|x+\(\dfrac{1}{2}\) |+|x+\(\dfrac{1}{6}\)| +|x+\(\dfrac{1}{12}\)|+|x+\(\dfrac{1}{20}\)|..........+|x+\(\dfrac{1}{110}\)|=11x
K
Khách
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23 tháng 10 2021
\(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{6}\right|+...+\left|x+\dfrac{1}{110}\right|=11x\left(đk:x\ge0\right)\)
\(\Leftrightarrow x+\dfrac{1}{2}+x+\dfrac{1}{6}+x+\dfrac{1}{12}+...+x+\dfrac{1}{110}=11x\)
\(\Leftrightarrow10x+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{10.11}\right)=11x\)
\(\Leftrightarrow x=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{11}\)
\(\Leftrightarrow x=1-\dfrac{1}{11}=\dfrac{10}{11}\left(tm\right)\)
27 tháng 1 2018
Với \(\forall x\) ta có :
+) \(\left|x+\dfrac{1}{2}\right|\ge0\)
+) \(\left|x+\dfrac{1}{6}\right|\ge0\)
..........................
+) \(\left|x+\dfrac{1}{110}\right|\ge0\)
\(\Leftrightarrow\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{6}\right|+.........+\left|x+\dfrac{1}{110}\right|\ge0\)
Mà \(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{6}\right|+........+\left|x+\dfrac{1}{110}\right|=11x\)
\(\Leftrightarrow11x\ge0\)
\(\Leftrightarrow x\ge0\)
Với \(x\ge0\) thì :
+) \(\left|x+\dfrac{1}{2}\right|=x+\dfrac{1}{2}\)
+) \(\left|x+\dfrac{1}{6}\right|=x+\dfrac{1}{6}\)
.....................................
+) \(\left|x+\dfrac{1}{110}\right|=x+\dfrac{1}{110}\)
\(\Leftrightarrow x+\dfrac{1}{2}+x+\dfrac{1}{6}+......+x+\dfrac{1}{110}=11x\)
\(\Leftrightarrow11x+\left(\dfrac{1}{2}+\dfrac{1}{6}+........+\dfrac{1}{110}\right)=11x\)
\(\Leftrightarrow0x=\dfrac{1}{2}+\dfrac{1}{6}+....+\dfrac{1}{110}\) (vô lí)
\(\Leftrightarrow x\in\varnothing\)
H9
HT.Phong (9A5)
CTVHS
30 tháng 3 2023
\(\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-9x+20}+\dfrac{1}{x^2-11x+30}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}+\dfrac{1}{x-5}-\dfrac{1}{x-6}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-6}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{x-6-x+2}{\left(x-2\right)\left(x-6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{4}{\left(x-2\right)\left(x-6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow32=\left(x-2\right)\left(x-6\right)\)
\(\Leftrightarrow32=x^2-8x+12\)
\(\Leftrightarrow x^2+8x-20=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=10\end{matrix}\right.\)
NH
30 tháng 3 2023
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
LP
\(\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-9x+20}+\dfrac{1}{x^2-11x+30}=\dfrac{1}{8}\)
1
ND
30 tháng 3 2018
ĐKXĐ: x khác 2;3;4;5;6
\(\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2-11x+30}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}+\dfrac{1}{x-5}-\dfrac{1}{x-6}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x-6}-\dfrac{1}{x-2}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{x+6-x+2}{\left(x-2\right)\left(x-6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{4}{\left(x-2\right)\left(x-6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow32=x^2-8x+12\)
\(\Leftrightarrow x^2+8x-20=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=10\end{matrix}\right.\)
16 tháng 6 2022
24:
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
\(\Leftrightarrow\left(x+2\right)\left(x+6\right)=8\left(x+6\right)-8\left(x+2\right)\)
\(\Leftrightarrow x^2+8x+12=8x+48-8x-16=32\)
=>(x+10)(x-2)=0
=>x=-10 hoặc x=2
25: \(\Leftrightarrow\dfrac{\left(x+1\right)^2+1}{x+1}+\dfrac{\left(x+4\right)^2+4}{x+4}=\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+3\right)^2+3}{x+3}\)
\(\Leftrightarrow x+1+\dfrac{1}{x+1}+x+4+\dfrac{4}{x+4}=x+2+\dfrac{2}{x+2}+x+3+\dfrac{3}{x+3}\)
\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{4}{x+4}=\dfrac{2}{x+2}+\dfrac{3}{x+3}\)
\(\Leftrightarrow x+5=0\)
hay x=-5
TV
3
28 tháng 2 2022
\(\Leftrightarrow\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+3}-\dfrac{1}{x+6}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{x+6-x-3}{\left(x+3\right)\left(x+6\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow x^2+9x+18=54\)
\(\Leftrightarrow x^2+9x-36=0\)
=>(x+12)(x-3)=0
=>x=-12 hoặc x=3
28 tháng 2 2022
\(ĐKXĐ:x\ne-3,-4,-5,-6\)
\(\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{18}\\ \Leftrightarrow\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{18}\\ \Leftrightarrow\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{18}\\ \Leftrightarrow\dfrac{1}{x+3}-\dfrac{1}{x+6}=\dfrac{1}{18}\\ \Leftrightarrow\dfrac{x+6-x-3}{\left(x+3\right)\left(x+6\right)}=\dfrac{1}{18}\\ \Leftrightarrow\dfrac{3}{x^2+9x+18}=\dfrac{1}{18}\\ \Leftrightarrow x^2+9x+18=54\)
\(\Leftrightarrow x^2+9x-36=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-12\left(tm\right)\end{matrix}\right.\)
ND
25 tháng 2 2018
a.
\(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)
\(\Leftrightarrow x\left(x+1\right).\left(x-1\right)\left(x+2\right)-24=0\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)-24=0\)
Đặt \(a=x^2+x-1\) , ta có pt:
\(\left(a+1\right)\left(a-1\right)-24=0\)
\(\Leftrightarrow a^2-1-24=0\)
\(\Leftrightarrow a^2-25=0\)
\(\Leftrightarrow\left(a-5\right)\left(a+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=5\\a=-5\end{matrix}\right.\)
*Với a = 5 ta được:
\(x^2+x-1=5\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow x^2+3x-2x-6=0\)
\(\Leftrightarrow\left(x^2+3x\right)-\left(2x+6\right)=0\)
\(\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
*Với a = -5 ta được:
\(x^2+x-1=-5\)
\(\Leftrightarrow x^2+x+4=0\)
\(\Leftrightarrow x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{15}{4}=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\) ( loại)
Vậy pt có tập nghiệm là: \(s=\left\{-3;2\right\}\)
25 tháng 2 2018
c)(ĐKXĐ: x khác 30;29)
\(\Leftrightarrow\dfrac{x-29}{30}-1+\dfrac{x-30}{29}-1=\dfrac{29}{x-30}-1+\dfrac{30}{x-29}-1\)
\(\Leftrightarrow\dfrac{x-59}{30}+\dfrac{x-59}{29}=\dfrac{x-59}{30-x}+\dfrac{x-59}{29-x}\)
\(\Leftrightarrow x=59\)(tm) or \(\dfrac{1}{30}+\dfrac{1}{29}-\dfrac{1}{30-x}-\dfrac{1}{29-x}=0\)
\(\Leftrightarrow\dfrac{-x}{30\left(30-x\right)}+\dfrac{-x}{29\left(29-x\right)}=0\)
\(\Leftrightarrow x=0\)(tm) or \(\dfrac{1}{30\left(30-x\right)}+\dfrac{1}{29\left(29-x\right)}=0\)
\(\Leftrightarrow1741-59x=0\)
\(\Leftrightarrow x=\dfrac{1741}{59}\left(tm\right)\)
Vậy S={0;\(\dfrac{1741}{59}\);59}
Bạn tham khảo nhé!
Câu hỏi của Hoàng Văn Cam - Toán lớp 6 - Học toán với OnlineMath