`#3107.101107`
`2.`
`a)`
\(\dfrac{2}{3}-\dfrac{5}{2}x=-\dfrac{13}{3}\\ \dfrac{5}{2}x=\dfrac{2}{3}-\left(-\dfrac{13}{3}\right)\\ \dfrac{5}{2}x=\dfrac{2}{3}+\dfrac{13}{3}\\ \dfrac{5}{2}x=\dfrac{15}{3}\\ x=5\div\dfrac{5}{2}\\ x=2\)
`b)`
\(2\cdot\left|3-2x\right|+\dfrac{1}{2}=\dfrac{5}{2}\\ 2\left|3-2x\right|=\dfrac{5}{2}-\dfrac{1}{2}\\ 2\left|3-2x\right|=\dfrac{4}{2}\\ 2\left|3-2x\right|=2\\ \left|3-2x\right|=1\)
TH1: `3 - 2x = 1`
`2x = 3 - 1`
`2x = 2`
`x = 1`
TH2: `3 - 2x = -1`
`2x = 3 - (-1)`
`2x = 4`
`x = 2`
Vậy, `x \in {1; 2}`
`c)`
\(x^2\cdot\left(2^x-6\right)-2x^2=0\\ x^2\left(2^x-6\right)=2x^2\\ 2^x-6=2\\ 2^x=8\\ 2^x=2^3\\ x=3\)
Vậy, `x = 3.`
a.
\(\dfrac{2}{3}-\dfrac{5}{2}x=\dfrac{-13}{3}\)
\(\dfrac{5}{2}x=\dfrac{2}{3}-\dfrac{-13}{3}\)
\(\dfrac{5}{2}x=5\)
\(x=5:\dfrac{5}{2}\)
\(x=2\)
b.
\(2.\left|3-2x\right|+\dfrac{1}{2}=\dfrac{5}{2}\)
\(2.\left|3-2x\right|=\dfrac{5}{2}-\dfrac{1}{2}\)
\(2.\left|3-2x\right|=2\)
\(\left|3-2x\right|=1\)
\(3-2x=1\) hoặc \(3-2x=-1\)
\(2x=2\) hoặc \(2x=4\)
\(x=1\) hoặc \(x=2\)
c.
\(x^2.\left(2^x-6\right)-2x^2=0\)
\(x^2.\left(2^x-8\right)=0\)
\(x^2=0\) hoặc \(2^x-8=0\)
\(x=0\) hoặc \(2^x=8=2^3\)
\(x=0\) hoặc \(x=3\)
