2

cái khoảng trống là jv

x= { -6;-5;-4;-3;-2;-1;0;1;2;3}

\(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3\)

\(\Leftrightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)

\(\Leftrightarrow\left(2x-15\right)^3.\left(2x-16\right).\left(2x-14\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=15\\2x=16\\x=14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)

`(2x-15)^5 =(2x-15)^3`

`<=> (2x-15)^5 -(2x-15)^3 =0`

`<=>(2x-15)^3*[(2x-15)^2-1]=0`

`@TH1 : (2x-15)^3=0 => 2x-15 =0`

`=> 2x=15 => x=15/2`

`@TH2 : (2x-15)^2 -1 =0`

`=> [(2x-15=1),(2x-15=-1):}`

`=> [(2x=1+15=16),(2x=-1+15=14):}`

`=>[(x=8),(x=7):}`

Vậy `x in {15/2;8;7}`

\(a,86-\left(86+x\right)=10\\ =>86+x=86-10\\ =>86+x=76\\ =>x=76-86\\ =>x=-10\\ b,10-x=7-\left(-8\right)\\ =>10-x=15\\ =>x=10-15\\ =>x=-5\\ c,\left(-24+x\right)-43=15\\ =>-24+x=15+43\\ =>-24+x=58\\ =>x=58-\left(-24\right)\\ =>x=82\\ d,\left(-x+6\right)+12=48\\ =>-x+6=48-12\\ =>-x+6=36\\ =>-x=36-6\\ =>-x=30\\ =>x=-30 \)

a. 86 - ( 86 + x ) = 10

             86 + x = 86 - 10

             86 + x = 76

                     x = 76 - 86

                     x = -10

b. 10 - x = 7 - ( -8 )

    10 - x = 15

           x = 10 - 15

           x = -5

c. ( -24 + x ) - 43 = 15

                -24 + x = 15 + 43

                -24 + x = 58

                         x = 58 - ( -24 )

                         x = 82

d. ( -x + 6 ) + 12 = 48

                 -x + 6 = 48 - 12

                 -x + 6 = 36

                       -x = 36 - 6

                       -x = 30

Tui ko bik câu d lm đúng hay ko đâu nha!!!

tôi đang cần gấp

a1+a2+a3+...+a2021=0
a=0 vì a1,a2,a3,...= a.1,a.2,a.3,... nên a=0 thì kết quả sẽ bằng 0
a1+a2=a3+a4=.....=a2021+a1=1
a2021= a.2021= 0.2021=0
a1=a.1=1.1= 1
\(\Rightarrow\) a2021+a1=a.2021+a.1=0.2021+1.1=0+1=1