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trước tiên bạn nên đưa về dạng tổng hai bình phương
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a) 2x-5y+4y+2x
=4x+y
Tai x=3 y=-12 thi
4x3+(-12)=12-12=0
b)3x+4y-2x-3y
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Bài 2: Tìm x
a) x2 - 6x + 5 = 0
<=> x2 - x - 5x + 5 = 0
<=> x(x - 1) - 5(x - 1) = 0
<=> (x - 1)(x - 5) = 0
<=> \(\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
Vậy x ={1; 5}
b) x2 - 2x - 24 = 0
<=> x2 + 4x - 6x - 24 = 0
<=> x(x + 4) - 6(x + 4) = 0
<=> (x + 4)(x - 6) = 0
<=> \(\left[{}\begin{matrix}x+4=0\\x-6=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=-4\\x=6\end{matrix}\right.\)
Vậy x ={-4; 6}
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2.
a. \(x^2-6x+5=0\)
\(\Leftrightarrow\left(x^2-x\right)-\left(5x-5\right)=0\)
\(\Leftrightarrow x\left(x-1\right)-5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
b. \(x^2-2x-24=0\)
\(\Leftrightarrow\left(x^2-6x\right)+\left(4x-24\right)=0\)
\(\Leftrightarrow x\left(x-6\right)+4\left(x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\x-6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=6\end{cases}}\)
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a)( 6x - 2)2 ( 5x - 2)2 - 2( 6x - 2 )( 5x - 2 )
=(6x-2)2-2(6x-2)(5x-2)+(5x-2)2
=[(6x-2)-(5x-2)]2
=(6x-2-5x+2)2
=X2
b) ( x2 + 3x + 1)2 - 2( x2 + 3x + 1)( 3x + 1) + ( 9x2 - 6x + 1)
=( x2 + 3x + 1)2 - 2( x2 + 3x + 1)( 3x + 1)+[(3x)2-2.3x.1+12]
=( x2 + 3x + 1)2 - 2( x2 + 3x + 1)( 3x + 1)+(3x+1)2
=[( x2 + 3x + 1)-( 3x + 1)]2
=( x2 + 3x + 1- 3x - 1)2
=(x2)2
=x4
Biểu thức không có giá trị nhỏ nhất. Bạn xem lại đề.