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trong quá trình bạn xem bài mk thấy chỗ nào sai dấu thì sửa giùm mk nha trong quá trình làm mk cx có thể sai sót nhầm lẫn nha
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trong quá trình bạn xem bài mk thấy chỗ nào sai dấu thì sửa giùm mk nha trong quá trình làm mk cx có thể sai sót nhầm lẫn nha
b, \(\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x+3}{97}+1+\frac{x+4}{96}+1\)
\(\frac{x+200}{99}+\frac{x+200}{98}=\frac{x+200}{97}+\frac{x+200}{96}\)
\(\frac{x+200}{99}+\frac{x+200}{98}-\frac{x+200}{97}-\frac{x+200}{96}=0\)
\(\left(x+200\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)
mà\(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\ne0\)
==> x+200=0
<=>x=-200
Vậy nghiệm của phương trình là x=-200
c, \(\frac{109-x}{91}+1+\frac{107-x}{93}+1+\frac{105-x}{95}+1+\frac{103-x}{97}+1=0\)
\(\frac{200-x}{91}+\frac{200-x}{93}+\frac{200-x}{95}+\frac{200-x}{97}=0\)
\(\left(200-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)
mà \(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\ne0\)
==>200-x=0
<=>x=200
vậy nghiệm của pt là x=200
\(\Leftrightarrow\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+3}+....+\frac{1}{x+5}-\frac{1}{x+6}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+6}=\frac{1}{8}\)
\(\Leftrightarrow\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{4}{32}\)
\(\Rightarrow x^2+8x+12=32\)
\(\Leftrightarrow x^2+8x-20=0\)
Đến đây đơn giản rồi nhé
\(\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}=0\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+6}=0\)
\(\Leftrightarrow\frac{1}{x+2}=\frac{1}{x+6}\)
\(\Leftrightarrow x+6=x+2\)
\(\Leftrightarrow x-x=2-6\)
\(\Leftrightarrow0x=-4\)
=> PT vô nghiệm
ĐK: \(x\ne-2;-3;-4;-5;-6\)
\(\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+6}=\frac{1}{8}\)
\(\Leftrightarrow\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{1}{8}\Leftrightarrow\left(x+2\right)\left(x+6\right)=32\)
\(\Leftrightarrow x^2+8x-20=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\)
\(...\Leftrightarrow\frac{1}{\left(x+2\right) \left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+6}=\frac{1}{18}\Leftrightarrow\frac{x+6}{\left(x+2\right)\left(x+6\right)}-\frac{x+2}{\left(x+2\right)\left(x+6\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\frac{1}{18}\Rightarrow\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{1}{18}\)
\(\Rightarrow\left(x+2\right)\left(x+6\right)=72\)
=> \(x^2+8x-60=0\)
Phân tich đa thức thành nhân tử để tìm x
pt <=> 1/(x+2).(x+3) + 1/(x+3).(x+4) + 1/(x+4).(x+5) + 1/(x+5).(x+6) = 1/8
<=> 1/x+2 - 1/x+3 + 1/x+3 - 1/x+4 + 1/x+4 - 1/x+5 + 1/x+5 - 1/x+6 = 1/8
<=> 1/x+2 - 1/x+6 = 1/8
<=> (x+6-x-2)/(x+2).(x+6) = 1/8
<=> 4/(x+2).(x+6) = 1/8
<=>(x+2).(x+6) = 4 : 1/8 = 32
<=>x^2 + 8x + 12 = 32
<=> x^2+8x+12-32=0
<=>x^2+8x-20=0
<=>(x-2).(x+10)=0
<=> x-2 =0 hoặc x+10 = 0
<=> x=2 hoặc x=-10
giang sinh an lanh $%###Xuyen gam cu chuoi###%$