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Bài 2:
a: Ta có: \(x^2+4x+7\)
\(=x^2+4x+4+3\)
\(=\left(x+2\right)^2+3\ge3\forall x\)
Dấu '=' xảy ra khi x=-2
![](https://rs.olm.vn/images/avt/0.png?1311)
1/
a, \(A=4x^2-4x+5=4x^2-4x+1+4=\left(2x-1\right)^2+4\ge4\)
Dấu "=" xảy ra khi x=1/2
Vậy Amin=4 khi x=1/2
b, \(B=3x^2+6x-1=3\left(x^2+2x+1\right)-4=3\left(x+1\right)^2-4\ge-4\)
Dấu "=" xảy ra khi x=-1
Vậy Bmin = -4 khi x=-1
2/
a, \(A=10+6x-x^2=-\left(x^2-6x+9\right)+19=-\left(x-3\right)^2+19\le19\)
Dấu "=" xảy ra khi x=3
Vậy Amax = 19 khi x=3
b, \(B=7-5x-2x^2=-2\left(x^2-\frac{5}{2}x+\frac{25}{16}\right)+\frac{31}{8}=-2\left(x-\frac{5}{4}\right)^2+\frac{31}{8}\le\frac{31}{8}\)
Dấu "=" xảy ra khi x=5/4
Vậy Bmax = 31/8 khi x=5/4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-4x+1=x^2-2\cdot x\cdot2+4-4+1=\left(x-2\right)^2-4+1\)
\(=\left(x-2\right)^2-3\) \(\forall x\in Z\)
\(\Rightarrow A_{min}=-3khix=2\)
\(a,A=x^2-4x+1=x^2-2.2.x+2^2-3=\left(x-2\right)^2-3\ge-3\)
dấu = xảy ra khi x-2=0
=> x=2
Vậy MinA=-3 khi x=2
\(b,B=5-8x-x^2=-\left(x^2+8x+5\right)=-\left(x^2+2.4.x+4^2\right)+9=-\left(x+4\right)^2+9\le9\)
dấu = xảy ra khi x+4=0
=> x=-4
Vậy MaxB=9 khi x=-4
\(c,C=5x-x^2=-\left(x^2-5x\right)=-\left(x^2-\frac{2.x.5}{2}+\frac{25}{4}\right)+\frac{25}{4}=-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\)
dấu = xảy ra khi \(x-\frac{5}{2}=0\)
=> x=\(\frac{5}{2}\)
Vậy Max C=\(\frac{25}{4}\)khi x=\(\frac{5}{2}\)
\(E=\frac{1}{x^2+5x+14}=\frac{1}{x^2+\frac{2.x.5}{2}+\frac{25}{4}+\frac{31}{4}}=\frac{1}{\left(x+\frac{5}{2}\right)^2+\frac{31}{4}}\)
\(\left(x+\frac{5}{2}\right)^2+\frac{31}{4}\ge\frac{31}{4}\)
dấu = xảy ra khi \(x+\frac{5}{2}=0\)
=> x\(=-\frac{5}{2}\)
vì tử thức >0,mẫu thức nhỏ nhất và lớn hơn 0 => E lớnnhất khi mẫu thức nhỏ nhất
Vậy \(MaxE=\frac{31}{4}\)khi x\(=-\frac{5}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =-x^2+6x-4
=-(x^2-6x+4)
=-(x^2-6x+9-5)
=-(x-3)^2+5<=5
Dấu = xảy ra khi x=3
b: =3(x^2-5/3x+7/3)
=3(x^2-2*x*5/6+25/36+59/36)
=3(x-5/6)^2+59/12>=59/12
Dấu = xảy ra khi x=5/6
c: \(=-\left(x-3\right)^2+2\left|x-3\right|\)
\(=-\left[\left(\left|x-3\right|\right)^2-2\left|x-3\right|+1-1\right]\)
\(=-\left(\left|x-3\right|-1\right)^2+1< =1\)
Dấu = xảy ra khi x=4 hoặc x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=3x^2-6x+1=3x^2-6x+3-2=3\times\left(x^2-2x+1\right)-2=3\times\left(x-1\right)^2-2\)
\(3\times\left(x-1\right)^2\ge0\Rightarrow3\times\left(x-1\right)^2-2\ge-2\)
\(MinB=-2\Leftrightarrow x=1\)
\(A=-5x^2-4x+13=-5\times\left(x^2+\frac{4}{5}x-\frac{13}{5}\right)=-5\times\left(x^2+2\times x\times\frac{2}{5}+\frac{4}{25}-\frac{4}{25}-\frac{13}{5}\right)=-5\times\left[\left(x+\frac{2}{5}\right)^2-\frac{69}{25}\right]\)
\(\left(x+\frac{2}{5}\right)^2\ge0\Rightarrow\left(x+\frac{2}{5}\right)^2-\frac{69}{25}\ge-\frac{69}{25}\Rightarrow-5\times\left[\left(x+\frac{2}{5}\right)^2-\frac{69}{25}\right]\le\frac{69}{5}\)
\(M\text{ax}A=\frac{69}{5}\Leftrightarrow x=-\frac{2}{5}\)
\(B=-x^2-10x+8=-x^2-10x-25+33=33-\left(x+5\right)^2\)
\(\left(x+5\right)^2\ge0\Rightarrow33-\left(x+5\right)^2\le33\)
\(M\text{ax}B=33\Leftrightarrow x=-5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
\(M=x^2-3x+5\)
\(M=x^2-2.\frac{3}{2}x+\frac{9}{4}+\frac{11}{4}\)
\(M=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\)
Dấu = xảy ra khi \(x-\frac{3}{2}=0\Rightarrow x=\frac{3}{2}\)
Vậy Min M = 11/4 khi x=3/2
b)\(N=2x^2+3x\)
\(N=2\left(x^2+\frac{3}{2}x\right)\)
\(N=2\left(x^2+2.\frac{3}{4}x+\frac{9}{16}\right)-\frac{9}{8}\)
\(N=2\left(x+\frac{3}{4}\right)^2-\frac{9}{8}\ge-\frac{9}{8}\)
Dấu = xảy ra khi \(x+\frac{3}{4}=0\Rightarrow x=-\frac{3}{4}\)
Vậy MIn N = -9/8 khi x=-3/4
c)Tự làm nha
Ta có : x2 - 3x + 5
= x2 - 2.x.\(\frac{3}{2}\) + \(\frac{3}{2}^2\) + \(\frac{11}{4}\)
= \(\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\)
Vì \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\in R\)
Nên : \(\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\) \(\ge\frac{11}{4}\forall x\in R\)
Vậy GTNN của biểu thức là : \(\frac{11}{4}\) khi \(x=\frac{3}{2}\)
\(A=\frac{2}{-5x^2+3x+2}=\frac{2}{\left(-5x^2+3x-\frac{9}{20}\right)+\frac{49}{20}}\)
\(A=\frac{2}{-5\left(x^2-\frac{3}{5}+\frac{9}{100}\right)+\frac{49}{20}}=\frac{2}{-5\left(x-\frac{3}{10}\right)^2+\frac{49}{20}}\ge\frac{2}{\frac{49}{20}}=\frac{40}{49}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(-5\left(x-\frac{3}{10}\right)^2=0\)\(\Leftrightarrow\)\(x=\frac{3}{10}\)
Vậy GTNN của \(A\) là \(\frac{40}{49}\) khi \(x=\frac{3}{10}\)
\(B=\frac{5}{5x^2+4x+1}=\frac{5}{\left(5x^2+4x+\frac{4}{5}\right)+\frac{1}{5}}\)
\(B=\frac{5}{5\left(x^2+\frac{4}{5}x+\frac{4}{25}\right)+\frac{1}{5}}=\frac{5}{5\left(x+\frac{2}{5}\right)^2+\frac{1}{5}}\le\frac{5}{\frac{1}{5}}=25\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(5\left(x+\frac{2}{5}\right)^2=0\)\(\Leftrightarrow\)\(x=\frac{-2}{5}\)
Vậy GTLN của \(B\) là \(25\) khi \(x=\frac{-2}{5}\)
Chúc bạn học tốt ~
a) Ta có: A bé nhất khi \(-5x^2+3x+2\) lớn nhất
Ta có: \(-5x^2+3x+2=\left(-5x^2+3x-\frac{9}{20}\right)+\frac{49}{20}\)
\(=-5\left(x^2-2.\frac{3}{10}+\frac{9}{100}\right)=-5\left(x-\frac{3}{10}\right)^2+\frac{49}{20}\le\frac{49}{20}\)
Do đó \(A=\frac{2}{-5\left(x-\frac{3}{10}\right)^2+\frac{49}{20}}\le\frac{40}{49}\)
Dấu "=" xảy ra \(\Leftrightarrow-5\left(x-\frac{3}{10}\right)^2=0\Leftrightarrow x=\frac{3}{10}\)
Vậy \(A_{max}=\frac{40}{49}\Leftrightarrow x=\frac{3}{10}\)
b) Để B lớn nhất thì \(5x^2+4x+1\) bé nhất.Ta có:
\(5x^2+4x+1=\left(5x^2+4x\right)+1\)
\(=5\left(x^2+\frac{4}{5}x\right)+1=5\left(x^2+2.\frac{4}{10}+\frac{4}{25}\right)+\frac{1}{5}\)
\(=5\left(x+\frac{2}{5}\right)^2+\frac{1}{5}\ge\frac{1}{5}\)
Do đó \(B=\frac{5}{5\left(x+\frac{2}{5}\right)^2}\le\frac{5}{\frac{1}{5}}=25\)
Dấu "=" xảy ra \(\Leftrightarrow5\left(x+\frac{2}{5}\right)^2=0\Leftrightarrow x=-\frac{2}{5}\)
Vậy \(B_{max}=25\Leftrightarrow x=-\frac{2}{5}\)