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Ta có : \(\dfrac{5x-150}{50}+\dfrac{5x-102}{49}+\dfrac{5x-56}{48}+\dfrac{5x-12}{47}+\dfrac{5x-660}{46}=0\)
\(\Leftrightarrow\dfrac{5x-150}{50}-1+\dfrac{5x-102}{49}-2+\dfrac{5x-56}{48}-3+\dfrac{5x-12}{47}-4+\dfrac{5x-660}{46}+10=0\)
\(\Leftrightarrow\dfrac{5x-200}{50}+\dfrac{5x-200}{49}+\dfrac{5x-200}{48}+\dfrac{5x-200}{47}+\dfrac{5x-200}{46}=0\)
\(\Leftrightarrow\left(5x-200\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}\right)=0\)
\(\Leftrightarrow5x-200=0\)
\(\Leftrightarrow x=40\)
Vậy ...
Ta có: \(\dfrac{5x-150}{50}+\dfrac{5x-102}{49}+\dfrac{5x-56}{48}+\dfrac{5x-12}{47}+\dfrac{5x-660}{46}=0\)
\(\Leftrightarrow\dfrac{5x-150}{50}-1+\dfrac{5x-102}{49}-2+\dfrac{5x-56}{48}-3+\dfrac{5x-12}{47}-4+\dfrac{5x-660}{46}+10=0\)
\(\Leftrightarrow\dfrac{5x-200}{50}+\dfrac{5x-200}{49}+\dfrac{5x-200}{48}+\dfrac{5x-200}{47}+\dfrac{5x-200}{46}=0\)
\(\Leftrightarrow\left(5x-200\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}\right)=0\)
mà \(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}>0\)
nên 5x-200=0
\(\Leftrightarrow5x=200\)
hay x=40
Vậy: S={40}
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+...+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)
=>\(\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+...+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
=>1/x+2-1/x+6=1/8
=>\(\dfrac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\)
=>x^2+8x+12=32
=>x^2+8x-20=0
=>(x+10)(x-2)=0
=>x=-10 hoặc x=2
b) \(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x^2+4x+5x+20}+\dfrac{1}{x^2+5x+6x+30}+\dfrac{1}{x^2+6x+7x+42}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x\left(x+4\right)+5\left(x+4\right)}+\dfrac{1}{x\left(x+5\right)+6\left(x+5\right)}+\dfrac{1}{x\left(x+6\right)+7\left(x+6\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{x+7}{\left(x+4\right)\left(x+7\right)}-\dfrac{x+4}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\left(x+4\right)\left(x+7\right)=54\)
\(\Leftrightarrow x^2+11x+28-54=0\)
\(\Leftrightarrow x^2-2x+13x-26=0\)
\(\Leftrightarrow x\left(x-2\right)+13\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Leftrightarrow\) x - 2 = 0 hoặc x + 13 = 0
\(\Leftrightarrow\) x = 2 hoặc x = -13
Vậy x = 2 hoặc x = -13.
\(\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-9x+20}+\dfrac{1}{x^2-11x+30}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}+\dfrac{1}{x-5}-\dfrac{1}{x-6}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-6}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{x-6-x+2}{\left(x-2\right)\left(x-6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{4}{\left(x-2\right)\left(x-6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow32=\left(x-2\right)\left(x-6\right)\)
\(\Leftrightarrow32=x^2-8x+12\)
\(\Leftrightarrow x^2+8x-20=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=10\end{matrix}\right.\)
bạn xem lại nhé
cái này là pt có chứa ẩn ở mẫu nên phải có điều kiện, đối chiếu điều kiện và từ dòng có pt chứa ẩn ở mẫu sang dòng có pt đưa dc về dạng ax+b=0 thì dùng dấu ''=>'' nhé
b) \(\dfrac{5x-150}{50}+\dfrac{5x-102}{49}+\dfrac{5x-56}{48}+\dfrac{5x-12}{47}+\dfrac{5x-660}{46}=0\)
\(\Leftrightarrow\dfrac{5x-150}{50}-1+\dfrac{5x-102}{49}-2+\dfrac{5x-56}{48}-3+\dfrac{5x-12}{47}-4+\dfrac{5x-660}{46}+10=0\)
\(\Leftrightarrow\dfrac{5x-200}{50}+\dfrac{5x-200}{49}+\dfrac{5x-200}{48}+\dfrac{5x-200}{47}+\dfrac{5x-200}{46}=0\)
\(\Leftrightarrow\left(5x-200\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}\right)=0\)
\(\Leftrightarrow5x-200=0\)
\(\Leftrightarrow x=40\)
b)
\(\dfrac{5x-150}{50}+\dfrac{5x-102}{49}+\dfrac{5x-56}{48}+\dfrac{5x-12}{47}+\dfrac{5x-660}{46}=0\)
\(\Rightarrow\left(\dfrac{5x-150}{50}-1\right)+\left(\dfrac{5x-102}{49}-2\right)+\left(\dfrac{5x-56}{48}-3\right)+\left(\dfrac{5x-12}{47}-4\right)\)
\(+\left(\dfrac{5x-660}{46}+10\right)=0\)
\(\Rightarrow\dfrac{5x-200}{50}+\dfrac{5x-200}{49}+\dfrac{5x-200}{48}+\dfrac{5x-200}{47}+\dfrac{5x-200}{46}=0\)
\(\Rightarrow\left(5x-200\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}\right)=0\)
\(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}\ne0\)
\(\Rightarrow5x-200=0\Rightarrow x=40\)
a: \(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
=>\(\dfrac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)
=>(x+4)(x+7)=54
=>x^2+11x+28-54=0
=>(x+13)(x-2)=0
=>x=-13 hoặc x=2
b: \(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-...+\dfrac{1}{x+4}-\dfrac{1}{x+5}=\dfrac{1}{3}\)
=>\(\dfrac{x+5-x-1}{\left(x+5\right)\left(x+1\right)}=\dfrac{1}{3}\)
=>x^2+6x+5=12
=>x^2+6x-7=0
=>(x+7)(x-1)=0
=>x=-7 hoặc x=1
1) \(\left(\dfrac{1}{2}x+3\right)\left(x^2-4x-6\right)\)
\(=\dfrac{1}{2}x^3-2x^2-3x+3x^2-12x-18\)
\(=\dfrac{1}{2}x^3+x^2-15x-18\)
2) \(\left(6x^2-9x+15\right)\left(\dfrac{2}{3}x+1\right)\)
\(=4x^3+6x^2-6x^2-9x+10x+15\)
\(=4x^3+x+15\)
3) Ta có: \(\left(3x^2-x+5\right)\left(x^3+5x-1\right)\)
\(=3x^5+15x^2-3x^2-x^4-5x^2+x+5x^3+25x-5\)
\(=3x^5-x^4+5x^3+10x^2+26x-5\)
4) Ta có: \(\left(x-1\right)\left(x+1\right)\left(x-2\right)\)
\(=\left(x^2-1\right)\left(x-2\right)\)
\(=x^3-2x^2-x+2\)
\(\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-9x+20}+\dfrac{1}{x^2-11x+30}=\dfrac{1}{8}\)
ĐKXĐ: x khác 2;3;4;5;6
\(\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2-11x+30}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}+\dfrac{1}{x-5}-\dfrac{1}{x-6}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x-6}-\dfrac{1}{x-2}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{x+6-x+2}{\left(x-2\right)\left(x-6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{4}{\left(x-2\right)\left(x-6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow32=x^2-8x+12\)
\(\Leftrightarrow x^2+8x-20=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=10\end{matrix}\right.\)