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Ta có 4M = 4a2 + 4ab + 4b2 - 12a - 12b + 8052
= (4a2 + 4ab + b2) - 6(2a + b) + 9 + 3b2 - 6b + 3 + 8040
= (2a + b)2 - 6(a + b) + 9 + 3(b2 - 2b + 1) + 8040
= (2a + b - 3)2 + 3(b - 1)2 + 8040 \(\ge\)8040
=> Min 4M = 8040
=> Min M = 2010
Dấu "=" xảy ra <=> \(\hept{\begin{cases}2a+b-3=0\\b-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=1\\b=1\end{cases}}\Leftrightarrow a=b=1\)
Vạy Min M = 2010 <=> a = b = 1
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![](https://rs.olm.vn/images/avt/0.png?1311)
bài này dễ ẹt ak
nhưng giúp mình bài này đi
chotam giac abc . co canh bc=12cm, duong cao ah=8cm
a> tinh s tam giac abc
b> tren canh bc lay diem e sao cho be=3/4bc. tinh s tam giac abe va s tam giac ace ( bằng nhiều cách )
c> lay diem chinh giua cua canh ac va m . tinh s tam giac ame
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Ta có : \(a+b=2\)
\(\Rightarrow\)\(a = 2 -b\)
\(A = 2a^2 +3b^2 +3ab\)
\(A = 2a^2 + 3b. (a+b)\)
\(A = 2. (2-b)^2+3b. (2-b+b)\)
\(A = 2. ( b^2 -4b+4)+6b\)
\(A = 2b^2 -8b+8+6b\)
\(A = 2b^2 -2b+8\)
\(A = 2. ( b ^2 -b+4)\)
\(A=2. (b^2 -2.b.{1\over2}+({1\over2})^2-({1\over2})^2+4)\)
\(A = 2. [ (b -{1\over2})^2-{15\over4}]\)
\(A =2. (b-{1\over2})^2 + {15\over2}\)\(\ge\)\({15\over2}\)
\(Min A ={15\over2}\)\(\Leftrightarrow\)\(a = {3\over2};b={1\over2}\)
Ta có : a+b=2→b=2−a
→P=2a2+3b2+3ab=2a2+3b(a+b)=2a2+3b.2=2a2+6b=2a2+6(2−a)=2a2−6a+12
→P=2(a2−3a)+12
→P=2(a2−2a.32+94)+152
→P=2(a−32)2+152≥152
→GTNNP=152
Dấu = xảy ra khi a−32=0
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(a^2+b^2+1-2ab-2a+2b\right)+\frac{1}{2}\left(4b^2-4b+1\right)+2008\)
\(A=\left(a-b-1\right)^2+\frac{1}{2}\left(2b-1\right)^2+2008\ge2008\)
\(A_{min}=2008\) khi \(\left\{{}\begin{matrix}a=\frac{3}{2}\\b=\frac{1}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta dễ dàng chứng minh:
\(0< a,b,c\le\frac{3}{2}\)
Áp dụng BDT cô si cho ba số dương ta có:
\(\left(\frac{3}{2}-a\right)+\left(\frac{3}{2}-b\right)+\left(\frac{3}{2}-c\right)\ge3\sqrt[3]{\frac{3}{2}-a)(\frac{3}{2}-b)(\frac{3}{2}-c)}\)
\(\Leftrightarrow\left(\frac{1}{2}\right)^3\ge\frac{3}{2}-a)(\frac{3}{2}-b)(\frac{3}{2}-c)\)
\(\Leftrightarrow\frac{1}{8}\ge\frac{27}{8}-\frac{9}{4}\left(a+b+c\right)+\frac{3}{2}\left(ab+bc+ac\right)-abc\)
\(\Leftrightarrow\frac{1}{8}\ge-\frac{27}{8}+\frac{3}{2}\left(ab+bc+ac\right)-abc\)
\(\Leftrightarrow4abc\ge-14+6\left(ab+bc+ac\right)\)
\(\Leftrightarrow3a^2+3b^2+3c^2+4abc\ge13\)
:< rồi để căn nó mệt người mik đặt hem:P
Ta có: \(\hept{\begin{cases}\sqrt{a}=a\\\sqrt{b}=b\end{cases}}\)
\(P=a^2-2ab+3b^2-2a+1\)
\(\Leftrightarrow3P=3a^2-6ab+9b^2-6a+3\)
\(\Leftrightarrow3P=\left(x-3b\right)^2+2\left(a-\frac{3}{2}\right)^2-\frac{3}{2}\ge-\frac{3}{2}\)
\(\Rightarrow P\ge\frac{1}{2}\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}a=\frac{3}{2}\\b=\frac{1}{2}\end{cases}}\) hay \(\hept{\begin{cases}a=\frac{9}{4}\\b=\frac{1}{4}\end{cases}}\)
Đặt \(\sqrt{a}=u;\sqrt{b}=v\left(u,v\ge0\right)\)
Lúc đó \(P=u^2-2uv+3v^2-2u+1\)
\(\Rightarrow3P=3u^2-6uv+9v^2-6u+3\)
\(=\left(u^2-6uv+9v^2\right)+2\left(u^2-6u+\frac{9}{4}\right)-\frac{3}{2}\)
\(=\left(u-3v\right)^2+2\left(u-\frac{3}{2}\right)^2-\frac{3}{2}\ge-\frac{3}{2}\)
\(\Rightarrow P\ge\frac{-1}{2}\)
(Dấu "=" khi \(\hept{\begin{cases}u=\frac{3}{2}\\v=\frac{1}{2}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\sqrt{a}=\frac{3}{2}\\\sqrt{b}=\frac{1}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}a=\frac{9}{4}\\b=\frac{1}{4}\end{cases}}\))