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Ta có 5(2x – 5) = x(2x – 5)
ó 5(2x – 5) – x(2x – 5) = 0 ó (2x – 5)(5 – x) = 0
ó 2 x - 5 = 0 5 - x = 0 ó 2 x = 5 5 = x ó x = 5 2 x = 5
Vậy x = 5; x = 5 2
Đáp án cần chọn là: B
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\(1)\)
\(A=-2x^2+x-5\)
\(-2A=\left(4x^2-2x+\frac{1}{4}\right)+\frac{39}{4}\)
\(-2A=\left(2x-\frac{1}{2}\right)^2+\frac{39}{4}\ge\frac{39}{4}\)
\(A=\frac{\left(2x-\frac{1}{2}\right)^2+\frac{39}{4}}{-2}\le\frac{39}{4}:\left(-2\right)=\frac{-39}{8}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\left(2x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow\)\(x=\frac{1}{4}\)
Vậy GTLN của \(A\) là \(\frac{-39}{8}\) khi \(x=\frac{1}{4}\)
Chúc bạn học tốt ~
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bài 2 đề ntn à
\(2\times2^2\times2^3\times...\times2^x=32768\)
\(2^{1+2+3+...+x}=2^{15}\)
\(\Rightarrow1+2+3+...+x=15\)
(x+1)x=30
x=5
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x ( 5 - 2x ) + 2x ( x - 1 ) = 45
=) 5x - 2x2 + 2x2 - 2x = 45
=) 3x = 45
=) x = 15
Vậy x = 15
<=> 5x -2x2 + 2x2 - 2x = 45
<=> ( 5x - 2x ) + ( -2x2 + 2x2 )
<=> 3x = 45 => x = 9