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Bài 5:
a) \(A=x^2-4x+9=\left(x^2-4x+4\right)+5=\left(x-2\right)^2+5\ge5\)
\(minA=5\Leftrightarrow x=2\)
b) \(B=x^2-x+1=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(minB=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}\)
c) \(C=2x^2-6x=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)
\(minC=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\)
Bài 4:
a) \(M=4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)
\(maxM=7\Leftrightarrow x=2\)
b) \(N=x-x^2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
\(maxN=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)
c) \(P=2x-2x^2-5=-2\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{9}{2}=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le-\dfrac{9}{2}\)
\(maxP=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{1}{2}\)
1.
a.\(\Leftrightarrow7x-5x=3+12\)
\(\Leftrightarrow2x=15\Leftrightarrow x=\dfrac{15}{2}\)
b.\(\Leftrightarrow6x-10-7x-7=2\)
\(\Leftrightarrow x=-19\)
c.\(\Leftrightarrow1-3x=4x-3\)
\(\Leftrightarrow7x=2\Leftrightarrow x=\dfrac{2}{7}\)
d.\(\Leftrightarrow8x^2-4x+12x-6-8x^2-8x-2=12\)
\(\Leftrightarrow-2=12\left(voli\right)\)
2.a) \(8x^2-4x=0\Rightarrow4x\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}4x=0\\2x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)
b) \(5x\left(x-3\right)+7\left(x-3\right)=0\Rightarrow\left(x-3\right)\left(5x+7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\5x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-1.4\end{matrix}\right.\)
c) \(2x^2=x\Rightarrow2x^2-x=0\Rightarrow x\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\2x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=0.5\end{matrix}\right.\)
d) \(x^3=x^5\Rightarrow x^3-x^5=0\Rightarrow x^3\left(1-x^2\right)=0\\ \Rightarrow x^3\left(1-x\right)\left(1+x\right)=0\Rightarrow\left[{}\begin{matrix}x^3=0\\1-x=0\\1+x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
e) \(x^2\left(x+1\right)+2x\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x^2+2x\right)=0\Rightarrow\left(x+1\right)x\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)
g. \(x\left(2x-3\right)-2\left(3-2x\right)=0\)
\(\Rightarrow x\left(2x-3\right)+2\left(2x-3\right)=0\\ \Rightarrow\left(2x-3\right)\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-3=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1.5\\x=-2\end{matrix}\right.\)
\(a,-x^2+2x+5=-\left(x^2-2x-5\right)=-\left(x^2-2x+1-6\right)=-\left(x-1\right)^2+6\le6\)
dấu'=' xảy ra<=>x=1=>Max A=6
\(b,B=-x^2-y^2+4x+4y+2=-x^2+4x-4-y^2+4x-4+10\)
\(=-\left(x^2-4x+4\right)-\left(y^2-4x+4\right)+10\)
\(=-\left(x-2\right)^2-\left(y-2\right)^2+10=-\left[\left(x-2\right)^2+\left(y-2\right)^2\right]+10\le10\)
dấu"=" xảy ra<=>x=y=2=>Max B=10
\(c,C=x^2+y^2-2x+6y+12=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\)
dấu'=' xảy ra<=>x=1,y=-3=>MinC=2
a) \(\left(2a-b\right)\left(b+4a\right)+2a\left(b-3a\right)\)
\(=2ab+8a^2-b^2-4ab+2ab-6a^2\)
\(=\left(2ab+2ab-4ab\right)+\left(8a^2-6a^2\right)-b^2\)
\(=2a^2-b^2\)
b) \(\left(3a-2b\right).\left(2a-3b\right)-6a\left(a-b\right)\)
\(=6a^2-9ab-4ab+6b^2-6a^2+6ab\)
\(=\left(6a^2-6a^2\right)-\left(9ab+4ab-6ab\right)+6b^2\)
\(=-7ab+b^2\)
c) \(5b\left(2x-b\right)-\left(8b-x\right)\left(2x-b\right)\)
\(=10bx-5b^2-\left(16bx-8b^2-2x^2+bx\right)\)
\(=10bx-5b^2-16bx+8b^2+2x^2-bx\)
\(=\left(10bx-16bx-bx\right)-\left(5b^2-8b^2\right)+2x^2\)
\(=-7bx+3b^2+2x^2\)
d) \(2x\left(a+15x\right)+\left(x-6a\right)\left(5a+2x\right)\)
\(=2ax+30x^2+5ax+2x^2-30a^2-12ax\)
\(=\left(2ax+5ax-12ax\right)+\left(30x^2+2x^2\right)-30a^2\)
\(=-5ax+32x^2-30a^2\)
a: =2ab+8a^2-b^2-4ab+2ab-6a^2
=2a^2-b^2
b: =6a^2-9ab-4ab+6b^2-6a^2+6ab
=-7ab+6b^2
c: =10bx-5b^2-16bx+8b^2+2x^2-xb
=3b^2+2x^2-7xb
d: =2xa+30x^2+5ax+2x^2-30a^2-12ax
=32x^2-30a^2-5ax
a: Ta có: \(A=-x^2+4x+3\)
\(=-\left(x^2-4x+4-7\right)\)
\(=-\left(x-2\right)^2+7\le7\forall x\)
Dấu '=' xảy ra khi x=2
b: Ta có: \(B=-x^2+x\)
\(=-\left(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}\right)\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
Thật ra cách làm dạng bài này cũng gần giống như bài tìm gtnn bạn vừa hỏi, chỉ khác ở chỗ đặt dấu âm ra ngoài để tìm được gtln thôi.
a: \(=\dfrac{\left(x+1\right)\left[\left(3x-2\right)-\left(2x+5\right)\left(x-1\right)\right]}{x+1}\)
=3x-2-2x^2+2x-5x+5
=-2x^2+3
b: \(=\left(2x+1-3+x\right)^2=\left(3x-2\right)^2=9x^2-12x+4\)
c: =x^3-3x^2+3x-1-x^3-1+9x^2-1
=6x^2+3x-3
\(a,\left[\left(3x-2\right)\left(x+1\right)-\left(2x+5\right)\left(x^2-1\right)\right]:\left(x+1\right)\)
\(=\left[\left(3x-2\right)\left(x+1\right)-\left(2x+5\right)\left(x-1\right)\left(x+1\right)\right]:\left(x+1\right)\)
\(=\left[\left(x+1\right)\left(3x-2-\left(2x+5\left(x-1\right)\right)\right)\right]:\left(x+1\right)\)
\(=\left[\left(x+1\right)\left(3x-2-2x^2+2x-5x+5\right)\right]:\left(x+1\right)\)
\(=\left[\left(x+1\right)\left(-2x^2+3\right)\right].\dfrac{1}{x+1}\)
\(=-2x^2+3\)
\(b,\left(2x+1\right)^2-2\left(2x+1\right)\left(3-x\right)\)
\(=\left(2x+1\right)\left[\left(2x+1\right)-2\left(3-x\right)\right]\)
\(=\left(2x+1\right)\left(2x+1-6+2x\right)\)
\(=\left(2x+1\right)\left(4x-5\right)\)
\(c,\left(x-1\right)^3-\left(x+1\right)\left(x^2-x+1\right)-\left(3x+1\right)\left(1-3x\right)\)
\(=x^3-3x^2+3x-1-x^3-1-\left(3x-9x^2+1-3x\right)\)
\(=-3x^2+3x-2-3x+9x^2-1+3x\)
\(=6x^2+3x-3\)