Bài 8 : Tìm GTNN của biểu thức:
F= ( x - 1 )2 + ( x - 3 )2
Bài 9 : Tìm GTLN của biểu thức:
A= 4 - x2 + 2x
B= 10x - 23 - x2
C= -x2 + 6x
a) Rút gọn A
b) Với giá trị x;y nguyên dương nào thỏa mãn x + 2y = 14 nhận giá trị nguyên dương.
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\(1,\\ a,A=4x^2\left(-3x^2+1\right)+6x^2\left(2x^2-1\right)+x^2\\ A=-12x^4+4x^2+12x^2-6x^2+x^2=-x^2=-\left(-1\right)^2=-1\\ b,B=x^2\left(-2y^3-2y^2+1\right)-2y^2\left(x^2y+x^2\right)\\ B=-2x^2y^3-2x^2y^2+x^2-2x^2y^3-2x^2y^2\\ B=-4x^2y^3-4x^2y^2+x^2\\ B=-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^3-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^2+\left(0,5\right)^2\\ B=\dfrac{1}{8}-\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{8}\)
\(2,\\ a,\Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ b,\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3=8=-2^3\\ \Leftrightarrow x=2\\ c,\Leftrightarrow4x^2\left(4x-2\right)-x^3+8x^2=15\\ \Leftrightarrow16x^3-8x^2-x^3+8x^2=15\\ \Leftrightarrow15x^3=15\\ \Leftrightarrow x^3=1\Leftrightarrow x=1\)
\(a,\\ A=25x^2-10x+11\\ =\left(5x\right)^2-2.5x.1+1^2+10\\ =\left(5x+1\right)^2+10\ge10\forall x\in R\\ Vậy:min_A=10.khi.5x+1=0\Leftrightarrow x=-\dfrac{1}{5}\\ B=\left(x-3\right)^2+\left(11-x\right)^2\\ =\left(x^2-6x+9\right)+\left(121-22x+x^2\right)\\ =x^2+x^2-6x-22x+9+121=2x^2-28x+130\\ =2\left(x^2-14x+49\right)+32\\ =2\left(x-7\right)^2+32\\ Vì:2\left(x-7\right)^2\ge0\forall x\in R\\ Nên:2\left(x-7\right)^2+32\ge32\forall x\in R\\ Vậy:min_B=32.khi.\left(x-7\right)=0\Leftrightarrow x=7\\Tương.tự.cho.biểu.thức.C\)
b:
\(D=-25x^2+10x-1-10\)
\(=-\left(25x^2-10x+1\right)-10\)
\(=-\left(5x-1\right)^2-10< =-10\)
Dấu = xảy ra khi x=1/5
\(E=-9x^2-6x-1+20\)
\(=-\left(9x^2+6x+1\right)+20\)
\(=-\left(3x+1\right)^2+20< =20\)
Dấu = xảy ra khi x=-1/3
\(F=-x^2+2x-1+1\)
\(=-\left(x^2-2x+1\right)+1=-\left(x-1\right)^2+1< =1\)
Dấu = xảy ra khi x=1
a) ĐKXĐ:
\(\left\{{}\begin{matrix}x^2-9\ne0\\x+3\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\ne-3\end{matrix}\right.\Leftrightarrow x\ne\pm3\)
b) \(A=\dfrac{x+15}{x^2-9}-\dfrac{2}{x+3}\)
\(A=\dfrac{x+15}{\left(x+3\right)\left(x-3\right)}-\dfrac{2\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)
\(A=\dfrac{x+15-2x+6}{\left(x+3\right)\left(x-3\right)}\)
\(A=\dfrac{21-x}{\left(x+3\right)\left(x-3\right)}\)
c) Thay x = - 1 vào A ta có:
\(A=\dfrac{21-\left(-1\right)}{\left(-1+3\right)\left(-1-3\right)}=\dfrac{21+1}{2\cdot-4}=\dfrac{22}{-8}=-\dfrac{11}{4}\)
a)
Ta có:
\(A=x^2-2x-1=x^2-2x+1-2=\left(x-1\right)^2-2\)
\(\ge0-2=-2\)
Vậy \(A_{min}=-2\), đạt được khi và chỉ khi \(x-1=0\Leftrightarrow x=1\)
b)\(B=4x^2+4x+8=4x^2+4x+1+7\)
\(=\left(2x+1\right)^2+7\ge0+7=7\)
Vậy \(B_{min}=7\), đạt được khi và chỉ khi \(2x+1=0\Leftrightarrow x=\dfrac{-1}{2}\)
c)
Ta có:
\(C=3x-x^2+2=2-\left(x^2-3x\right)\)
\(=2+\dfrac{9}{4}-\left(x^2-2x.\dfrac{3}{2}+\dfrac{9}{4}\right)\)
\(=\dfrac{17}{4}-\left(x-\dfrac{3}{2}\right)^2\le\dfrac{17}{4}-0=\dfrac{17}{4}\)
Vậy \(C_{max}=\dfrac{17}{4}\), đạt được khi và chỉ khi \(x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)
d) Ta có:
\(D=-x^2-5x=-\left(x^2+5x\right)=\dfrac{25}{4}-\left(x^2+2x.\dfrac{5}{2}+\dfrac{25}{4}\right)\)
\(=\dfrac{25}{4}-\left(x+\dfrac{5}{2}\right)^2\le\dfrac{25}{4}-0=\dfrac{25}{4}\)
Vậy \(D_{max}=\dfrac{25}{4}\), đạt được khi và chỉ khi \(x+\dfrac{5}{2}=0\Leftrightarrow x=-\dfrac{5}{2}\)
e) Ta có:
\(E=x^2-4xy+5y^2+10x-22y+28\)
\(=x^2+4y^2+5^2-4xy+10x-20y+y^2-2y+1+2\)
\(=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\)
\(\ge0+0+2=2\)
Vậy \(E_{min}=2\), đạt được khi và chỉ khi \(x-2y+5=y-1=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)
Bài 1:
a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{4}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\dfrac{4\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)
\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{\left(x+3\right)\left(2x+10\right)+3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)
2:
a: \(9x^2-1=\left(3x\right)^2-1=\left(3x-1\right)\left(3x+1\right)\)
b: \(2\left(x-1\right)+x^2-x\)
\(=2\left(x-1\right)+x\left(x-1\right)\)
\(=\left(x-1\right)\left(x+2\right)\)
c: \(3x^2+14x-5\)
\(=3x^2+15x-x-5\)
\(=3x\left(x+5\right)-\left(x+5\right)=\left(x+5\right)\left(3x-1\right)\)
3:
a: \(2x\left(x-1\right)-2x^2=4\)
=>\(2x^2-2x-2x^2=4\)
=>-2x=4
=>x=-2
b: \(x\left(x-3\right)-\left(x+2\right)\left(x-1\right)=5\)
=>\(x^2-3x-\left(x^2+x-2\right)=5\)
=>\(x^2-3x-x^2-x+2=5\)
=>-4x=3
=>x=-3/4
c: \(4x^2-25+\left(2x+5\right)^2=0\)
=>\(\left(2x-5\right)\left(2x+5\right)+\left(2x+5\right)^2=0\)
=>\(\left(2x+5\right)\left(2x-5+2x+5\right)=0\)
=>4x(2x+5)=0
=>\(\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)
a: \(P=\dfrac{x^2-x-18+2x+6-4x+12}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2-3x}{\left(x-3\right)\left(x+3\right)}=\dfrac{x}{x+3}\)
b: P=2/3
=>x/(x+3)=2/3
=>3x=2x+6
=>x=6(nhận)
c: P nguyên
=>x chia hết cho x+3
=>x+3-3 chia hết cho x+3
=>x+3 thuộc {1;-1;2;-2}
=>x thuộc {-2;-4;-1;-5}
Bài 1:
a: Sửa đề \(x^3y-2x^2y+xy\)
\(=y\left(x^3-2x^2+x\right)\)
\(=x\cdot y\cdot\left(x^2-2x+1\right)\)
\(=xy\left(x-1\right)^2\)
b: Sửa đề: \(x^2-9-2xy+y^2\)
\(=\left(x^2-2xy+y^2\right)-9\)
\(=\left(x-y\right)^2-9\)
\(=\left(x-y-3\right)\left(x-y+3\right)\)
Bài 2:
a: ĐKXĐ: \(x\notin\left\{3;-3;-1\right\}\)
b: \(A=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\right):\left(2-\dfrac{x+5}{x+3}\right)\)
\(=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}-\dfrac{x^2-1}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{2x+6-x-5}{x+3}\)
\(=\dfrac{x\left(x-3\right)-2\left(x+3\right)-x^2+1}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{x+3}{x+1}\)
\(=\dfrac{x^2-3x-2x-6-x^2+1}{x-3}\cdot\dfrac{1}{x+1}\)
\(=\dfrac{-5x-5}{\left(x-3\right)\left(x+1\right)}=-\dfrac{5\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}=-\dfrac{5}{x-3}\)
c: \(x^2-x-2=0\)
=>\(\left(x-2\right)\left(x+1\right)=0\)
=>\(\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-1\left(loại\right)\end{matrix}\right.\)
Thay x=2 vào A, ta được:
\(A=\dfrac{-5}{2-3}=\dfrac{-5}{-1}=5\)
Bài 8:
\(F=x^2-2x+1+x^2-6x+9=2x^2-8x+10\\ F=2\left(x^2-4x+4\right)+2=2\left(x-2\right)^2+2\ge2\\ F_{min}=2\Leftrightarrow x=2\)
Bài 9:
\(A=-x^2+2x-1+5=-\left(x-1\right)^2+5\le5\\ A_{max}=5\Leftrightarrow x=1\\ B=-x^2+10x-25+2=-\left(x-5\right)^2+2\le2\\ B_{max}=2\Leftrightarrow x=5\\ C=-x^2+6x-9+9=-\left(x-3\right)^2+9\le9\\ C_{max}=9\Leftrightarrow x=3\)