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\(\frac{x^2-8}{x^2-16}=\frac{1}{x+4}+\frac{1}{x-4}\)
\(\Rightarrow\frac{x^2-8}{\left(x+4\right)\left(x-4\right)}=\frac{x-4}{\left(x+4\right)\left(x-4\right)}+\frac{x+4}{\left(x-4\right)\left(x+4\right)}\)
\(\Rightarrow x^2-8=x-4+x+4\)
\(\Rightarrow x^2-8=2x\)
\(\Rightarrow x^2-2x-8=0\)
\(\Delta=b^2-4ac=\left(-2\right)^2-4.1.\left(-8\right)=4+32=36>0\)
phương trình có 2 nghiệm phân biệt : \(x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{2+\sqrt{36}}{2}=\frac{2+6}{2}=\frac{8}{2}=4\)
\(x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{2-\sqrt{36}}{2}=\frac{2-6}{2}=\frac{-4}{2}=\left(-2\right)\)
đặt \(\hept{\begin{cases}x+\frac{1}{x}=a\\y+\frac{1}{y}=b\\z+\frac{1}{z}=c\end{cases}}\)=> \(\hept{\begin{cases}x^2+\frac{1}{x^2}=a^2-2\\y^2+\frac{1}{y^2}=b^2-2\\z^2+\frac{1}{z^2}=c^2-2\end{cases}}\)
thay vào đề ta đc: \(\hept{\begin{cases}a+b+c=\frac{51}{4}\\a^2+b^2+c^2-6=\frac{771}{16}=>a^2+b^2+c^2=\frac{867}{16}\end{cases}}\)
mình chưa học giải hpt nên đến đây k biết lm đc nữa k
=))
Lời giải:
ĐKXĐ: $x\neq \pm 4$
PT \(\Leftrightarrow \frac{x^2-8}{x^2-16}=\frac{x-4+x+4}{(x-4)(x+4)}=\frac{2x}{x^2-16}\)
\(\Rightarrow x^2-8=2x\)
\(\Leftrightarrow x^2-2x-8=0\)
\(\Leftrightarrow (x-1)^2-9=0\Leftrightarrow (x-1)^2-3^2=0\)
\(\Leftrightarrow (x-4)(x+2)=0\Rightarrow \left[\begin{matrix} x=4\\ x=-2\end{matrix}\right.\)
khó quá làm sao mà trả lời đc
Vắt óc đi
tự đầu mình vắt óc mà suy nghĩ
Nguyễn Kim Ngân Selina Joyce Không nàm thì cut
\(\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}+\frac{32}{1+x^{32}}\)
\(=\frac{2+2x^2+2-2x^2}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}+\frac{32}{1+x^{32}}\)
\(=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}+\frac{32}{1+x^{32}}\)
\(=\frac{4+4x^4+4-4x^4}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}+\frac{32}{1+x^{32}}\)
\(=\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}+\frac{32}{1+x^{32}}\)
\(=\frac{8+8x^8+8-8x^8}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{16}{1+x^{16}}+\frac{32}{1+x^{32}}\)
\(=\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}}+\frac{32}{1+x^{32}}\)
\(=\frac{32}{1-x^{32}}+\frac{32}{1+x^{32}}\)
\(=\frac{64}{1+x^{64}}\)
\(\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}+\frac{32}{1+x^{32}}\)
\(=\left(\frac{2}{1-x^2}+\frac{2}{1+x^2}\right)+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}+\frac{32}{1+x^{32}}\)
\(=\frac{2\left(1+x^2\right)+2\left(1-x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(+\frac{32}{1+x^{32}}\)
\(=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}+\frac{32}{1+x^{32}}\)
\(=\left(\frac{4}{1-x^4}+\frac{4}{1+x^4}\right)+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}+\frac{32}{1+x^{32}}\)
\(=\frac{4\left(1+x^4\right)+4\left(1-x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}+\frac{32}{1+x^{32}}\)
\(=\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}+\frac{32}{1+x^{32}}\)
\(=\left(\frac{8}{1-x^8}+\frac{8}{1+x^8}\right)+\frac{16}{1+x^{16}}+\frac{32}{1+x^{32}}\)
\(=\frac{8\left(1-x^8\right)+8\left(1+x^8\right)}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{16}{1+x^{16}}+\frac{32}{1+x^{32}}\)
\(=\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}}+\frac{32}{1+x^{32}}\)
\(=\left(\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}}\right)+\frac{32}{1+x^{32}}\)
\(=\frac{16\left(1+x^{16}\right)+16\left(1-x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}+\frac{32}{1+x^{32}}\)
\(=\frac{32}{1-x^{32}}+\frac{32}{1+x^{32}}\)
\(=\frac{32\left(1+x^{32}\right)+32\left(1-x^{32}\right)}{\left(1-x^{32}\right)\left(1+x^{32}\right)}\)
\(=\frac{64}{1-x^{64}}\)