\(x\times2,4+x:10=10\)

\(\Rightarrow x\times2,4+x\times0,1=10\)

\(\Rightarrow x\times\left(2,4+0,1\right)=10\)

\(\Rightarrow x\times2,5=10\)

\(\Rightarrow x=10:2,5\)

\(\Rightarrow x=4\)

x * 2,4 + x : 10  = 10

x * 2,4 + x * 0,1 = 10

x * ( 2,4 + 0,1)   = 10

x *      2,5          = 10

x                       = 10 : 2,5

x                       =      4

X x 24/10 + X x 1/10 = 10

X x (24/10+1/10) = 10

X x 25/10 = 10

X=10:25/10 

X=4

X x 24/10 + X x 1/10 = 10

X x (24/10+1/10) = 10

X x 25/10 = 10

X=10:25/10 

X=4

Gọi biểu thức trên là A

Ta có:

2A = (\(\dfrac{1}{2.4}\)+\(\dfrac{1}{4.6}\)+...+\(\dfrac{1}{x.\left(x+2\right)}\)).2

2A = \(\dfrac{2}{2.4}\)+\(\dfrac{2}{4.6}\)+...+\(\dfrac{2}{x\left(x+2\right)}\)

2A = \(\dfrac{1}{2}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{6}\)+...+\(\dfrac{1}{x}\)-\(\dfrac{1}{x+2}\)

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

mà A = \(\dfrac{1}{10}\)(đề bài)

nên 2A = \(\dfrac{2}{10}\) hay \(\dfrac{1}{2}\) - \(\dfrac{1}{x+2}\) = \(\dfrac{2}{10}\)

                     suy ra \(\dfrac{1}{x+2}\) = \(\dfrac{1}{2}\)-\(\dfrac{2}{10}\)=\(\dfrac{3}{10}\) 

1) (-10)-|5-x|=(-12)

|5-x|=(-10)-(-12)

|5-x|=2

\(\Rightarrow\left[{}\begin{matrix}5-x=2\\5-x=\left(-2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=7\end{matrix}\right.\)

Vậy:............

1. ... \(\Leftrightarrow\left(x-2\right)^2-\left(x^2-9\right)=6\Leftrightarrow\left(x-2\right)^2-\left(x\right)^2=-3\Leftrightarrow\left(x-2-x\right)\left(x-2+x\right)=-3\)

\(\Leftrightarrow-2\left(2x-2\right)=-3\Leftrightarrow4x-4=3\Leftrightarrow4x=7\Leftrightarrow x=\frac{7}{4}\)

2. ... \(\Leftrightarrow4\left(x-3\right)^2-\left(4x^2-1\right)=10\Leftrightarrow\left[2\left(x-3\right)\right]^2-\left(2x\right)^2=9\)

\(\Leftrightarrow\left(2x-6-2x\right)\left(2x-6+2x\right)=9\Leftrightarrow-6\left(4x-6\right)=9\Leftrightarrow-24x=-27\Leftrightarrow x=\frac{9}{8}\)

3. ... \(\Leftrightarrow\left(x-4\right)^2-\left(x^2-4\right)=6\Leftrightarrow\left(x-4\right)^2-\left(x\right)^2=2\Leftrightarrow\left(x-4-x\right)\left(x-4+x\right)=2\)

\(\Leftrightarrow-4\left(2x-4\right)=2\Leftrightarrow-8x=-14\Leftrightarrow x=\frac{7}{4}\)

4. ... \(\Leftrightarrow9\left(x+1\right)^2-\left(9x^2-4\right)=10\Leftrightarrow\left[3\left(x+1\right)\right]^2-\left(3x\right)^2=6\)

\(\Leftrightarrow\left(3x+3-3x\right)\left(3x+3+3x\right)=6\Leftrightarrow3\left(6x+3\right)=6\Leftrightarrow6x=-1\Leftrightarrow x=-\frac{1}{6}\)

Ta có : (x - 2)² - (x - 3) (x + 3) = 6 

<=> (x² - 4x + 4) - (x² - 3²) = 6

=> x² - 4x + 4 - x² + 9 = 6

=> -4x + 13 = 6

=> -4x = -7

=> x = \(\frac{7}{4}\)

=>