x : \(\frac{1}{4}\)= 8
Những câu hỏi liên quan
\(\frac{x+4}{x-1}+\frac{x-4}{x+1}=\frac{x-8}{x+2}+\frac{x+8}{x-2}-\frac{8}{3}\)
\(ĐKXĐ:x\ne1;-1;2;-2\)
\(\frac{\left(x+4\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{\left(x-4\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{\left(x-8\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x+8\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{8}{3}\)
\(\Leftrightarrow\frac{x^2+x+4x+4+x^2-x-4x+4}{x^2-1}=\frac{x^2-2x-8x+16+x^2+2x+8x+16}{x^2-4}-\frac{8}{3}\)
\(\Leftrightarrow\frac{2x^2+8}{x^2-1}=\frac{2x^2+32}{x^2-4}-\frac{8}{3}\)
\(\Leftrightarrow\frac{2x^2+8}{x^2-1}=\frac{3\left(2x^2+32\right)}{3\left(x^2-4\right)}-\frac{8\left(x^2-4\right)}{3\left(x^2-4\right)}\)
\(\Leftrightarrow\frac{2x^2+8}{x^2-1}=\frac{9x^2+96-8x^2+32}{3\left(x^2-4\right)}\)
\(\Leftrightarrow\frac{2x^2+8}{x^2-1}=\frac{x^2+128}{3\left(x^2-4\right)}\)
\(\Leftrightarrow3\left(x^2-4\right)\left(2x^2+8\right)=\left(x^2+128\right)\left(x^2-1\right)\)
\(\Leftrightarrow9x^4+24x^2-24x^2-96=x^4-x^2+128x^2-128\)
\(\Leftrightarrow9x^4+24x^2-24x^2-x^4+x^2+128x^2=-128+96\)
\(\Leftrightarrow8x^4+129x^2=-32\)
\(\Leftrightarrow8x^4+129x^2+32=0\)
\(\Leftrightarrow x=\frac{1}{2}\left(tmđkxđ\right)\)
bạn sai r bạn ơi cái chỗ chuyển vế dòng tương đương số 8 : x^4 - x^2 + 128x^2 - 128 đáng ra sau khi chuyển px là -x^4 +x^2 - 128x^2 + 128 chứ sao lại là x^4 + x^2 + 128x^2 +128
\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
Tính:\(\frac{1}{x}+\frac{1}{x+1}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}+\frac{32}{1+x^{32}}\)
Thu gọn : \(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(bt=\frac{1\left(1+x\right)}{\left(1-x\right)\left(1+x\right)}+\frac{1\left(1-x\right)}{\left(1+x\right)\left(1-x\right)}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2\left(1+x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{2\left(1-x^2\right)}{\left(1+x^2\right)\left(1-x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}}\)
\(=\frac{32}{1-x^{32}}\)
Chúc bạn làm bài tốt
Đúng 0
Bình luận (0)
tính:
\(\frac{1}{1-x}+\frac{1}{x+1}+\frac{2}{x^2+1}+\frac{4}{x^4+1}+\frac{8}{x^8+1}+\frac{16}{x^{16}+1}\)
tính A=\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\)
Xem chi tiết
Quy đồng tí là ra.. :>>
\(A=\frac{1+x+1-x}{\left(1-x\right)\left(1+x\right)}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\)
\(A=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\)
\(A=\frac{2+2x^2+2-2x^2}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\)
\(A=\frac{4+4x^4+4-4^2x}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}\)
\(A=\frac{8}{1-x^8}+\frac{8}{1+x^8}=\frac{8+8x^8+8-8x^8}{\left(1-x^8\right)\left(1+x^8\right)}=\frac{16}{1-x^{16}}\)
Chúc bạn học tốt ~
Đúng 0
Bình luận (0)
Giải phương trình:1,left(x^2-x+1right)^4+5x^46left(x^2-x+1right)^42,frac{x+4}{x-1}+frac{x-4}{x+1}frac{x-8}{x+2}+frac{x+8}{x-2}+frac{8}{3}3,left|x-2015right|^{2015}+left|x-2016right|^{2016}14,frac{5}{2x-3}-frac{1}{x+2}frac{5}{x-6}-frac{7}{2x-1}5,left(x+2008right)^4+left(x+2009right)^4frac{1}{8}
Đọc tiếp
Giải phương trình:
1,\(\left(x^2-x+1\right)^4+5x^4=6\left(x^2-x+1\right)^4\)
2,\(\frac{x+4}{x-1}+\frac{x-4}{x+1}=\frac{x-8}{x+2}+\frac{x+8}{x-2}+\frac{8}{3}\)
3,\(\left|x-2015\right|^{2015}+\left|x-2016\right|^{2016}=1\)
4,\(\frac{5}{2x-3}-\frac{1}{x+2}=\frac{5}{x-6}-\frac{7}{2x-1}\)
5,\(\left(x+2008\right)^4+\left(x+2009\right)^4=\frac{1}{8}\)
tớ ko bt lm abc , tớ lm d thôi nha , thứ lỗi
\(\frac{5}{2x-3}-\frac{1}{x+2}=\frac{5}{x-6}-\frac{7}{2x-1}\)
\(\frac{3x+13}{2x^2+x-6}=\frac{5}{x-6}+\frac{7}{1-2x}\)
\(\frac{3x+13}{\left(x+2\right)\left(2x-3\right)}=\frac{3x+37}{\left(x-6\right)\left(2x-1\right)}\)
\(\frac{10-9x}{-4x^3+32x^2-51x+18}=0\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{10}{9}\end{cases}}\)
Bài 1: Thưch hiện phép tính:
\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
= 1+x+1--x/1-x^2 +2/1+x^2+....+16/1+x^26
= 2/1-x^2+2/1+x^2+....+16/1+x^16
= ........
= 16/1-x^16 + 16/1+x^16
= 16+16x^16+16-16x^16/1-x^32
= 32/1-x^32
k mk nha
Đúng 0
Bình luận (0)
ĐKXĐ: \(x\ne\pm1\)
\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}}\)
\(=\frac{32}{1-x^{32}}\)
Đúng 0
Bình luận (0)
Rút gọn biểu thức
A =\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{1+x+1-x}{\left(1+x\right)\left(1-x\right)}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2+2x^2+2-2x^2}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4+4x^4+4-4x^4}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{8+8x^8+8-8x^8}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{16}{1+x^{16}}\)
\(=\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}}\)
\(=\frac{16+16x^{16}+16-16x^{16}}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)
\(=\frac{32}{1-x^{32}}\)
Rút gọn A = \(\frac{1}{-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)