 1) = 95

9 × x + 5 = 95

9 × x = 95 - 5

9 × x = 90

x = 90 : 9

x = 10

$\left(x+9\right)+\left(x-2\right)+\left(x+7\right)+\left(x-4\right)+\left(x+5\right)+\left(x-6\right)+\left(x+3\right)+\left(x-8\right)+\left(x+1\right)=95$

$x+9+x-2+x+7+x-4+x+5+x-6+x+3+x-8+x+1=95$

$\left(x\times 9\right)+\left(9-2+7-4+5-6+3-8+1\right)=95$

$x\times 9+5=95$

$x\times 9=95-5$

$x\times 9=90$

$x=90:9$

$x=10$

a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)

\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)

\(\Leftrightarrow24x=-13\)

hay \(x=-\dfrac{13}{24}\)

a)4/9 : x/6 = 2/3

           x/6 = 4/9 : 2/3

          x/6  =   2/3 = 4/6

vậy x = 4

b)x/5 x 3/7 = 9/35

   x/5          = 9/35 : 3/7

   x/5          = 3/5

vậy x = 3

rảnh à

100

\(\left(x+9\right)+\left(x-8\right)+\left(x+7\right)+\left(x-6\right)+\left(x+5\right)+\left(x-4\right)+\left(x+3\right)+\left(x-2\right)+\left(x+1\right)=95.9\\ =>x+9+x-8+x+7+x-6+x+5+x-4+x+3+x-2+x+1=855\\ =>\left(x+x+x+x+x+x+x+x+x\right)+\left(9-8+7-6+5-4+3-2+1\right)=855\\ =>9x+5=855\\ =>9x=855-5\\ =>9x=850\\ =>x=\dfrac{850}{9}\)

\(\left(x+9\right)+\left(x-8\right)+...+\left(x-2\right)+\left(x+1\right)\)

\(=x+9+x-8+...+x-2+x+1\)

\(=\left(x+9+x-8\right)+...+\left(x+5\right)+...+\left(x-2+x+1\right)\)

(Ta gộp 4 số vào 1 tổng, riêng (x+5) là ta giữ nguyên)

\(=\left(2x-1\right)+...+\left(x+5\right)+...+\left(2x-1\right)\)

\(=4\left(2x-1\right)+\left(x+5\right)\)

\(=8x-4+x-5\)

\(=9x-9\)        (1)

Từ bài toán trên, ta có:

\(\left(x+9\right)+\left(x-8\right)+...+\left(x-2\right)+\left(x+1\right)=95,9\)

Từ (1)

\(\Leftrightarrow9x-9=95,9\)

\(9x=95,9-9\)

\(x=86,9:9\)

\(x=9,6\left(5\right)\)

x-4/9=13/20+2/3

a) Ta có: \(x+\dfrac{1}{3}=\dfrac{2}{6}\)

\(\Leftrightarrow x+\dfrac{1}{3}=\dfrac{1}{3}\)

hay x=0

Vậy: x=0

b) Ta có: \(x-\dfrac{1}{4}=\dfrac{1}{-2}\)

\(\Leftrightarrow x-\dfrac{1}{4}=\dfrac{-1}{2}\)

\(\Leftrightarrow x=\dfrac{-1}{2}+\dfrac{1}{4}=\dfrac{-2}{4}+\dfrac{1}{4}=\dfrac{-1}{4}\)

Vậy: \(x=-\dfrac{1}{4}\)

c) Ta có: \(\dfrac{-1}{6}=\dfrac{3}{2}x\)

\(\Leftrightarrow x=\dfrac{-1}{6}:\dfrac{3}{2}=\dfrac{-1}{6}\cdot\dfrac{2}{3}\)

hay \(x=\dfrac{-1}{9}\)

Vậy: \(x=\dfrac{-1}{9}\)

\(a.x=\dfrac{1}{3}-\dfrac{1}{3}\)

\(x=0\)

\(b.x-\dfrac{1}{4}=\dfrac{-1}{2}\)

\(x=\dfrac{-1}{2}+\dfrac{1}{4}\)

\(x=\dfrac{-1}{4}\)

c. \(\dfrac{-1}{6}=\dfrac{3}{2x}\)

\(-2x=18\)

\(x=-9\)

d. \(\dfrac{4}{5}=\dfrac{-12}{9-x}\)

\(4.\left(9-x\right)=-60\)

\(9-x=-15\)

\(x=24\)

\(e.\dfrac{x+1}{3}=\dfrac{3}{x+1}\)

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

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

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

f.\(\dfrac{x-1}{-4}=\dfrac{-4}{x-1}\)

\(\left(x-1\right)^2=16\)

\(\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)