1. Tìm GTNN :
a) A=x^2+x
b) B=9x^2-6x+7
c) C=25x^2+30x-1
d) D=5x^2-2x-1
2. Tìm GTLN :
a) A=2-x-x^2
b) B=-8+3x-x^2
c)C=2015+7x-x^2
d) D=19-x-11x^2
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a: ĐKXĐ: \(\left[{}\begin{matrix}x\ge3\\x\le2\end{matrix}\right.\)
b: ĐKXĐ: \(\left[{}\begin{matrix}x>\dfrac{2\sqrt{14}}{7}\\x< -\dfrac{2\sqrt{14}}{7}\end{matrix}\right.\)
c: ĐKXĐ: \(x=\dfrac{1}{3}\)
d: ĐKXĐ: \(-\dfrac{2}{3}< x\le\sqrt{3}\)
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
Tìm GTLN:
\(A=-x^2+6x-15\)
\(=-\left(x^2-6x+15\right)\)
\(=-\left(x^2-2.x.3+9+6\right)\)
\(=-\left(x+3\right)^2-6\le0\forall x\)
Dấu = xảy ra khi:
\(x-3=0\Leftrightarrow x=3\)
Vậy Amax = - 6 tại x = 3
Tìm GTNN :
\(A=x^2-4x+7\)
\(=x^2+2.x.2+4+3\)
\(=\left(x+2\right)^2+3\ge0\forall x\)
Dấu = xảy ra khi:
\(x+2=0\Leftrightarrow x=-2\)
Vậy Amin = 3 tại x = - 2
Các câu còn lại làm tương tự nhé... :)
`A=x^2-2x+5`
`=x^2-2x+1+4`
`=(x-1)^2+4>=4`
Dấu "=" `<=>x=1`
`B=4x^2+4x+3`
`=4x^2+4x+1+2`
`=(2x+1)^2+2>=2`
Dấu "=" xảy ra khi `x=-1/2`
`C=9x^2-6x+7`
`=9x^2-6x+1+6`
`=(3x-1)^2+6>=6`
Dấu '=' xảy ra khi `x=1/3`
`D=5x^2+3x+8`
`=5(x^2+3/5x)+8`
`=5(x^2+3/5x+9/100-9/100)+8`
`=5(x+3/10)^2+151/20>=151/20`
Dấu "=" xảy ra khi `x=-3/10`
\(A=x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4\)
Ta có: \(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2+4\ge4\Rightarrow A_{min}=4\) khi \(x=1\)
\(B=4x^2+4x+3=4x^2+4x+1+2=\left(2x+1\right)^2+2\)
Ta có: \(\left(2x+1\right)^2\ge0\Rightarrow\left(2x+1\right)^2+2\ge2\Rightarrow B_{min}=2\) khi \(x=-\dfrac{1}{2}\)
\(C=9x^2-6x+7=9x^2-6x+1+6=\left(3x-1\right)^2+6\)
Ta có: \(\left(3x-1\right)^2\ge0\Rightarrow\left(3x-1\right)^2+6\ge6\Rightarrow C_{min}=6\) khi \(x=\dfrac{1}{3}\)
\(D=5x^2+3x+8\Rightarrow5\left(x^2+2.x.\dfrac{3}{10}+\dfrac{9}{100}\right)+\dfrac{151}{20}=5\left(x+\dfrac{3}{10}\right)^2+\dfrac{151}{20}\)
Ta có: \(5\left(x+\dfrac{3}{10}\right)^2\ge0\Rightarrow5\left(x+\dfrac{3}{10}\right)^2+\dfrac{151}{20}\ge\dfrac{151}{20}\)
\(\Rightarrow D_{min}=\dfrac{151}{20}\) khi \(x=-\dfrac{3}{10}\)
Bài 1:
\(A=-x^2-2x+9\)
\(A=-\left(x^2+2x-9\right)\)
\(A=-\left(x^2+2x+1-10\right)\)
\(A=-\left(x+1\right)^2+10\)
Vì \(-\left(x+1\right)^2\le0\) với mọi x
\(\Rightarrow-\left(x+1\right)^2+10\le10\)
\(\Rightarrow Amax=10\Leftrightarrow x=-1\)
\(B=-9x^2+6x+25\)
\(B=-\left(9x^2-6x-25\right)\)
\(B=-\left[\left(3x\right)^2-2.3x+1-26\right]\)
\(B=-\left(3x-1\right)^2+26\)
Vì \(-\left(3x-1\right)^2\le0\) với mọi x
\(\Rightarrow-\left(3x-1\right)^2+26\le26\)
\(\Rightarrow Bmax=26\Leftrightarrow3x-1=0\Rightarrow x=\dfrac{1}{3}\)
\(C=-x^2+x+1\)
\(C=-\left(x^2-x-1\right)\)
\(C=-\left(x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}-1\right)\)
\(C=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}\)
Vì \(-\left(x-\dfrac{1}{2}\right)^2\le0\) với mọi x
\(\Rightarrow-\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\)
\(\Rightarrow Cmax=\dfrac{5}{4}\Leftrightarrow x=\dfrac{1}{2}\)
\(D=-2x^2+3x+1\)
\(D=-2\left(x^2-\dfrac{3}{2}x-\dfrac{1}{2}\right)\)
\(D=-2\left(x^2-2.x\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{9}{16}-\dfrac{1}{2}\right)\)
\(D=-2\left(x-\dfrac{3}{4}\right)^2+\dfrac{17}{8}\)
Vì \(-2\left(x-\dfrac{3}{4}\right)^2\le0\) với mọi x
\(\Rightarrow-2\left(x-\dfrac{3}{4}\right)^2+\dfrac{17}{8}\le\dfrac{17}{8}\)
\(\Rightarrow Dmax=\dfrac{17}{8}\Leftrightarrow x=\dfrac{3}{4}\)
\(E=-25x^2-10x+7\)
\(E=-\left(25x^2+10x-7\right)\)
\(E=-\left[\left(5x\right)^2+2.5x+1-8\right]\)
\(E=-\left(5x+1\right)^2+8\)
Vì \(-\left(5x+1\right)^2\le0\) với mọi x
\(\Rightarrow-\left(5x+1\right)^2+8\le8\)
\(\Rightarrow Emax=8\Leftrightarrow5x+1=0\Rightarrow x=-\dfrac{1}{5}\)
Bài 2:
\(A=9x^2+6x+4\)
\(A=\left(3x\right)^2+2.3x+1+3\)
\(A=\left(3x+1\right)^2+3\)
Vì \(\left(3x+1\right)^2\ge0\) với mọi x
\(\Rightarrow\left(3x+1\right)^2+3\ge3\)
\(\Rightarrow Amin=3\Leftrightarrow x=-\dfrac{1}{3}\)
\(B=4x^2+4x+12\)
\(B=\left(2x\right)^2+2.2x+1+11\)
\(B=\left(2x+1\right)^2+11\)
Vì \(\left(2x+1\right)^2\ge0\) với mọi x
\(\Rightarrow\left(2x+1\right)^2+11\ge11\)
\(\Rightarrow Bmin=11\Leftrightarrow x=-\dfrac{1}{2}\)
\(C=x^2+x+3\)
\(C=x^2+2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+3\)
\(C=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\)
Vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)
\(\Rightarrow Cmin=\dfrac{11}{4}\Leftrightarrow x=-\dfrac{1}{2}\)
\(D=2x^2+3x+1\)
\(D=2\left(x^2+\dfrac{3}{2}x+\dfrac{1}{2}\right)\)
\(D=2\left(x^2+2.x.\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{9}{16}+\dfrac{1}{2}\right)\)
\(D=2\left(x+\dfrac{3}{4}\right)^2-\dfrac{1}{8}\)
Vì \(2\left(x+\dfrac{3}{4}\right)^2\ge0\) với mọi x
\(\Rightarrow2\left(x+\dfrac{3}{4}\right)^2-\dfrac{1}{8}\ge-\dfrac{1}{8}\)
\(\Rightarrow Dmin=-\dfrac{1}{8}\Leftrightarrow x=-\dfrac{3}{4}\)
\(E=64x^2+16x+3\)
\(E=\left(8x\right)^2+2.8x+1+2\)
\(E=\left(8x+1\right)^2+2\)
Vì \(\left(8x+1\right)^2\ge0\) với mọi x
\(\Rightarrow\left(8x+1\right)^2+2\ge2\)
\(\Rightarrow Emin=2\Leftrightarrow x=-\dfrac{1}{8}\)