cho a>_3,tim gtri nho nhat cua S=a+\(\frac{1}{a}\)
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\(A=x^2-5x+12\\ A=x^2-5x+\dfrac{25}{4}+\dfrac{23}{4}\\ A=\left(x^2-5x+\dfrac{25}{4}\right)+\dfrac{23}{4}\\ A=\left[x^2-2\cdot x\cdot\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]+\dfrac{23}{4}\\ A=\left(x-\dfrac{5}{2}\right)^2+\dfrac{23}{4}\\ Do\text{ }\left(x-\dfrac{5}{2}\right)^2\ge0\forall x\\ \Rightarrow A=\left(x-\dfrac{5}{2}\right)^2+\dfrac{23}{4}\ge\dfrac{23}{4}\forall x\\ \text{Dấu "=" xảy ra khi : }\\ \left(x-\dfrac{5}{2}\right)^2=0\\ \Leftrightarrow x-\dfrac{5}{2}=0\\ \Leftrightarrow x=\dfrac{5}{2}\\ \text{Vậy }A_{\left(Min\right)}=\dfrac{23}{4}\text{ }khi\text{ }x=\dfrac{5}{2}\)
\(B=2x^2-14x+5\\ \\ A=2x^2-14x+\dfrac{49}{2}-\dfrac{39}{2}\\ A=\left(2x^2-14x+\dfrac{49}{2}\right)-\dfrac{39}{2}\\ A=2\left(x^2-7x+\dfrac{49}{4}\right)-\dfrac{39}{2}\\ A=\left[x^2-2\cdot x\cdot\dfrac{7}{2}+\left(\dfrac{7}{2}\right)^2\right]-\dfrac{39}{2}\\ A=\left(x-\dfrac{7}{2}\right)^2-\dfrac{39}{2}\\ Do\text{ }\left(x-\dfrac{7}{2}\right)^2\ge0\forall x\\ \Rightarrow A=\left(x-\dfrac{7}{2}\right)^2-\dfrac{39}{2}\ge-\dfrac{39}{2}\forall x\\ \text{Dấu "=" xảy ra khi : }\\ \left(x-\dfrac{7}{2}\right)^2=0\\ \Leftrightarrow x-\dfrac{7}{2}=0\\ \Leftrightarrow x=\dfrac{7}{2}\\ \text{Vậy }B_{\left(Min\right)}=-\dfrac{39}{2}\text{ }khi\text{ }x=\dfrac{7}{2}\)
\(B=2x^2-14x+5\\ B=2x^2-14x+\dfrac{49}{2}-\dfrac{39}{2}\\ B=\left(2x^2-14x+\dfrac{49}{2}\right)-\dfrac{39}{2}\\ B=2\left(x^2-7x+\dfrac{49}{4}\right)-\dfrac{39}{2}\\ B=2\left[x^2-2\cdot x\cdot\dfrac{7}{2}+\left(\dfrac{7}{2}\right)^2\right]-\dfrac{39}{2}\\ B=2\left(x-\dfrac{7}{2}\right)^2-\dfrac{39}{2}\\ \)
Do \(\left(x-\dfrac{7}{2}\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-\dfrac{7}{2}\right)^2\ge0\forall x\)
\(\Rightarrow B=2\left(x-\dfrac{7}{2}\right)^2-\dfrac{39}{2}\ge-\dfrac{39}{2}\forall x\)
Dấu \("="\) xảy ra khi :
\(\left(x-\dfrac{7}{2}\right)^2=0\\ \Leftrightarrow x-\dfrac{7}{2}=0\\ \Leftrightarrow x=\dfrac{7}{2}\)
Vậy \(B_{\left(Min\right)}=-\dfrac{39}{2}\) khi \(x=\dfrac{7}{2}\)
Do máy bị lỗi nên câu B bị trục trặc.
Mk xin lỗi.
\(A=\frac{3n+3}{n-3}\left(n\ne3\right)\)
\(A=\frac{3\left(n-3\right)+12}{n-3}=3+\frac{12}{n-3}\)
A có GTLN khi \(\frac{12}{n-3}\)nhỏ nhất => n-3 nhỏ nhất
=> n-3=1
=> n=4
A có GTNN khi \(\frac{12}{n-3}\)lớn nhất => n-3 lớn nhất
=> n-3 =12
=> n=15
A= 6n-1 chia hết cho 3n-2
=> 3(6n-10) chia hết cho 3n-2
=> 18n-10 chia hết cho 3n-2
=> 6(3n-2) -2 chia hết cho 3n-2
=> 2 chia hết cho 3n-2
=> 3n-2E{-1; -2; 1;2}
=> 3nE{ 1; 0; 3; 4}
=> nE{ 0; 1}
\(A=\frac{2010x+2680}{x^2+1}\Leftrightarrow A\left(x^2+1\right)=2010x+2680\Leftrightarrow Ax^2-2010x+\left(A-2680\right)=0\)
Để A có nghiệm, ta phải có \(\Delta'\ge0\Leftrightarrow-A^2+2680A+1005^2\ge0\Leftrightarrow-335\le A\le3015\)
Vậy Min A = -335 \(\Leftrightarrow x=-3\)
M=x2+y2-x+6y+10
=(x-1/2)2+(y+3)3+3/4
Ta thấy:(x-1/2)2>=0
(y+3)3>=0
=>(x-1/2)2+(y+3)>=0
=>(x-1/2)2+(y+3)+3/4>=0+3/4=3/4
Dấu "="<=>x=1/2 hoặc y=-3
Vậy...
1, Tinh gtri nho nhat cua bieu thuc A = \(\left|x-1\right|+\left|x-2\right|+..+\left|x-2020\right|\)
GTLN :
\(A=\frac{x+1}{x^2+x+1}=\frac{\left(x^2+x+1\right)-x^2}{x^2+x+1}=1-\frac{x^2}{x^2+x+1}\)
Vì \(\frac{x^2}{x^2+x+1}=\frac{x^2}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}\ge0\forall x\) nên \(A=1-\frac{x^2}{x^2+x+1}\le1\forall x\) có GTLN là 1
GTNN :
\(A=\frac{x+1}{x^2+x+1}=\frac{-\frac{1}{3}x^2-\frac{1}{3}x-\frac{1}{3}+\frac{1}{3}x^2+\frac{4}{3}x+\frac{4}{3}}{x^2+x+1}=\frac{-\frac{1}{3}\left(x^2+x+1\right)+\frac{1}{3}\left(x+2\right)^2}{x^2+x+1}\)
\(=-\frac{1}{3}+\frac{\frac{1}{3}\left(x+2\right)^2}{x^2+x+1}=-\frac{1}{3}+\frac{\left(x+2\right)^2}{3\left(x^2+x+1\right)}\ge-\frac{1}{3}\) có GTNN là \(-\frac{1}{3}\)
Lạnh Lạnh đúng vì;
\(a+\frac{1}{a}=a+\frac{9}{a}-\frac{8}{a}\ge2\sqrt{a.\frac{9}{a}}-\frac{8}{3}=2.3-\frac{8}{3}=\frac{10}{3}\)
Dấu "=" xảy ra khi và chỉ khi a=3
vậy min S=10/3 tại a=3
=10/3 .. bạn nghĩ sao