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M = \(\left(\dfrac{4}{x-4}-\dfrac{4}{x+4}\right).\dfrac{x^2+8x+16}{32}\)
= \(\left(\dfrac{4\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}-\dfrac{4\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}\right).\dfrac{\left(x+4\right)^2}{32}\)
= \(\left(\dfrac{4x+16-4x+16}{\left(x-4\right)\left(x+4\right)}\right).\dfrac{\left(x+4\right)^2}{32}\)
= \(\dfrac{32}{x^2-16}.\dfrac{\left(x+4\right)^2}{32}\)
= \(\dfrac{\left(x+4\right)^2}{x^2-16}\) \(=\dfrac{x+4}{x-4}\)
\(=\left(\dfrac{1}{1-x}+\dfrac{1}{1+x}\right)+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{8}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{32}{1-x^{32}}\)
a) \(\dfrac{2x}{x^2+2xy}+\dfrac{y}{xy-2y^2}+\dfrac{4}{x^2-4y^2}\)
\(=\dfrac{2x}{x\left(x+2y\right)}+\dfrac{y}{y\left(x-2y\right)}+\dfrac{4}{\left(x-2y\right)\left(x+2y\right)}\) MTC: \(xy\left(x-2y\right)\left(x+2y\right)\)
\(=\dfrac{2x.y\left(x-2y\right)}{xy\left(x+2y\right)\left(x-2y\right)}+\dfrac{y.x\left(x+2y\right)}{xy\left(x-2y\right)\left(x+2y\right)}+\dfrac{4.xy}{xy\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2xy\left(x-2y\right)+xy\left(x+2y\right)+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{2x^2y-4xy^2+x^2y+2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{3x^2y-2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
b) \(\dfrac{1}{x-y}+\dfrac{3xy}{y^3-x^3}+\dfrac{x-y}{x^2+xy+y^2}\)
\(=\dfrac{1}{x-y}-\dfrac{3xy}{x^3-y^3}+\dfrac{x-y}{x^2+xy+y^2}\)
\(=\dfrac{1}{x-y}-\dfrac{3xy}{\left(x-y\right)\left(x^2+xy+y^2\right)}+\dfrac{x-y}{x^2+xy+y^2}\) MTC: \(\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=\dfrac{x^2+xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}-\dfrac{3xy}{\left(x-y\right)\left(x^2+xy+y^2\right)}+\dfrac{\left(x-y\right)\left(x-y\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{\left(x^2+xy+y^2\right)-3xy+\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{x^2+xy+y^2-3xy+x^2-2xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2x^2-4xy+2y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2\left(x^2-2xy+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2\left(x-y\right)}{x^2+xy+y^2}\)
bn kt lại đề giúp mk , mk nghĩ mấu phải là x2 - 1 ; x4 - 1 ; x16 - 1
Sửa đề
\(\dfrac{1}{x-1}-\dfrac{1}{x+1}-\dfrac{2}{x^2+1}-\dfrac{4}{x^4+1}-\dfrac{8}{x^8+1}-\dfrac{16}{x^{16}+1}\)
\(=\dfrac{2}{x^2-1}-\dfrac{2}{x^2+1}-\dfrac{4}{x^4+1}-\dfrac{8}{x^8+1}-\dfrac{16}{x^{16}+1}\)
\(=\dfrac{4}{x^4-1}-\dfrac{4}{x^4+1}-\dfrac{8}{x^8+1}-\dfrac{16}{x^{16}+1}\)
\(=\dfrac{8}{x^8-1}-\dfrac{8}{x^8+1}-\dfrac{16}{x^{16}+1}\)
\(=\dfrac{16}{x^{16}-1}-\dfrac{16}{x^{16}+1}=\dfrac{32}{x^{32}-1}\)
Bài 2 .
a) \(\dfrac{2x}{x^2+2xy}+\dfrac{y}{xy-2y^2}+\dfrac{4}{x^2-4y^2}\)
\(=\dfrac{2x}{x\left(x+2y\right)}+\dfrac{y}{y\left(x-2y\right)}+\dfrac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2xy\left(x-2y\right)+xy\left(x+2y\right)+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{2x^2y-2xy^2+x^2y+2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{3x^2y+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
b) Sai đề hay sao ý
c) \(\dfrac{2x+y}{2x^2-xy}+\dfrac{16x}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)
\(=\dfrac{2x+y}{x\left(2x-y\right)}+\dfrac{-16x}{\left(2x-y\right)\left(2x+y\right)}+\dfrac{2x-y}{x\left(2x+y\right)}\)
\(=\dfrac{\left(2x+y\right)^2-16x^2+\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{4x^2+4xy+y^2-16x^2+4x^2-4xy+y^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{-8x^2}{x\left(2x-y\right)\left(2x+y\right)}\)
d) \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
.....
\(=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{32}{1-x^{32}}\)
a) \(\dfrac{5x-1}{3x+2}=\dfrac{5x-7}{3x-1}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\dfrac{5x-1}{3x+2}=\dfrac{5x-7}{3x-1}\)
\(=\dfrac{5x-1-5x+7}{3x+2-3x+1}\)
\(=\dfrac{-1+7}{2+1}\)
\(=\dfrac{6}{3}\)
\(=2\)
Với \(\dfrac{5x-1}{3x+2}=2\)
\(\Rightarrow5x-1=2\left(3x+2\right)\)
\(\Rightarrow5x-1-2\left(3x+2\right)=0\)
\(\Rightarrow5x-1-6x-4=0\)
\(\Rightarrow-x-5=0\)
\(\Rightarrow x=-5\)
\(\dfrac{x+14}{86}+1+\dfrac{x+15}{85}+1+\dfrac{x+16}{84}+1+\dfrac{x+17}{83}+\dfrac{x+16}{4}=4\)
\(\dfrac{x+100}{86}+\dfrac{x+100}{85}+\dfrac{x+100}{84}+\dfrac{x+100}{83}=4-\dfrac{x+16}{4}\)
\(\left(x+100\right)\left(\dfrac{1}{86}+\dfrac{1}{85}+\dfrac{1}{84}+\dfrac{1}{83}\right)=-x\)
Mk giải đế đây rùi bạn tự giải nốt đi
a: \(=\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}:\left(\dfrac{1}{x+1}+\dfrac{x}{x-1}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}:\dfrac{x-1+x^2+x+2}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{x^2+2x+1}=\dfrac{4x}{x^2+2x+1}\)
b: \(=\dfrac{x+2}{-\left(x-2\right)}\cdot\dfrac{\left(x-2\right)^2}{4x^2}\cdot\left(\dfrac{2}{2-x}-\dfrac{4}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\dfrac{x^2-2x+4}{2-x}\right)\)
\(=\dfrac{-\left(x+2\right)\left(x-2\right)}{4x^2}\cdot\left(\dfrac{2}{2-x}-\dfrac{4}{\left(x+2\right)\left(2-x\right)}\right)\)
\(=\dfrac{-\left(x+2\right)\left(x-2\right)}{4x^2}\cdot\dfrac{2x+4-4}{\left(2-x\right)\left(x+2\right)}\)
\(=\dfrac{2x}{4x^2}=\dfrac{1}{2x}\)
\(\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}=-\dfrac{16}{x^2-4}\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x+2\right)-\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=-\dfrac{16}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{x^2+2x+2x+4-\left(x^2-2x-2x+4\right)+16}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{x^2+4x+4-x^2+2x+2x-4+16}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{8x+16}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{8\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{8}{x-2}=0\)
\(\Leftrightarrow8=0\) (vô lý)
S=\(\left\{\varnothing\right\}\)
Giải:
ĐKXĐ: \(x\ne\pm2\)
\(\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}=\dfrac{-16}{x^2-4}\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}-\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}=\dfrac{-16}{x^2-4}\)
\(\Rightarrow\left(x+2\right)^2-\left(x-2\right)^2=-16\)
\(\Leftrightarrow x^2+4x+4-\left(x^2-4x+4\right)=-16\)
\(\Leftrightarrow x^2+4x+4-x^2+4x-4=-16\)
\(\Leftrightarrow8x=-16\)
\(\Leftrightarrow x=-2\) (không thỏa mãn)
Vậy ...
\(\dfrac{x}{x+4}+\dfrac{4}{x-4}-\dfrac{32}{x^2-16}\)
\(=\dfrac{x\left(x-4\right)+4\left(x+4\right)-32}{\left(x+4\right).\left(x-4\right)}\)
\(=\dfrac{x^2-4x+4x+16-32}{\left(x+4\right).\left(x-4\right)}\)
\(=\dfrac{x^2-16}{x^2-16}\)
\(=1\)
Ta có: \(\dfrac{x}{x+4}+\dfrac{4}{x-4}-\dfrac{32}{x^2-16}\)
\(=\dfrac{x\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}+\dfrac{4\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}-\dfrac{32}{x^2-16}\)
\(=\dfrac{x^2-4x+4x+16-32}{\left(x-4\right)\left(x+4\right)}\)
\(=\dfrac{x^2-16}{\left(x-4\right)\left(x+4\right)}\)
\(=\dfrac{\left(x-4\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=1\)