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a/ \(2x+\frac{1}{7}=\frac{1}{3}\)
=> \(2x=\frac{1}{3}-\frac{1}{7}=\frac{7}{21}-\frac{3}{21}\)
=> \(2x=\frac{4}{21}\)
=> \(x=\frac{4}{21}:2=\frac{4}{21}.\frac{1}{2}=\frac{2}{21}\)
b/ \(3\left(x-\frac{1}{2}\right)=\frac{4}{9}\)
=> \(x-\frac{1}{2}=\frac{4}{9}:3=\frac{4}{9}.\frac{1}{3}\)
=> \(x-\frac{1}{2}=\frac{4}{27}\)
=> \(x=\frac{4}{27}+\frac{1}{2}=\frac{8}{54}+\frac{27}{54}=\frac{35}{54}\)
c/ \(\left(x-5\right)^2+4=68\)
=> \(\left(x-5\right)^2=68-4=64\)
=> \(\left[{}\begin{matrix}x-5=8\\x-5=-8\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=8+5=13\\x=-8+5=-3\end{matrix}\right.\)
d/ \(\left(\left|x\right|-\frac{1}{2}\right)\left(2x+\frac{3}{2}\right)=0\)
=> \(\left[{}\begin{matrix}\left|x\right|-\frac{1}{2}=0\\2x+\frac{3}{2}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left|x\right|=0+\frac{1}{2}=\frac{1}{2}\\2x=0-\frac{3}{2}=-\frac{3}{2}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\\x=-\frac{3}{2}:2=-\frac{3}{2}.\frac{1}{2}=-\frac{3}{4}\end{matrix}\right.\)
e) \(5x+2=3x+8\)
=> \(5x-3x=8-2=6\)
=> \(2x=6\)
=> \(x=6:2=3\)
f/ \(26-\left(5-2x\right)=27\)
=> \(5-2x=26-27=-1\)
=> \(2x=5-\left(-1\right)=5+1=6\)
=> \(x=6:2=3\)
g/ \(\left(4x-8\right)-\left(2x-6\right)=4\)
=> \(4x-8-2x+6=4\)
=> \(\left(4x-2x\right)+\left(-8+6\right)=4\)
=> \(2x+-2=4\)
=> \(2x=4+2=6\)
=> \(x=6:2=3\)
h/ \(\left(x+3\right)^3:3-1=-10\)
=> \(\left(x+3\right)^3:3=-10+1=-9\)
=> \(\left(x+3\right)^3=-9.3=-27\)
=> \(x+3=-3\)
=> \(x=-3-3=-6\)
2x ( 3y -2) + ( 3y - 2 ) = -55
=> ( 3y-1) ( 2x+1) =-55
=> 2x+1 = \(\frac{-55}{3y-2}\)(1)
Để x là số nguyên thì 3y-2 \(\in\)Ư(-55) ={ 1; 5; 11; 55; -1; -5; -11; -55}
Ta có: 3y -2 =1 => 3y = 3 => y= 1 thay vào (1) ta được x= 28
3y-2 = 5 => 3y = 7 => y= 7/3 (loại)
3y-2= 11 => 3y = 13 => y= 13/3 ( loại)
3y -2 = 55 => 3y = 57 => y= 19 thay vào ( 1) ta được x= -1
3y-2= -1 => 3y= 1 => y= 1/3 loại
3y-2 = -5 => 3y = -3 => y= -1 thay vào ( 1) ta được x=5
3y-2 = -11 => 3y = -9 => y= -3 thay vào ( 1) ta được x= 2
3y-2= -55 => 3y = -53 => y= -53/3 loại
Vậy.....
- B=(1/2).(2/3).(3/4)....(2010/2011).(2011/2012)
B=(1.2.3....2011)/(2.3.4....2012)
B=1/2012
\(a)\)
\(2^{2x-1}+6.2^8=14.2^8\)
\(\Leftrightarrow2^{2x-1}=14.2^8-6.2^8\)
\(\Leftrightarrow2^{2x-1}=8.2^8\)
\(\Leftrightarrow2^{2x-1}=2^3.2^8\)
\(\Leftrightarrow2x-1=11\)
\(\Leftrightarrow2x=11+1\)
\(\Leftrightarrow x=\frac{12}{2}\)
\(\Leftrightarrow x=6\)
\(b)\)
\(A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)\left(1-\frac{1}{25}\right)...\left(1-\frac{1}{196}\right)\left(1-\frac{1}{225}\right)\)
\(\Leftrightarrow A=\frac{3}{4}.\frac{8}{9}...\frac{224}{225}\)
\(\Leftrightarrow A=\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{14.16}{15.15}\)
\(\Leftrightarrow A=\frac{1.2.3.4...14.16}{2.2.3.3...25.25}\)
\(\Leftrightarrow A=\frac{1.2.3...14}{2.3.4...15}.\frac{3.4.5...16}{2.3.4...15}\)
\(\Leftrightarrow A=\frac{1}{15}.\frac{16}{2}\)
\(\Leftrightarrow A=\frac{8}{15}\)