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\(\left(2x-3\right)^2=16\)
\(\Rightarrow\left(2x-3\right)^2=4^2\)
\(\Rightarrow2x-3=4\)
\(\Rightarrow2x=4+3\)
\(\Rightarrow2x=7\)
\(\Rightarrow x=\frac{7}{2}\)

(2x- 3)2 = 16
=>\(\hept{\begin{cases}2x-3=4\\2x-3=-4\end{cases}}\)
=>\(\hept{\begin{cases}2x=7\\2x=-1\end{cases}}\)
=>\(\frac{ }{\hept{\begin{cases}x=3,5\\x=-0,5\end{cases}}}\)

Ta có : \(\left|2x-1\right|-\frac{1}{2}=\frac{1}{3}\)
=> \(\left|2x-1\right|=\frac{5}{6}\)
TH1 : \(2x-1\ge0\left(x\ge\frac{1}{2}\right)\)
=> \(\left|2x-1\right|=2x-1=\frac{5}{6}\)
=> \(x=\frac{\frac{5}{6}+1}{2}=\frac{11}{12}\)( TM )
TH2 : \(2x-1< 0\left(x< \frac{1}{2}\right)\)
=> \(\left|2x-1\right|=1-2x=\frac{5}{6}\)
=> \(x=\frac{\frac{5}{6}-1}{-2}=\frac{1}{12}\) ( TM )
Vậy phương trình trên có tập nghiệm là \(S=\left\{\frac{1}{12};\frac{11}{12}\right\}\)
\(|2x-1|-\frac{1}{2}=\frac{1}{3}\)
\(|2x-1|=\frac{1}{3}+\frac{1}{2}\)
\(|2x-1|=\frac{2+3}{6}\)
\(|2x-1|=\frac{5}{6}\)
\(\Rightarrow2x-1=\frac{5}{6}\) hoặc \(2x-1=-\frac{5}{6}\)
\(TH1:2x-1=\frac{5}{6}\)
\(2x=\frac{5}{6}+1\)
\(2x=\frac{5+6}{6}\)
\(2x=\frac{11}{6}\)
\(x=\frac{11}{6}:2\)
\(x=\frac{11}{6}.\frac{1}{2}\)
\(x=\frac{11}{12}\)
\(TH2:2x-1=-\frac{5}{6}\)
\(2x=-\frac{5}{6}+1\)
\(2x=\frac{-5+6}{6}\)
\(2x=\frac{1}{6}\)
\(x=\frac{1}{6}:2\)
\(x=\frac{1}{6}.\frac{1}{2}\)
\(x=\frac{1}{12}\)
Vậy \(x\in\left\{\frac{11}{12};\frac{1}{12}\right\}\)

a) (x+2) + (x+3) + (x+5) = 25
3x + 10 = 25
3x = 15
x = 5
b) 62 - 3.(x+2) = 52.2
62 - 3.(x+2) = 50
3.(x+2) = 12
x+2 = 4
x = 2
c) 25 - (2x+3) = 16
25 - 2x - 3 = 16
22 - 2x = 16
2x =6
x = 3

\(\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{9999}{10000}\)
\(=\frac{1\cdot3}{2^2}\cdot\frac{2\cdot4}{3^2}\cdot\frac{3\cdot5}{4^2}\cdot...\cdot\frac{99\cdot101}{100^2}\)
\(=\frac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(99\cdot101\right)}{2^2\cdot3^2\cdot4^2\cdot...\cdot100^2}\)
\(=\frac{\left(1\cdot2\cdot3\cdot...\cdot99\right)\left(3\cdot4\cdot5\cdot101\right)}{\left(2\cdot3\cdot4\cdot...\cdot100\right)\left(2\cdot3\cdot4\cdot...\cdot100\right)}\)
\(=2\cdot101=202\)
\(= \frac{1 \cdot 3}{2^{2}} \cdot \frac{2 \cdot 4}{3^{2}} \cdot \frac{3 \cdot 5}{4^{2}} \cdot . . . \cdot \frac{99 \cdot 101}{10 0^{2}}\)
\(= \frac{\left(\right. 1 \cdot 3 \left.\right) \left(\right. 2 \cdot 4 \left.\right) \left(\right. 3 \cdot 5 \left.\right) . . . \left(\right. 99 \cdot 101 \left.\right)}{2^{2} \cdot 3^{2} \cdot 4^{2} \cdot . . . \cdot 10 0^{2}}\)
\(= \frac{\left(\right. 1 \cdot 2 \cdot 3 \cdot . . . \cdot 99 \left.\right) \left(\right. 3 \cdot 4 \cdot 5 \cdot 101 \left.\right)}{\left(\right. 2 \cdot 3 \cdot 4 \cdot . . . \cdot 100 \left.\right) \left(\right. 2 \cdot 3 \cdot 4 \cdot . . . \cdot 100 \left.\right)}\)
\(= 2 \cdot 101 = 202\)
( 2x - 3 )2 = 16
( 2x - 3 )2 = 42 hoặc (-4)2
=> 2x - 3 = 4 hoặc 2x - 3 = -4
2x = 4+3 2x = -4 + 3
2 x = 7 2x = -1
x = 7/2 x = -1/2
~ Ko bt đúng ko, ko chắc lắm ~
( 2x -3 )2 = 16
( 2x - 3 )2 = 24
2x - 3 = 2
2x =2 + 3
2x = 5
x 5:2
==> x = ko có gt x
k nha bn