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A=100 (nhập biểu thức vào máy tính rồi gán x=98)
Pp gán : solve 98 =
1. x( x - 3 ) + y( y - 3 ) + 2xy - 35
= x2 - 3x + y2 - 3y + 2xy - 35
= ( x2 + 2xy + y2 ) - ( 3x + 3y ) - 35
= ( x + y )2 - 3( x + y ) - 35
= 52 - 3.5 - 35
= 25 - 15 - 35 = -25
2. 4x2 + y2 + 8x - 4xy - 4y + 100
= ( 4x2 - 4xy + y2 + 8x - 4y + 4 ) + 96
= [ ( 4x2 - 4xy + y2 ) + ( 8x - 4y ) + 4 ] + 96
= [ ( 2x - y )2 + 2.( 2x - y ).2 + 22 ] + 96
= ( 2x - y + 2 )2 + 96
= ( 4 + 2 )2 + 96
= 62 + 96 = 36 + 96 = 132
Để N nguyên thì \(3x^2-4x-17⋮x+2\)
\(3x^2+6x-10x-20+3⋮x+2\)
\(3x\left(x+2\right)-10\left(x+2\right)+3⋮x+2\)
\(\left(x+2\right)\left(3x-10\right)+3⋮x+2\)
Dễ thấy \(\left(x+2\right)\left(3x-10\right)⋮x+2\)
\(\Rightarrow3⋮x+2\)
\(\Rightarrow x+2\inƯ\left(3\right)=\left\{1;3;-1;-3\right\}\)
\(\Rightarrow x\in\left\{-1;1;-5;-3\right\}\)
Vậy......
\(x\left(x-2\right)+x-2=0\\ \Leftrightarrow x\left(x-2\right)+\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
\(x^2-2x+1=9\\ \Leftrightarrow\left(x-1\right)^2=9\\ \Leftrightarrow\left[{}\begin{matrix}x-1=-3\\x-1=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\)
\(7x^2=2x\\ \Leftrightarrow7x^2-2x=0\\ \Leftrightarrow x\left(7x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\7x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{7}\end{matrix}\right.\)
\(x^2-6x=8\\ \Leftrightarrow x^2-6x-8=0\\ \left(x^2-6x+9\right)-17=0\\ \Leftrightarrow\left(x-3\right)^2-\sqrt{17^2}=0\\ \Leftrightarrow\left(x-3-\sqrt{17}\right)\left(x-3+\sqrt{17}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3-\sqrt{17}=0\\x-3+\sqrt{17}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3+\sqrt{17}\\x=3-\sqrt{17}\end{matrix}\right.\)
Ta có:\(x^2+4x+10=\left(x^2+2\cdot2\cdot x+2^2\right)+6=\left(x+2\right)^2+6\)
\(\Rightarrow\frac{3}{x^2+4x+10}=\frac{3}{\left(x+2\right)^2+6}\)
Do \(\left(x+2\right)^2\ge0\Rightarrow\left(x+2\right)^2+6\ge6\)
\(\Rightarrow\frac{3}{\left(x+2\right)^2+6}\le\frac{3}{6}=\frac{1}{2}\)
Dấu "=" xảy ra khi và chỉ khi:
\(\left(x+2\right)^2=0\Leftrightarrow x=-2\)
Vậy \(A_{min}=\frac{1}{2}\Leftrightarrow x=-2\)
\(A=x^4-98x^3+x^3-98x^2+x^2-98x+x+2\)
\(=x^3\left(x-98\right)+x^2\left(x-98\right)+x\left(x-98\right)+x+2\)
\(=0+0+0+98+2\left(x=98\right)\)
=100
x = 98 => 97 = x - 1
Ta có :
\(A=x^4-\left(x-1\right)x^3-\left(x-1\right)x^2-\left(x-1\right)x+2\)
\(A=x^4-x^4+x^3-x^3+x^2-x^2+x+2\)
\(A=0+0+0+x+2\)
\(A=x+2\)
Thay x = 98 vào A ta có :
\(A=98+2=100\)