a)(x+2).(x+3)-(x-2).(x+5)=10

  ( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10

 x^2 +3x+2x+6-x^2 -5x+2x+10-10=0

 2x+6=0

2x=-6

x=-3

a, \(\dfrac{1}{7}\) x ( \(\dfrac{4}{3}\))² - \(\dfrac{1}{7}\) : \(\dfrac{9}{11}\) 

= \(\dfrac{1}{7}\) x  \(\dfrac{16}{9}\)   -   \(\dfrac{1}{7}\) x \(\dfrac{11}{9}\)

= \(\dfrac{1}{7}\) x ( \(\dfrac{16}{9}-\dfrac{11}{9}\))

= \(\dfrac{1}{7}\) x \(\dfrac{5}{9}\)

= \(\dfrac{5}{63}\)

b, 2x + \(\dfrac{1}{4}\) = \(\dfrac{3}{5}\)

    2x         = \(\dfrac{3}{5}-\dfrac{1}{4}\)

   2x          = \(\dfrac{7}{20}\)

     x          =  \(\dfrac{7}{20}:2\)

     x          = \(\dfrac{7}{40}\)

a, \(x\) - \(\dfrac{5}{7}\) = \(\dfrac{1}{9}\) 

    \(x\)        = \(\dfrac{1}{9}\) + \(\dfrac{5}{7}\)

    \(x\)        = \(\dfrac{52}{63}\)

b, \(\dfrac{2x}{5}\) = \(\dfrac{6}{2x+1}\)

     2\(x\).(2\(x\) + 1) = 30

     4\(x^2\)+ 2\(x\)  - 30 = 0

  4\(x^2\) + 12\(x\) - 10\(x\) - 30 = 0

   (4\(x^2\) + 12\(x\)) - (10\(x\) + 30) =0

   4\(x\).(\(x\) + 3) - 10.(\(x\) +3) = 0

        2 (\(x\) + 3).(2\(x\) -  5) = 0

            \(\left[{}\begin{matrix}x+3=0\\2x-5=0\end{matrix}\right.\)

             \(\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)

Vậy \(x\) \(\in\) {-3; \(\dfrac{5}{2}\)}

c, \(\dfrac{11}{8}\) + \(\dfrac{13}{6}\) = \(\dfrac{85}{x}\)

     \(\dfrac{85}{x}\)         = \(\dfrac{85}{24}\)

       \(x\)           = 24

Tìm x : 

a) | x + 12x | = 2x

=> \(\orbr{\begin{cases}13x=2x\\13x=-2x\end{cases}}\)

=>  \(\orbr{\begin{cases}11x=0\\15x=0\end{cases}}\)

=>  \(x=0\)

b) 3x − |x + 1| = 1

=> |x + 1|  = 3x -1

=>\(\orbr{\begin{cases}x+1=3x-1\\x+1=1-3x\end{cases}}\)

=>  \(\orbr{\begin{cases}2x=2\\4x=0\end{cases}}\)

=>  \(\orbr{\begin{cases}x=1\\x=0\end{cases}}\)

c) |2x + 3| = x + 1

=> \(\orbr{\begin{cases}2x+3=x+1\\2x+3=-x-1\end{cases}}\)

=>  \(\orbr{\begin{cases}x=-2\\3x=-4\end{cases}}\)

=> \(\orbr{\begin{cases}x=-2\\x=-\frac{4}{3}\end{cases}}\)

b) 3x - |x + 1| = 1

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

ĐK : \(x\ge\frac{1}{3}\)

Khi đó (1) <=> \(\orbr{\begin{cases}x+1=3x-1\\x+1=-3x+1\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=2\\4x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\left(\text{loại}\right)\\x=1\end{cases}}\)

Vậy x = 1

c) ĐK : x + 1\(\ge0\Rightarrow x\ge-1\)

Khi đó |2x + 3| = x + 1

<=> \(\orbr{\begin{cases}2x+3=x+1\\2x+3=-x-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{4}{3}\end{cases}}\left(\text{loại}\right)\)

Vậy \(x\in\varnothing\)

d) ||x + 9| + 11| = 2x + 11 (1)

ĐK : \(2x+11\ge0\Rightarrow x\ge-\frac{5}{2}\)

Khi đó (1) <=> \(\orbr{\begin{cases}\left|x+9\right|+11=2x+11\\\left|x+9\right|+11=-2x-11\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left|x+9\right|=2x\\\left|x+9\right|=-2x-22\end{cases}}\)

Khi |x + 9| = 2x (x \(\ge0\))

<=> \(\orbr{\begin{cases}x+9=2x\\x+9=-2x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\left(tm\right)\\x=-3\left(\text{loại}\right)\end{cases}}\)

Khi |x + 9| = -2x - 22 ( \(-\frac{5}{2}\le x\le-11\))

<=> \(\orbr{\begin{cases}x+9=-2x-22\\x+9=2x+22\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{31}{3}\\x=-13\end{cases}}\left(\text{loại}\right)}\)

Vậy x = 9 

đúng. Mik học qua lớp anh rồi lên mik biết

Xét cấp số cộng 1, 6, 11, ..., 96.

Ta có: 96 = 1 + 5(n − 1) ⇒ n = 20

Suy ra

Và 2x.20 + 970 = 1010

Từ đó x = 1

a) x – 2 = -6 + 17

x – 2 = 11

x = 11 + 2 = 13

b) x + 2 = -9 – 11

x + 2 = -20

x = -20 – 2 = -22

c) 2x + 5 = x – 1

2x – x = -1 – 5

x = -6

d) | x – 4 | = | -81 |

x – 4 = 81 hoặc x – 4 = -81

x = 81 + 4 hoặc x = -81 + 4

x = 85 hoặc x = -77

a) x - 8 - (12 - 2x) = -20

=> x - 8 - 12 + 2x = -20

=> (x + 2x) + (-8 - 12) = -20

=> 3x - 20 = -20

=> 3x = 0 => x = 0

b) -27 + (x + 8) - ( +11) = 2

=> -27 + x + 8 - 11 = 2

=> -27 + x = 2 + 11 - 8

=> -27 + x = 5

=> x = 5 - (-27) = 32

c) -2x - 16 = -2 - (3x + 9)

=> -2x - 16 = -2 - 3x - 9

=> -2x - 16 + 2 + 3x + 9 = 0

=> (-2x + 3x) + (-16 + 2 + 9) = 0

=> x - 5 = 0

=> x = 5

\(a,x-8-\left(12-2x\right)=-20\)

\(x-8-12+2x=-20\)

\(x+2x-8-12=-20\)

\(3x-20=-20\)

\(3x=-20+20\)

\(3x=0\)

\(x=0\)

\(b,-27+\left(x+8\right)-\left(+11\right)=2\)

\(-27+x+8-11=2\)

\(x-27+8-11=2\)

\(x-30=2\)

\(x=2+30\)

\(x=32\)

\(c,-2x-16=2-\left(3x+9\right)\)

\(-2x-16=2-3x-9\)

\(-2x+3x=2-9+16\)

\(x=9\)

Học tốt

a) x - 8 - ( 12 - 2x ) = -20

x - 8 - 12 + 2x = -20

x + 2x = -20 + 8 + 12

3x = 0

x = 0 \(\left(x\in Z,tm\right)\)

Vậy x=0

b) -27 + ( x + 8 ) - ( +11 ) = 2

-27 + x + 8 - 11 = 2

x = 2 + 27 - 8 + 11

x = 32 \(\left(x\in Z,tm\right)\) 

Vậy x = 32

c) -2x - 16 = 2 - ( 3x + 9 )

-2x - 16 = 2 - 3x - 9

-2x + 3x = 2 - 9 +16

x = 9 \(\left(x\in Z,tm\right)\)

Vậy x = 9