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\(2x-x^0=3^5:3^3\\ 2x-1=3^{5-3}\\ 2x-1=3^2\\ 2x-1=9\\ 2x=9+1\\ 2x=10\\ x=\dfrac{10}{2}\\ x=5\)
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\(\left(\dfrac{9}{16}\right)^5\cdot x=\left(\dfrac{27}{64}\right)^3\)
\(\Leftrightarrow\left(\dfrac{3}{4}\right)^{10}\cdot x=\left(\dfrac{3}{4}\right)^9\)
\(\Rightarrow x=\left(\dfrac{3}{4}\right)^9:\left(\dfrac{3}{4}\right)^{10}\)
\(\Rightarrow x=\left(\dfrac{3}{4}\right)^{-1}\)
\(\Rightarrow x=\dfrac{4}{3}\)
Vậy x=4/3
(\(\dfrac{9}{16}\))5\(\times\) \(x\) = (\(\dfrac{27}{64}\))3
\(x\) = (\(\dfrac{27}{64}\))3 : (\(\dfrac{9}{16}\))5
\(x\) = (\(\dfrac{3^3}{2^6}\))3: (\(\dfrac{3^2}{2^4}\))5
\(x\) = \(\dfrac{3^9}{2^{18}}\) : \(\dfrac{3^{10}}{2^{20}}\)
\(x\) = \(\dfrac{3^9}{2^{18}}\) \(\times\) \(\dfrac{2^{20}}{3^{10}}\)
\(x\) = \(\dfrac{2^2}{3}\)
\(x\) = \(\dfrac{4}{3}\)
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Ta có : A = 2x + 5y
Thay giá trị x = 3; y = 4 vào A, ta được:
A = 2.3 + 5.4
A = 6 + 20
A = 26
Vậy A = 26
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P=\(2x^2+5x\)
=\(2\left(x^2+\frac{5}{2}x\right)\)
=\(2\left(x^2+2x.\frac{5}{4}+\frac{25}{16}-\frac{25}{16}\right)\)
= \(2\left(x+\frac{5}{4}\right)^2-\frac{25}{8}\)
de P nhan gia tri duong thi
\(2\left(x+\frac{5}{4}\right)^2>\frac{25}{8}\)
<=> \(\left(x+\frac{5}{4}\right)^2>\frac{25}{16}\)
<=> \(\orbr{\begin{cases}x+\frac{5}{4}>\frac{5}{4}\\x+\frac{5}{4}< \frac{-5}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x>0\\x< \frac{-5}{2}\end{cases}}}\)
vay voi x>0 hoac x< -5/2 thi P dat gia tri duong
Chuc ban hoc tot