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\(\frac{x^{30}+x^{28}+x^{26}+x^{24}+...+x^4+x^2+1}{x^{28}+x^{24}+x^{20}+...+x^8+x^4+1}=\frac{\left(x^{30}+x^{26}+x^{22}+...+x^2\right)+\left(x^{28}+x^{24}+...+x^4+1\right)}{x^{28}+x^{24}+x^{20}+...+x^4+1}\)
\(=\frac{x^2\left(x^{28}+x^{24}+...+x^4+1\right)+\left(x^{28}+x^{24}+...+x^4+1\right)}{x^{28}+x^{24}+...+x^4+1}\)
\(=\frac{\left(x^2+1\right)\left(x^{28}+x^{24}+...+x^4+1\right)}{x^{28}+x^{24}+...+x^4+1}\)
\(=x^2+1\)
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Xét \(x\ne1\)
Đặt \(y=x^4\).\(M=x^{28}+x^{24}+...+x^4+1\)
\(M=y^7+y^6+...+y^2+y+1\)\(\Rightarrow Ay=y^8+y^7+...+y^2+y\)
\(\Rightarrow M\left(y-1\right)=y^8-1\Rightarrow M=\frac{y^8-1}{y-1}=\frac{x^{32}-1}{x^4-1}\)
Tương tự \(N=x^{30}+x^{28}+...+x^2+1=\frac{\left(x^2\right)^{16}-1}{x-1}=\frac{x^{32}-1}{x-1}\)
\(A=\frac{M}{N}=\frac{\frac{x^{32}-1}{x^4-1}}{\frac{x^{32}-1}{x^2-1}}=\frac{x^2-1}{x^4-1}=\frac{1}{x^2+1}\)
Thay số vô tính ra A.
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ta có x2+5x+4
=x2+x+4x+4
=(x2+x)+(4x+4)
=x(x+1)+4(x+1)
=(x+1)(x+4)
tương tự ta đc
x2+11x+28=(x+4)(x+7)
x2+17x+70=(x+7)(x+10)
x2+23x+130=(x+10)(x+13)
=>\(\dfrac{1}{\left(x+1\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+7\right)}+\dfrac{1}{\left(x+7\right)\left(x+10\right)}+\dfrac{1}{\left(x+10\right)\left(x+13\right)}=\dfrac{4}{13}\)\(\dfrac{3}{\left(x+1\right)\left(x+4\right)}+\dfrac{3}{\left(x+4\right)\left(x+7\right)}+\dfrac{3}{\left(x+7\right)\left(x+10\right)}+\dfrac{3}{\left(x+10\right)\left(x+11\right)}=\dfrac{4}{13}\)=>\(\dfrac{1}{x+1}-\dfrac{1}{x+4}+\dfrac{1}{x+4}+....+\dfrac{1}{x+13}=\dfrac{4}{13}\)
=>\(\dfrac{1}{x+1}-\dfrac{1}{x+13}=\dfrac{4}{13}\)
=>\(\dfrac{13\left(x+13\right)}{13\left(x+1\right)\left(x+13\right)}-\dfrac{13\left(x+1\right)}{13\left(x+1\right)\left(x+13\right)}=\dfrac{4\left(x+1\right)\left(x+13\right)}{13\left(x+1\right)\left(x+13\right)}\)
=> 13(x+13)-13(x+1)=4(x+1)(x+13)
=> 13[(x+13)-(x+1)]=(4x+4)(x+13)
=>13(x+13-x-1)=4x2+52x+4x+52
=13.12=4x2+56x+52
=>4x2+56x+52=156
=>4x2+56x-104=0
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a)
\(\dfrac{1}{x+1}+\dfrac{2}{x^3-x^2-x+1}+\dfrac{3}{x^2-1}=0\) (\(x\ne\pm1\))
\(\Rightarrow\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)^2}+\dfrac{2}{\left(x+1\right)\left(x-1\right)^2}+\dfrac{3\left(x-1\right)}{\left(x+1\right)\left(x-1\right)^2}=0\)
\(\Rightarrow\dfrac{x^2-2x+1+2+3x-3}{\left(x+1\right)\left(x-1\right)^2}=0\)
\(\Rightarrow\dfrac{x^2+x-2}{\left(x+1\right)\left(x-1\right)^2}=0\)
\(\Rightarrow x^2-x+2=0\)
\(\Rightarrow\left(x-1\right)\left(x+2\right)=0\)
=> Th1 :
x- 1 =0
=> x = 1 ( hư cấu vì không thỏa mãn ĐK )
Th2 :
x+2 = 0
=> x = -2 ( hợp lí )
Vậy nghiệm của phương trình là x = -2
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\(\dfrac{x+1}{29}+\dfrac{x+3}{28}=\dfrac{x+5}{27}+\dfrac{x+7}{26}\)
<=>\(\dfrac{x+1}{29}+2+\dfrac{x+3}{28}+2=\dfrac{x+5}{27}+2+\dfrac{x+7}{26}+2\)
<=>\(\dfrac{x+59}{29}+\dfrac{x+59}{28}=\dfrac{x+59}{27}+\dfrac{x+59}{26}\)
<=>\(\left(x+59\right)\left(\dfrac{1}{29}+\dfrac{1}{28}-\dfrac{1}{27}-\dfrac{1}{26}\right)=0\)
vì 1/29+1/28-1/27-1/26 khác 0 =>x+59=0<=>x=-59
vậy....
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a.
\(\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-11x+28}+\dfrac{1}{x^2-19x+84}=\dfrac{1}{4}\\ \Rightarrow\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-7\right)}+\dfrac{1}{\left(x-7\right)\left(x-12\right)}=\dfrac{1}{4}\\ \Rightarrow\dfrac{1}{x-3}-\dfrac{1}{x-12}=\dfrac{1}{4}\\ \Rightarrow\dfrac{-9}{\left(x-3\right)\left(x-12\right)}=\dfrac{1}{4}\\ \Rightarrow x^2-15x+36=-36\\ \)
Tự giải tiếp