\(\left(3-x\right)^2+\frac{-9}{25}=\frac{2}{5}-\frac{8}{5}\)

\(\Rightarrow\left(3-x\right)^2+\frac{-9}{25}=\frac{-6}{5}\)

\(\Rightarrow\left(3-x\right)^2=\frac{-6}{5}+\frac{9}{25}\)

\(\Rightarrow\left(3-x\right)^2=\frac{-30}{25}+\frac{9}{25}\)

\(\Rightarrow\left(3-x\right)^2=\frac{-21}{25}\)

..đến đey thì có vấn đề =="

bạn ơi bài toán có sai đề bài ko nếu ko sai thì mình nghĩ lại

Mk đang tắc chỗ đó đấy

Dù sao cũng Thanks bạn nhé

\(\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right).x+\frac{4}{325}+1+\left(\frac{5}{324}+1\right)+\frac{349}{5}-4=0\)

\(\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right).x=-329\)

=>x=-329

 (x-3)^2 *(x+3) *(x-3) =

 (x+4)^2 -(x-2)^2 -2(x+2) *(x-2) =

a) \(\left(x-3\right)^2.\left(x+3\right).\left(x-3\right)\)

\(=\left(x-3\right).\left(x-3\right).\left(x+3\right).\left(x-3\right)\)

\(=\left(x-3\right)^3.\left(x+3\right)\)

\(=\left(3x-9\right).\left(x+3\right)\)

Phần b tương tự

    m = 1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99

    m = 1/3-1/99=32/99

   Sorry chị em ko làm đc câu b vì em mới học lớp 4

   k em ha

a) \(M=\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{97\times99}\)

\(\Rightarrow M=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

\(\Rightarrow M=\frac{1}{3}-\frac{1}{99}\)

\(\Rightarrow M=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)

b)  \(N=\frac{3}{5\times7}+\frac{3}{7\times9}+\frac{3}{9\times11}+...+\frac{3}{197\times199}\)

\(\Rightarrow N=3\times\left(\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+...+\frac{1}{197\times199}\right)\)

\(\Rightarrow N=3\times\left[2\times\left(\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+...+\frac{1}{197\times199}\right)\right]\)

\(\Rightarrow N=3\times\left(\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}+...+\frac{2}{197\times199}\right)\)

\(\Rightarrow N=3\times\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{199}\right)\)

\(\Rightarrow N=3\times\left(\frac{1}{5}-\frac{1}{199}\right)\)

\(\Rightarrow N=3\times\frac{194}{995}=\frac{582}{995}\)

----Chúc em học giỏi !----

a M=1/3-1/5+1/5-1/7+1/7-1/9+....+1/97-1/99

M=1/3-1/99

M= 33/99-1/99

M=32/99

b

N=3/2( 2/5x7+2/7x9+2/9x11+...+ 2/ 197x199)

N=3/2x(1/5-1/7+1/7-1/9+1/9-1/11+.....+ 1/197-1/199)

N=3/2x( 1/5-1/199)

N=3/2x194/995

N=291/995

chắc chắn đúng 100% nha bạn

Ta có : \(x-\frac{3}{1.5}-\frac{3}{5.9}-\frac{3}{9.13}-\frac{3}{13.17}-\frac{3}{17.21}=\frac{2}{7}\)

=> \(x-\left(\frac{3}{1.5}+\frac{3}{5.9}+\frac{3}{9.13}+\frac{3}{13.17}+\frac{3}{17.21}\right)=\frac{2}{7}\)

=> \(x-\frac{3}{4}\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}\right)=\frac{2}{7}\)

=> \(x-\frac{3}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{3}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}\right)=\frac{2}{7}\)

=> \(x-\frac{3}{4}\left(1-\frac{1}{21}\right)=\frac{2}{7}\)

=> \(x-\frac{3}{4}.\frac{20}{21}=\frac{2}{7}\)

=> \(x-\frac{5}{7}=\frac{2}{7}\)

=> x = 1

Vậy x = 1

a) 4/9 x ( 3/5 + 8/5 - 2/10 )

= 4/9 x 1

=4/9

b) ( 312 + 325 - 247 ) : 13

= 390 : 13

= 30

a)4/9 x 3/5 + 4/9 x 8/5 - 4/9 x 2/10 

=4/9x(3/5+8/5-2/10)

= 4/9x1

=4/9

b,312 : 13 + 325 : 13 - 247 : 13

=(312+325-247) : 13

=         390 : 3

= 130

a: Ta có: \(\left(x-\dfrac{2}{5}\right)\left(x+\dfrac{2}{7}\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{2}{5}\\x< -\dfrac{2}{7}\end{matrix}\right.\)

Q=(2^9.3+2^9.5):2^12

Đặt A=2^9.3+2^9.5

      A=2^9.(3+5)

      A=2^9.8

Mặt khác:8=2^3

=>A=2^9.2^3

     A=2^12

Theo đề bài ta có Q=(2^9.3+2^9.5):2^12

=>Q=2^12:2^12

    Q=1

Nhìn dài dòng thế thôi chứ đơn giản lắm.Nếu thấy đúng thì cho mình nhé!