\(\frac{4}{x-4}-\frac{x}{x+4}+\frac{32}{16-x^2}\)
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\(F=\frac{3}{2}\cdot x^4-\frac{1}{16}\cdot x^4+\frac{1}{32}\cdot x^4-\frac{1}{4}\cdot x^4\)
\(=x^4\left(\frac{3}{2}-\frac{1}{16}+\frac{1}{32}-\frac{1}{4}\right)\)
\(=\frac{32}{39}\cdot x^4\)
Vì \(x\ne0\Rightarrow x^4>0\)
=> \(\frac{32}{39}x^4>0\forall x\ne0\)
\(ĐKXĐ:x\ne0\)
\(\frac{x+2}{x^2+2x+4}+\frac{x-2}{x^2-2x+4}=\frac{32}{x\left(x^4+4x^2+16\right)}\)
\(\Leftrightarrow\frac{x+2}{x^2+2x+4}+\frac{x-2}{x^2-2x+4}-\frac{32}{x\left(x^4+4x^2+16\right)}=0\)
\(\Leftrightarrow\frac{x\left(x+2\right)\left(x^2-2x+4\right)+x\left(x-2\right)\left(x^2+2x+4\right)-32}{x\left(x+2x+4\right)\left(x^2-2x+4\right)}=0\)
\(\Leftrightarrow x\left(x^3+8\right)+x\left(x^3-8\right)-32=0\)
\(\Leftrightarrow x\left(x^3+8+x^3-8\right)-32=0\)
\(\Leftrightarrow2x^4-32=0\)
\(\Leftrightarrow x^4=16\)
\(\Leftrightarrow x=\pm2\)
Vậy tập nghiệm của phương trình là \(S=\left\{2;-2\right\}\)
a, \(\frac{x^{32}+x^{16}+1}{x^{16}+x^8+1}\)
\(=\frac{x^8+x^4+1}{x^4+x^2+1}\) Vậy phân thức \(a=\frac{x^8+x^4+1}{x^4+x^2+1}\)
P/s; Căn thức a, là phân số tối giản
b, \(\frac{x^8+3x^4+4}{x^4+x^2+2}\)
\(=\frac{x^4+3x^2+2}{x^2+x^1+1}\) Vậy căn thức \(b=\frac{x^4+3x^2+2}{x^2+x^1+1}\)
P/s; Căn thức b, có thể rút gọn được cho 2 và 4
Em ko chắc đâu nhé *-*
a) \(\frac{36\left(x-2\right)}{32-16x}=\frac{36\left(x-2\right)}{16\left(2-x\right)}=-\frac{36\left(2-x\right)}{16\left(2-x\right)}=-\frac{36}{16}=-\frac{9}{4}\)
b) \(\frac{3x^2-12x+12}{x^4-8x}=\frac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}=\frac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}=\frac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}=\frac{3x-6}{x^3+2x^2+4x}\)
c) \(\frac{7x^2+14x+7}{3x^2+3x}=\frac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}=\frac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\frac{7\left(x+1\right)}{3x}=\frac{7x+7}{3x}\)
d) \(\frac{x^4-5x^2+4}{x^4-10x^2+9}=\frac{x^4-x^2-4x^2+4}{x^4-x^2-9x^2+9}=\frac{x^2\left(x^2-1\right)-4\left(x^2-1\right)}{x^2\left(x^2-1\right)-9\left(x^2-1\right)}=\frac{\left(x^2-4\right)\left(x^2-1\right)}{\left(x^2-9\right)\left(x^2-1\right)}=\frac{\left(x-2\right)\left(x+2\right)}{\left(x-3\right)\left(x+3\right)}\)
e) \(\cdot\frac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}=\frac{x^3\left(x+1\right)+\left(x+1\right)}{x^4-x^3+x^2+x^2-x+1}=\frac{\left(x^3+1\right)\left(x+1\right)}{x^2\left(x^2-x+1\right)+\left(x^2-x+1\right)}=\frac{\left(x+1\right)^2\left(x^2-x+1\right)}{\left(x^2+1\right)\left(x^2-x+1\right)}=\frac{\left(x+1\right)^2}{x^2+1}=\frac{x^2+2x+1}{x^2+1}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}\)
\(=1-\frac{1}{32}\)
\(=\frac{31}{32}\)
có tử = 1 thì k bt à nha
Đặt: \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(\Rightarrow\)\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
\(\Rightarrow\)\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\)
\(\Rightarrow\)\(A=1-\frac{1}{32}=\frac{31}{32}\)
mà \(A=\frac{1}{x}\)
nên \(\frac{1}{x}=\frac{31}{32}\)
\(\Rightarrow\)\(x=\frac{32}{31}\)
Vậy...
\(\frac{4}{x-4}-\frac{x}{x+4}+\frac{32}{16-x^2}.\)
\(=\frac{4\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}-\frac{x\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}-\frac{32}{\left(x-4\right)\left(x+4\right)}\)
\(=\frac{4x+16-x^2+4x-32}{\left(x-4\right)\left(x+4\right)}\)
\(=\frac{-x^2+8x-16}{\left(x-4\right)\left(x+4\right)}=\frac{-\left(x-4\right)^2}{\left(x-4\right)\left(x+4\right)}=\frac{-\left(x-4\right)}{x+4}\)