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3|x - 7| - 3 = 32
<=> 3|x - 7| = 12
<=> |x - 7| = 4
<=> \(\orbr{\begin{cases}x-7=4\\x-7=-4\end{cases}}\)
<=> \(\orbr{\begin{cases}x=11\\x=3\end{cases}}\)
Vậy ...
\(3\left|x-7\right|-3=9\)
=> \(3\left|x-7\right|=9+3\)
=> \(3\left|x-7\right|=12\)
=> \(\left|x-7\right|=4\)
=> \(\orbr{\begin{cases}x+7=4\\x+7=-4\end{cases}}\)
=> \(\orbr{\begin{cases}x=-3\\x=-11\end{cases}}\)

1. a) \(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{1}{2}+\frac{1}{3}=\frac{9}{12}+\frac{6}{12}+\frac{4}{12}=\frac{19}{12}\)
b) \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)
\(=\frac{140}{27}-\frac{5}{27}+\frac{7}{23}+\frac{16}{23}+\frac{1}{2}\)
\(=\frac{135}{27}+\frac{23}{23}+\frac{1}{2}\)
\(=5+1+0,5=6,5\)
2) a) 1/2 + 2/3x = 1/4
=> 2/3x = 1/4 - 1/2
=> 2/3x = -1/4
=> x = -1/4 : 2/3
=> x = -3/8
b) 3/5 + 2/5 : x = 3 1/2
=> 3/5 + 2/5 : x = 7/2
=> 2/5 : x = 7/2 - 3/5
=> 2/5 : x = 29/10
=> x = 2/5 : 29/10
=> x = 4/29
c) x+4/2004 + x+3/2005 = x+2/2006 + x+1/2007
=> x+4/2004 + 1 + x+3/2005 + 1 = x+2/2006 + 1 + x+1/2007 + 1
=> x+2008/2004 + x+2008/2005 = x+2008/2006 + x+2008/2007
=> x+2008/2004 + x+2008/2005 - x+2008/2006 - x+2008/2007 = 0
=> (x+2008). (1/2004 + 1/2005 - 1/2006 - 1/2007) = 0
Vì 1/2004 + 1/2005 - 1/2006 - 1/2007 khác 0
Nên x + 2008 = 0 <=> x = -2008
Vậy x = -2008
1,a,\(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{2}{4}+\frac{1}{3}=\frac{5}{4}+\frac{1}{3}=\frac{15}{12}+\frac{4}{12}=\frac{19}{12}\)
b, \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}=\frac{140}{27}-\frac{5}{27}+\frac{7}{23}+\frac{16}{23}+\frac{1}{2}=\frac{135}{27}+\frac{23}{23}+\frac{1}{2}=5+1+\frac{1}{2}=\frac{13}{2}\)2,a,\(\frac{1}{2}+\frac{2}{3}.x=\frac{1}{4}\)
<=>\(\frac{2}{3}.x=-\frac{1}{2}\)
<=>\(x=-\frac{3}{4}\)
b,\(\frac{3}{5}+\frac{2}{5}\div x=3\frac{1}{2}\)
<=>\(\frac{2}{5x}=\frac{29}{10}\)
<=>\(x=\frac{29}{4}\)
c,\(\frac{x+4}{2004}+\frac{x+3}{2005}=\frac{x+2}{2006}+\frac{x+1}{2007}\)
<=> \(\frac{x+4}{2004}+1+\frac{x+3}{2005}+1=\frac{x+2}{2006}+1+\frac{x+1}{2007}+1\)
<=>\(\frac{x+2008}{2004}+\frac{x+2008}{2005}=\frac{x+2008}{2006}+\frac{x+2008}{2007}\)
<=>\(\left(x+2008\right)\left(\frac{1}{2004}+\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}\right)\)=0
<=>x+2008=0 vì cái ngoặc còn lại\(\ne0\)
<=>x=-2008
Vậy x=-2008
Bạn nhớ tk cho mình vì mình đã chăm chỉ làm hết bài bạn hỏi nha!

Theo đề ta có:
\(x-\frac{1}{7}=x-\frac{2}{3}\)
=> \(x=x-\frac{2}{3}+\frac{1}{7}\)
=> \(x=x-\frac{11}{21}\) (vô lý)
Vậy không tìm được x thỏa mãn đề bài. (nhớ k mình nha)

a) 23x+2 = 4x+5 = (22)x+5 = 22x+10
=> 3x + 2 = 2x + 10
=> 3x - 2x = 10 - 2
x = 8
b) 3-1.3x + 9.3x = 28
3x. ( 3-1 + 9) = 28
3x. 28/3 = 28
3x = 3 = 31
=> x = 1
\(2^{3x+2}=4^{x+5}\)
\(\Rightarrow2^{3x+2}=\left(2^2\right)^{x+5}\)
\(\Rightarrow3x+2=2\left(x+5\right)\)
\(\Rightarrow3x+2=2x+10\)
\(\Rightarrow3x-2x=10-2\Rightarrow x=8\)
\(3^{-1}.3^x+9.3^x=28\)
\(\Rightarrow\frac{1}{3}.3^x+9.3^x=28\)
\(\Rightarrow3^x.\left(9+\frac{1}{3}\right)=28\)
\(\Rightarrow3^x.\frac{28}{3}=28\)
\(\Rightarrow3^x=3\Rightarrow x=1\)
Chúc bạn học tốt.
3\(^{x+2}\) + 3\(^{x}\) - 23 = 949 - 2 x 3\(^{x}\)
3\(^2.\)3\(^{x}\) + 3\(^{x}\) + 2.3\(^{x}\) = 949 + 23
3\(^{x}\).(9 + 1 + 2) = 97
3\(^{x}\).(10 +2) = 97
3\(^{x}\).12 = 97
3\(^{x}\) = 97 : 12
3\(^{x}\) = 81
3\(^{x}\) = 3\(^4\)
\(x=4\)
Vậy \(x=4\)