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BD
4
PA
19 tháng 7 2016
a.
\(\frac{x-1}{x-5}=\frac{6}{7}\)
\(\left(x-1\right)\times7=6\times\left(x-5\right)\)
\(7x-7=6x-30\)
\(7x-6x=-30+7\)
\(x=-23\)
b.
\(\frac{x^2}{6}=\frac{24}{25}\)
\(x^2=\frac{24}{25}\times6\)
\(x^2=\frac{144}{25}\)
\(x^2=\left(\pm\frac{12}{5}\right)^2\)
\(x=\pm\frac{12}{5}\)
Vậy \(x=\frac{12}{5}\) hoặc \(x=-\frac{12}{5}\)
PP
22 tháng 7 2016
a,\(\left(x-\frac{7}{9}\right)^3=\left(\left(\frac{2}{3}\right)^2\right)^3\)
\(x-\frac{7}{9}=\frac{4}{9}\)
\(x=\frac{4}{9}+\frac{7}{9}\)
\(x=\frac{11}{9}\)
Vậy x=\(\frac{11}{9}\)
21 tháng 7 2019
\(\frac{2^{4-x}}{16^5}=32^6\)
=> \(\frac{2^{4-x}}{\left(2^4\right)^5}=\left(2^5\right)^6\)
=> \(\frac{2^{4-x}}{2^{20}}=2^{30}\)
=> \(2^{4-x}=2^{30}.2^{20}\)
=> \(2^{4-x}=2^{50}\)
=> 4 - x = 50
=> x = 4 - 50 = -46
\(\frac{3^{2x+3}}{9^3}=9^{14}\)
=> \(\frac{3^{2x+3}}{\left(3^2\right)^3}=\left(3^2\right)^{14}\)
=> \(\frac{3^{2x+3}}{3^6}=3^{28}\)
=> \(3^{2x+3}=3^{28}.3^6\)
=> \(3^{2x+3}=3^{34}\)
=> 2x + 3 = 34
=> 2x = 34 - 3
=> 2x = 31
=> x = 31/2
BT
27 tháng 3 2018
Ta có: \(\frac{3}{\left(x+2\right)\left(x+5\right)}=\frac{1}{x+2}-\frac{1}{x+5}\); \(\frac{5}{\left(x+5\right)\left(x+10\right)}=\frac{1}{x+5}-\frac{1}{x+10}\)
\(\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{1}{x+10}-\frac{1}{x+17}\);
=> Phương trình tương đương:
\(\frac{1}{x+2}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\frac{1}{x+2}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)<=> \(\frac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
<=> \(\frac{15}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
=> x=15
Đáp số: x=15
NL
2
20 tháng 4 2017
a. 2x-1+ 5.2x-1:2=7/32
=> 2x+1.(1+5/2)=7/32
=>2x+1.7/2=7/32
=> 2x+1=1/16=1/24
=> x+1=-4=>x=-5
KH
11 tháng 7 2016
a) \(\Leftrightarrow\frac{x+7}{2003}+1+\frac{x+4}{2006}+1-\frac{x-1}{2011}-1-\frac{x-5}{2015}-1=0\)
\(\Leftrightarrow\frac{x+2010}{2003}+\frac{x+2010}{2006}-\frac{x+2010}{2011}-\frac{x+2010}{2015}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2003}+\frac{1}{2006}-\frac{1}{2011}-\frac{1}{2015}\right)=0\)
\(\Leftrightarrow x+2010=0\) ( vì 1/2003 + 1/2006 -- 1/2011 -- 1/2015 \(\ne\)0)
\(\Leftrightarrow x=-2010\)
câu b làm tương tự (có gì không hiểu hỏi mk nha) >v<
25 tháng 9 2016
\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\frac{\left(x+17\right)-\left(x+2\right)}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\frac{15}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow x=15\)
Vậy x = 15
\(\frac{x-5}{2}=\frac{32}{x-5}\)
\(\Leftrightarrow\left(x-5\right).\left(x-5\right)=2.32\)
\(\Leftrightarrow\left(x-5\right)^2=64\)
\(\Leftrightarrow\left(x-5\right)^2=8^2=\left(-8\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=8\\x-5=-8\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=13\\x=-3\end{cases}}\)
Vậy ...
\(\frac{x-5}{2}=\frac{32}{x-5}\)
\(\Leftrightarrow(x-5)(x-5)=32\cdot2=64\)
\(\Leftrightarrow(x-5)^2=64\)
\(\Leftrightarrow(x-5)^2=8^2\)
\(\Leftrightarrow x-5=\pm8\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=8\\x-5=-8\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=13\\x=-3\end{cases}}\)