X x (1/2+1/4+1/8+1/16+1/32+1/64+1/128) = 127/128

X x 127/128                                            = 127/128

                                                        X   = 127/128 : 127/128

                                                        X   = 1

\(\frac{127}{128}\times x=1\)

\(\Rightarrow x=\frac{128}{127}\)

minh muon ban trinh bay ro rang hon

ko bít lm bài này ak! dễ mak

bày cho câu b nha!

umk

Không thích khai triển hằng đẳng thức bậc 5 thì có thể làm thế này, dễ hiểu dễ biến đổi:

\(sin^6x+cos^6x=\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=1-\dfrac{3}{4}sin^22x\)

\(=1-\dfrac{3}{4}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4x\right)=\dfrac{5}{8}+\dfrac{3}{8}cos4x\)

\(sin^4x+cos^4x=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=1-\dfrac{1}{2}sin^22x\)

\(=1-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4x\right)=\dfrac{3}{4}+\dfrac{1}{4}cos4x\)

\(sin^{10}x+cos^{10}x=\left(sin^6x+cos^6x\right)\left(sin^4x+cos^4x\right)-sin^4x.cos^4x\left(sin^2x+cos^2x\right)\)

\(=\left(\dfrac{5}{8}+\dfrac{3}{8}cos4x\right)\left(\dfrac{3}{4}+\dfrac{1}{4}cos4x\right)-\dfrac{1}{16}sin^42x\)

\(=\dfrac{15}{32}+\dfrac{3}{8}cos4x+\dfrac{3}{32}cos^24x-\dfrac{1}{16}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4x\right)^2\)

\(=\dfrac{15}{32}+\dfrac{3}{8}cos4x+\dfrac{3}{32}\left(\dfrac{1}{2}+\dfrac{1}{2}cos8x\right)-\dfrac{1}{64}\left(1-2cos4x+cos^24x\right)\)

\(=\dfrac{15}{32}+\dfrac{3}{8}cos4x+\dfrac{3}{64}+\dfrac{3}{64}cos8x-\dfrac{1}{64}+\dfrac{1}{32}cos4x-\dfrac{1}{64}\left(\dfrac{1}{2}+\dfrac{1}{2}cos8x\right)\)

\(=\dfrac{63}{128}+\dfrac{13}{32}cos4x+\dfrac{5}{128}cos8x\)

ko biết mk mới học lớp 5 nữa

 Khó quá thì nên hoc24 .vn 

Nếu đúng thì ấnĐúng 2 không những thees sau khi ấn sẽ may mắn cả năm

                    Đặt \(A=\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)

                    \(A=\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}+\frac{1}{2^8}\)

                \(2A=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}\)

               \(2A-A=\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}\right)-\left(\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}+\frac{1}{2^8}\right)\)

              \(A=\frac{1}{2^2}-\frac{1}{2^8}\)

           \(A=\frac{1}{4}-\frac{1}{256}=\frac{63}{256}\)

          \(\Rightarrow\frac{63}{256}.x=\frac{1}{512}=\frac{1}{2^9}\)

           \(\Rightarrow\frac{63}{2^8}.x=\frac{1}{2^9}\)

            \(\Rightarrow x=\frac{1}{2^9}:\frac{63}{2^8}=\frac{1}{2^9}.\frac{2^8}{63}=\frac{1}{2.63}=\frac{1}{126}\)

           Ủng hộ mk nha !!! ^_^

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)

\(2\times A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)

\(2\times A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\right)\)

\(A=1-\frac{1}{128}\)

\(A=\frac{127}{128}\)

\(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)

\(2\times B=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\)

\(B=1-\frac{1}{16}=\frac{15}{16}\)

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)

\(\Leftrightarrow4\times x+\frac{15}{16}=1\)

\(\Leftrightarrow4\times x=\frac{1}{16}\)

\(\Leftrightarrow x=\frac{1}{64}\)

a.1025

mk chịu câu b

*** mk nha