b)\(\left(x-8\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x-8=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=2\end{cases}}\)

c) \(\left(x+1\right)+\left(x+2\right)+...+\left(x+10\right)=9x+200\)

\(\Leftrightarrow\left(x+x+...+x\right)+\left(1+2+...+10\right)=9x+200\) (10 số hạng x)

\(\Leftrightarrow10x+55=9x+200\Leftrightarrow x+55=200\)

\(\Leftrightarrow x=145\)

các bạn ơi giúp mik ý đầu tiên đi

các bạn ơi giúp mik với mik đang cần gấp ai làm nhanh mik sẽ k giải chi tiết giúp mik nhé cảm ơn

(125^3 .7^5 -17^5 ) : 2001^2

(x-8) .(x-2) =0

(x+1) +(x+2)+...........+(x+10)=9.x+200 

a) ta có: A = 3^0 + 3^1 + 3^2 + ...+ 3^100

=> 3A = 3^1 + 3^2 + 3^3 + ...+ 3^101

=> 3A-A = 3^101 - 3^0

2A = 3^101 - 1

\(A=\frac{3^{101}-1}{2}\)

b) D = 1 - 5 + 5^2 - 5^3 + ...+ 5^98 - 5^99

=> 5D = 5 - 5^2 + 5^3 - 5^4+...+ 5^99 - 5^100

=> 5D+D = -5^100 + 1

6D = -5^100 + 1

\(D=\frac{-5^{100}+1}{6}\)

2x^2  - xy + 2x - y = 5 

=> (2 x^2 - xy) + (2x - y) = 5 

=> x (2 x - y) + (2x - y) = 5 

=> (x + 1 ) (2 x - y)= 5 

th1:

=> \(\hept{\begin{cases}x+1=5\\2x-y=1\end{cases}}\)    

=>  \(\hept{\begin{cases}x=4\\y=7\end{cases}}\)   

th2

=>\(\hept{\begin{cases}x+1=1\\2x-y=5\end{cases}}\)

=>  \(\hept{\begin{cases}x=0\\y=-5\end{cases}}\)    

th3

=>  \(\hept{\begin{cases}x+1=-1\\2x-y=-5\end{cases}}\)   

=>  \(\hept{\begin{cases}x=-2\\y=-1\end{cases}}\)       

th4

=>  \(\hept{\begin{cases}x+1=-5\\2x-y=-1\end{cases}}\)      

=>   \(\hept{\begin{cases}x=-6\\y=-11\end{cases}}\)         

:2x^2-xy+2x-y=5   

cái đề này có y mà

có tìm y ko?

6\(^2\)+ 64 : ( x - 1 ) = 52

36 + 64 : ( x - 1 ) =52

        64 ; ( x - 1 ) =64 : 52

                x - 1 = \(\frac{16}{13}\)

               x  = \(\frac{16}{13}\)+1

                x = \(\frac{29}{13}\)

HT

a)\(\dfrac{24}{36}\)=\(\dfrac{8}{12}\)

b)\(\dfrac{14}{56}\)=\(\dfrac{1}{4}\)

c)\(\dfrac{9}{24}\)=\(\dfrac{21}{56}\)

Chúc bạn học tốt!

a: x/12=24/36

nên x/12=2/3

=>x=2/3x12=8

b: 1/x=14/56

nên 1/x=1/14

hay x=14

c: x/56=9/24

=>x/56=3/8

hay x=21

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+.......+\dfrac{1}{x\cdot\left(x+1\right)}=\dfrac{122}{123}\)

\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{122}{123}\)

\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{122}{123}\)

\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{123}\)

\(\Leftrightarrow x=122\)

\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)

\(\Rightarrow A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)\)

\(\Rightarrow A=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\)

\(\Rightarrow A=\left(2+2^2\right)\left(1+2^2+...+2^{98}\right)\)

\(\Rightarrow A=6\left(1+2^2+...+2^{98}\right)⋮6\)