\(1< \frac{11}{x}\)\(< 2\)
K
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Những câu hỏi liên quan
5 tháng 6 2015
Vê trái:
\(=\frac{2}{\left(x-1\right)\left(x+1\right)}+\frac{4}{\left(x-2\right)\left(x+2\right)}+...+\frac{20}{\left(x-10\right)\left(x+10\right)}\)
\(=\frac{\left(x+1\right)-\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{\left(x+2\right)-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+...+\frac{\left(x+10\right)-\left(x-10\right)}{\left(x+10\right)\left(x-10\right)}\)
\(=\frac{1}{x-1}-\frac{1}{x+1}+\frac{1}{x-2}-\frac{1}{x+2}+...+\frac{1}{x-10}-\frac{1}{x+10}\)
\(=\left(\frac{1}{x-1}+\frac{1}{x-2}+...+\frac{1}{x-10}\right)-\left(\frac{1}{x+1}+\frac{1}{x+2}+...+\frac{1}{x+10}\right)\)
Vế phải:
\(=\frac{\left(x+1\right)-\left(x-10\right)}{\left(x-10\right)\left(x+1\right)}+\frac{\left(x+2\right)-\left(x-9\right)}{\left(x-9\right)\left(x+2\right)}+...+\frac{\left(x+10\right)-\left(x-1\right)}{\left(x-1\right)\left(x+10\right)}\)
\(=\frac{1}{x-10}-\frac{1}{x+1}+\frac{1}{x-9}-\frac{1}{x+2}+...+\frac{1}{x-1}-\frac{1}{x+10}\)
\(=\left(\frac{1}{x-1}+\frac{1}{x-2}+...+\frac{1}{x-10}\right)-\left(\frac{1}{x+1}+\frac{1}{x+2}+...+\frac{1}{x+10}\right)\) = vế phải
=> đpcm
CW
28 tháng 1 2017
a) \(\left(x-11\right)+\frac{3x}{x-11}=3+\frac{33}{x-11}\)
\(\Leftrightarrow x+\frac{3x}{x-11}-\frac{33}{x-11}=14\)
\(\Leftrightarrow x^2-11x+3x-33=14x-154\)
\(\Leftrightarrow x^2-22x+121=0\)
\(\Leftrightarrow\left(x-11\right)^2=0\Leftrightarrow x=11\)
Vậy .......
b) \(\frac{7-2x}{x-1}=\frac{1-4x}{x+2}\Leftrightarrow\left(7-2x\right)\left(x+2\right)=\left(1-4x\right)\left(x-1\right)\)
\(\Leftrightarrow7x-2x^2+14-4x=x-4x^2-1+4x\)
\(\Leftrightarrow2x^2=-15\)(vô lí)
Vậy pt vô nghiệm
c) \(\frac{3-2x}{x+1}=2+\frac{1-4x}{x-2}\)
\(\Leftrightarrow\left(3-2x\right)\left(x-2\right)=2\left(x+1\right)\left(x-2\right)+\left(1-4x\right)\left(x+1\right)\)
\(\Leftrightarrow3x-2x^2-6x+4x=2x^2+2x-4x-4+x-4x^2+1-4x\)
\(\Leftrightarrow6x=-3\Leftrightarrow x=-\frac{1}{2}\)
Vậy.........
(gửi trước 3 câu)
CW
28 tháng 1 2017
d) \(\frac{109x-4}{111x+1}-1=0\Leftrightarrow109x-4=111x+1\Leftrightarrow2x=-5\Leftrightarrow x=-\frac{5}{2}\)
Vậy x=-5/2
e) \(\frac{x^2-7}{x}=x-\frac{1}{2}\Leftrightarrow\frac{x^2-7}{x}-\frac{x^2}{x}=-\frac{1}{2}\Leftrightarrow-\frac{7}{x}=\frac{1}{2}\Leftrightarrow x=-14\)
f) \(\frac{x+1}{x+2}=3\Leftrightarrow x+1=3x+6\Leftrightarrow2x=7\Leftrightarrow x=\frac{7}{2}\)
G
3 tháng 3 2018
Câu 2 đây:
\(|x^2+|x-1||=x^2+2\)
\(\Rightarrow\orbr{\begin{cases}x^2+\left|x-1\right|=x^2+2\\x^2+\left|x-1\right|=-x^2-2\left(l\right)\end{cases}}\)
\(\Rightarrow\left|x-1\right|=2\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
KT
3 tháng 3 2018
a) \(M=\left(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+0,5}{1\frac{1}{6}-0,875+0,7}\right):\frac{2012}{2013}\)
\(=\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{2}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\right):\frac{2012}{2013}\)
\(=\left(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{2\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\right):\frac{2012}{2013}\)
\(=\left(\frac{2}{7}-\frac{2}{7}\right):\frac{2012}{2013}\)
\(=0\)
T
9 tháng 3 2018
a. \(\frac{x+1}{2}=\frac{8}{x+1}\)
\(\Leftrightarrow\left(x+1\right).\left(x+1\right)=8.2\)
\(\Leftrightarrow\left(x+1\right)^2=16\)
\(\Leftrightarrow\left(x+1\right)^2=2^4\)
\(\Leftrightarrow\left(x+1\right)=2^2\)
\(\Leftrightarrow\left(x+1\right)=4\)
\(\Leftrightarrow x=4-1=3\)
b. \(x:\left(9\frac{1}{2}-\frac{3}{2}\right)=\frac{0,4+\frac{2}{9}-\frac{2}{11}}{1,6+\frac{8}{9}-\frac{8}{11}}\)
\(\Leftrightarrow x:\left(\frac{10}{2}-\frac{3}{2}\right)=\frac{0,4+0,2-0,18}{1,6+0,8-0,72}\)
\(\Leftrightarrow x:\frac{7}{2}=\frac{\frac{21}{50}}{\frac{42}{25}}\)
\(\Leftrightarrow x=\frac{\frac{21}{50}}{\frac{42}{25}}.\frac{7}{2}\Leftrightarrow x=\frac{1}{4}.\frac{7}{2}=\frac{7}{8}\)
AK
9 tháng 3 2018
a ) \(\frac{x+1}{2}=\frac{8}{x+1}\)
\(\Rightarrow\left(x+1\right).\left(x+1\right)=2.8\)
\(\Rightarrow\left(x+1\right)^2=16\)
\(\Rightarrow\orbr{\begin{cases}x+1=4\\x+1=-4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=4-1\\x=-4-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)
Dấu " \(\orbr{\begin{cases}\\\end{cases}}\)là hoặc nha !!!
T
5 tháng 6 2019
#)Giải :
a) x + 2x + 3x + ... + 100x = - 213
=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213
=> 100x + 5049 = - 213
<=> 100x = - 5262
<=> x = - 52,62
T
5 tháng 6 2019
#)Giải :
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)
\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{2}{3}\)
28 tháng 4 2020
a) ĐKXĐ: \(x\notin\left\{\frac{1}{3};\frac{-11}{3}\right\}\)
Ta có: \(\frac{2}{\left(1-3x\right)\left(3x+11\right)}=\frac{1}{9x^2-6x+1}-\frac{3}{\left(3x+11\right)^2}\)
\(\Leftrightarrow\frac{2\left(1-3x\right)\left(3x+11\right)}{\left(1-3x\right)^2\cdot\left(3x+11\right)^2}=\frac{\left(3x+11\right)^2}{\left(1-3x\right)^2\cdot\left(3x+11\right)^2}-\frac{3\left(1-3x\right)^2}{\left(1-3x\right)^2\cdot\left(3x+11\right)^2}\)
\(\Leftrightarrow-18x^2-60x+22=9x^2+66x+121-3\left(1-6x+9x^2\right)\)
\(\Leftrightarrow-18x^2-60x+22-9x^2-66x-121+3\left(1-6x+9x^2\right)=0\)
\(\Leftrightarrow-27x^2-126x-99+3-18x+27x^2=0\)
\(\Leftrightarrow-144x-96=0\)
\(\Leftrightarrow-144x=96\)
hay \(x=\frac{-2}{3}\)(tm)
Vậy: \(x=\frac{-2}{3}\)
29 tháng 6 2018
a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne=\)
Nên x + 1 = 0 => x = -1
b) \(\frac{x+1}{14}+\frac{x+2}{13}=\frac{x+3}{12}+\frac{x+4}{11}\)
\(\Leftrightarrow\frac{x+1}{14}+1+\frac{x+2}{13}+1=\frac{x+3}{12}+1+\frac{x+4}{11}+1\)
\(\Leftrightarrow\frac{x+15}{14}+\frac{x+15}{13}=\frac{x+15}{12}+\frac{x+15}{11}\)
\(\Leftrightarrow\frac{x+15}{14}+\frac{x+15}{13}-\frac{x+15}{12}-\frac{x+15}{11}=0\)
\(\Leftrightarrow\left(x+15\right)\left(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\right)=0\)
Vì \(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\ne0\)
Nên x +15 = 0 => x = -15
29 tháng 6 2018
a,\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)=\left(x+1\right).\left(\frac{1}{13}+\frac{1}{14}\right)\)
\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)-\left(x+1\right).\left(\frac{1}{13}+\frac{1}{14}\right)=0\)
\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Vì \(\frac{1}{10}>\frac{1}{13};\frac{1}{11}>\frac{1}{14}\Rightarrow\frac{1}{10}+\frac{1}{11}>\frac{1}{13}+\frac{1}{14}\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}>\frac{1}{13}+\frac{1}{14}\)
\(\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}>0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
b, Bạn cộng thêm 1 vào \(\frac{x+1}{14};\frac{x+1}{13};\frac{x+1}{12};\frac{x+1}{11}\)Mội bên phân số 1 đơn vị rồi áp dụng như bài 1
1 < 11/10 < 2
11/10,11/9.11/8,11/7,11/6