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1.
Đặt \(x-2=t\ne0\Rightarrow x=t+2\)
\(B=\dfrac{4\left(t+2\right)^2-6\left(t+2\right)+1}{t^2}=\dfrac{4t^2+10t+5}{t^2}=\dfrac{5}{t^2}+\dfrac{2}{t}+4=5\left(\dfrac{1}{t}+\dfrac{1}{5}\right)^2+\dfrac{19}{5}\ge\dfrac{19}{5}\)
\(B_{min}=\dfrac{19}{5}\) khi \(t=-5\) hay \(x=-3\)
2.
Đặt \(x-1=t\ne0\Rightarrow x=t+1\)
\(C=\dfrac{\left(t+1\right)^2+4\left(t+1\right)-14}{t^2}=\dfrac{t^2+6t-9}{t^2}=-\dfrac{9}{t^2}+\dfrac{6}{t}+1=-\left(\dfrac{3}{t}-1\right)^2+2\le2\)
\(C_{max}=2\) khi \(t=3\) hay \(x=4\)
b: Ta có: \(B=-2x^2+4x+1\)
\(=-2\left(x^2-2x-\dfrac{1}{2}\right)\)
\(=-2\left(x^2-2x+1-\dfrac{3}{2}\right)\)
\(=-2\left(x-1\right)^2+3\le3\forall x\)
Dấu '=' xảy ra khi x=1
Lời giải:
1.
$x^3+3x^2-16x-48=(x^3+3x^2)-(16x+48)=x^2(x+3)-16(x+3)$
$=(x+3)(x^2-16)=(x+3)(x-4)(x+4)$
2.
$4x(x-3y)+12y(3y-x)=4x(x-3y)-12y(x-3y)=(x-3y)(4x-12y)=4(x-3y)(x-3y)=4(x-3y)^2$
3.
$x^3+2x^2-2x-1=(x^3-x^2)+(3x^2-3x)+(x-1)=x^2(x-1)+3x(x-1)+(x-1)$
$=(x-1)(x^2+3x+1)$
\(A=\dfrac{6x^2+21x+22}{x^2+4x+4}\)
\(=\dfrac{6\left(x^2+4x+4\right)-3x-2}{x^2+4x+4}\)
\(=6+\dfrac{-3x-2}{\left(x+2\right)^2}\)
\(=6+\dfrac{-3\left(x+2\right)+4}{\left(x+2\right)^2}\)
\(=6-\dfrac{3}{x+2}+\dfrac{4}{\left(x+2\right)^2}\)
-Đặt \(a=\dfrac{1}{x+2}\) thì:
\(A=6-3a+4a^2=\left(2a\right)^2-2.2a.\dfrac{3}{4}+\dfrac{9}{16}+\dfrac{87}{16}=\left(2a-\dfrac{3}{4}\right)^2+\dfrac{87}{16}\ge\dfrac{87}{16}\)
\(A_{min}=\dfrac{87}{16}\)\(\Leftrightarrow\left(2a-\dfrac{3}{4}\right)^2=0\Leftrightarrow2a-\dfrac{3}{4}=0\Leftrightarrow2a=\dfrac{3}{4}\)
\(\Leftrightarrow2.\dfrac{1}{x+2}=\dfrac{3}{4}\Leftrightarrow\dfrac{1}{x+2}=\dfrac{3}{8}\Leftrightarrow x+2=\dfrac{8}{3}\Leftrightarrow x=\dfrac{2}{3}\)
Bài 2 :
\(A=4x^2-2.2x.2+4+1\)
\(=\left(2x-2\right)^2+1\)
Thấy : \(\left(2x-2\right)^2\ge0\)
\(A=\left(2x-2\right)^2+1\ge1\)
Vậy \(MinA=1\Leftrightarrow x=1\)
\(B=\left(5x\right)^2-2.5x.1+1-4\)
\(=\left(5x-1\right)^2-4\)
Thấy : \(\left(5x-1\right)^2\ge0\)
\(\Rightarrow B=\left(5x-1\right)^2-4\ge-4\)
Vậy \(MinB=-4\Leftrightarrow x=\dfrac{1}{5}\)
\(C=\left(7x\right)^2-2.7x.2+4-5\)
\(=\left(7x-2\right)^2-5\)
Thấy : \(\left(7x-2\right)^2\ge0\)
\(\Rightarrow C=\left(7x-2\right)^2-5\ge-5\)
Vậy \(MinC=-5\Leftrightarrow x=\dfrac{2}{7}\)
\(1.\)
\(A=-x^2-10x+1=-\left(x^2+10x-1\right)\)
\(=-\left(x^2+2.5x+5^2-5^2-1\right)=-\left[\left(x+5\right)^2-26\right]\)
\(=-\left(x+5\right)^2+26\le26\) dấu "=" xảy ra<=>x=-5
\(B=-4x^2-6x-5=-4\left(x^2+\dfrac{6}{4}x+\dfrac{5}{4}\right)\)
\(=-4\left(x^2+2.\dfrac{3}{4}x+\dfrac{9}{16}+\dfrac{11}{16}\right)\)\(=-4\left[\left(x+\dfrac{3}{2}\right)^2+\dfrac{11}{6}\right]\le-\dfrac{11}{4}\)
\(C=-16x^2+8x-1=-16\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)\)
\(=-16\left(x^2-2.\dfrac{1}{4}x+\dfrac{1}{16}\right)=-16\left(x-\dfrac{1}{4}\right)^2\le0\)
dấu"=" xảy ra<=>x=1/4
Bài 1:
a: \(M=x^2-10x+3\)
\(=x^2-10x+25-22\)
\(=\left(x^2-10x+25\right)-22\)
\(=\left(x-5\right)^2-22>=-22\forall x\)
Dấu '=' xảy ra khi x-5=0
=>x=5
b: \(N=x^2-x+2\)
\(=x^2-x+\dfrac{1}{4}+\dfrac{7}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\forall x\)
Dấu '=' xảy ra khi x-1/2=0
=>x=1/2
c: \(P=3x^2-12x\)
\(=3\left(x^2-4x\right)\)
\(=3\left(x^2-4x+4-4\right)\)
\(=3\left(x-2\right)^2-12>=-12\forall x\)
Dấu '=' xảy ra khi x-2=0
=>x=2