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Bài giải
a, \(\frac{x+5}{2017}-\frac{x+5}{2018}+\frac{x+5}{2019}-\frac{x+5}{2020}=0\)
\(\left(x+5\right)\left(\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}\right)=0\)
Do \(\left(\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}\right)\ne0\)
\(\Rightarrow\text{ }x+5=0\)
\(x=0-5\)
\(=-5\)
Theo đề ta có :
\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\frac{\left(x+5\right)-\left(x+2\right)}{\left(x+2\right)\left(x+5\right)}+\frac{\left(x+10\right)-\left(x+5\right)}{\left(x+5\right)\left(x+10\right)}+\frac{\left(x+17\right)-\left(x+10\right)}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\frac{\left(x+17\right)-\left(x+2\right)}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\left(x+17\right)-\left(x+2\right)=x\)
\(\Rightarrow x=15\)
Để 1 phân số được xác định thì mẫu số của chúng phải khác 0
BÀI LÀM
ĐKXĐ: \(\left(x-1\right)\left(-2x+8\right)\ne0\)
\(\Leftrightarrow\)\(-2\left(x-1\right)\left(x-4\right)\ne0\)
\(\Leftrightarrow\) \(\orbr{\begin{cases}x-1\ne0\\x-4\ne0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=1\\x=4\end{cases}}\)
Vậy....
Bài làm:
Ta có: \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.....\frac{30}{62}.\frac{31}{64}=2^x\)
\(\Leftrightarrow\frac{1.2.3.....30.31}{2.2.2.3.2.4.....2.31.2.32}=2^x\)
\(\Leftrightarrow\frac{1}{2^{31}.2^5}=2^x\)
\(\Leftrightarrow\frac{1}{2^{36}}=2^x\)
\(\Rightarrow x=-36\)
Mình làm cho bạn 2 câu khó hơn còn mấy câu còn lại dungf phương pháp quy đồng rồi chuyển vế là tính được mà
c, <=> [(x-1)/2009 ]-1 +[ (x-2)/2008] -1 = [(x-3)/2007]-1 +[(x-4)/2006]-1
<=> (x-2010)/2009 + (x-2010)/2008 = (x-2010)/2007 + (x-2010)/2006
<=> (x-2010)*(1/2009+1/2008-1/2007-1/2006)=0
=> x-2010=0 => x=2010
d, TH1 : cả hai cùng âm
=>> 2X-4 <O => X< 2
Và 9-3x<0 =>> x> 3
=>> loại
Th2 cả hai cùng dương
2x-4>O => x>2
Và 9-3x>O => x<3
=>> 2<x<3 (tm)
a)
( 4x - 9 ) ( 2,5 + (-7/3) . x ) = 0
\(\Rightarrow\orbr{\begin{cases}4x-9=0\\2,5+\frac{-7}{3}x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{9}{4}\\x=\frac{15}{14}\end{cases}}\)
P/s: đợi xíu làm câu b
b) \(\frac{1}{x\left(x+1\right)}\cdot\frac{1}{\left(x+1\right)\left(x+2\right)}\cdot\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2015}\)
\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2015}\)
\(\frac{-1}{x+3}=\frac{1}{2015}\)
\(\Leftrightarrow x+3=-2015\)
\(\Leftrightarrow x=-2018\)
Vậy,.........
a) ta có: \(\frac{x+13}{2006}+\frac{x+2006}{13}+\frac{x+1}{2018}+3=0\)
\(\Rightarrow\frac{x+13}{2006}+1+\frac{x+2006}{13}+1+\frac{x+1}{2018}+1=0\)
\(\Rightarrow\frac{x+2019}{2006}+\frac{x+2019}{13}+\frac{x+2019}{2018}=0\)
\(\Rightarrow\left(x+2019\right)\left(\frac{1}{2006}+\frac{1}{13}+\frac{1}{2018}\right)=0\)
mà \(\frac{1}{2006}+\frac{1}{13}+\frac{1}{2018}>0\)
\(\Rightarrow x+2019=0\)
\(\Rightarrow x=-2019\)
b) \(\frac{4}{\left(x+3\right)\left(x+7\right)}+\frac{3}{\left(x+7\right)\left(x+10\right)}=\frac{x}{\left(x+3\right)\left(x+10\right)}\)
\(\Rightarrow\frac{\left(x+7\right)-\left(x+3\right)}{\left(x+3\right)\left(x+7\right)}+\frac{\left(x+10\right)-\left(x+7\right)}{\left(x+7\right)\left(x+10\right)}=\frac{x}{\left(x+3\right)\left(x+10\right)}\)
\(\Rightarrow\frac{1}{x+3}-\frac{1}{x+7}+\frac{1}{x+7}-\frac{1}{x+10}=\frac{x}{\left(x+3\right)\left(x+10\right)}\)
\(\Rightarrow\frac{1}{x+3}-\frac{1}{x+10}=\frac{x}{\left(x+3\right)\left(x+10\right)}\)
\(\Rightarrow\frac{7}{\left(x+3\right)\left(x+10\right)}=\frac{x}{\left(x+3\right)\left(x+10\right)}\)
\(\Rightarrow x=7\)