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

Mình quên yêu cầu bài 2: Tìm GTNN GTLN của x.

yêu cầu bài 2 Tìm giá trị min max của x

áp dụng bđt bunhia đê 

Câu 1:

\(y^2+yz+z^2=1-\frac{3x^2}{2}\)

\(\Leftrightarrow2y^2+2yz+2z^2=2-3x^2\)

\(\Leftrightarrow\left(y+z\right)^2+y^2+z^2+3x^2=2\)

\(\Leftrightarrow\left(y+z\right)^2+x^2+2x\left(y+z\right)+y^2+z^2+2x^2-2x\left(y+z\right)=2\)

\(\Leftrightarrow\left(x+y+z\right)^2+\left(x^2-2xy+y^2\right)+\left(x^2-2xz+z^2\right)=2\)

\(\Leftrightarrow\left(x+y+z\right)^2=2-\left(x-y\right)^2-\left(x-z\right)^2\)

\(\Leftrightarrow A^2=2-\left[\left(x-y\right)^2+\left(x-z\right)^2\right]\le2\forall x;y;z\)

\(\Leftrightarrow-\sqrt{2}\le A\le\sqrt{2}\)

Vậy \(A_{min}=-\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x=y=z\\x+y+z=-\sqrt{2}\end{matrix}\right.\)\(\Leftrightarrow x=y=z=\frac{-\sqrt{2}}{3}\)

\(A_{max}=\sqrt{2}\Leftrightarrow a=b=c=\frac{\sqrt{2}}{3}\)

Câu 2:

Áp dụng BĐT Cauchy-Schwarz:

\(P=\frac{1}{1+xy}+\frac{1}{1+yz}+\frac{1}{1+zx}\ge\frac{9}{3+xy+yz+zx}\ge\frac{9}{3+x^2+y^2+z^2}=\frac{9}{6}=\frac{3}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)

Câu 3:

\(P=\frac{ab\sqrt{c-2}+bc\sqrt{a-3}+ca\sqrt{b-4}}{abc}\) ( \(a\ge3;b\ge4;c\ge2\) )

\(P=\frac{\sqrt{c-2}}{c}+\frac{\sqrt{a-3}}{a}+\frac{\sqrt{b-4}}{b}\)

Áp dụng BĐT Cauchy:

\(P=\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{2}\cdot\sqrt{c-2}}{c}+\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}\cdot\sqrt{a-3}}{a}+\frac{1}{2}\cdot\frac{2\cdot\sqrt{b-4}}{b}\)

\(\le\frac{1}{\sqrt{2}}\cdot\frac{1}{2}\cdot\frac{2+c-2}{c}+\frac{1}{\sqrt{3}}\cdot\frac{1}{2}\cdot\frac{3+a-3}{a}+\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{4+b-4}{b}=\frac{1}{2}\cdot\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{2}\right)\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a=6\\b=8\\c=4\end{matrix}\right.\)

Câu 4:

Đặt \(\sqrt{x}=a;\sqrt{y}=b\left(a;b\ge0\right)\)

\(M=a^2-2ab+3b^2-2a+1\)

\(M=a^2-a\left(2b+2\right)+3b^2+1\)

\(\Delta=\left(2b+2\right)^2-4\left(3b^2+1\right)\)

\(=-8b^2+8b\)

\(=-8b\left(b+1\right)\ge0\)

Vì \(b\ge0\) nên \(-8b\left(b+1\right)\le0\)

Dấu "=" xảy ra \(\Leftrightarrow b=0\)

Khi đó \(M=a^2-2a+1=\left(a-1\right)^2\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow a=1\)

Vậy \(M_{min}=1\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)

b) Ta có \(A=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge\frac{\left(x+y+z\right)^2}{y+z+z+x+x+y}\)(BĐT Schwarz) 

\(=\frac{x+y+z}{2}=\frac{2}{2}=1\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}\frac{x^2}{y+z}=\frac{y^2}{z+x}=\frac{z^2}{x+y}\\x+y+z=2\end{cases}}\Leftrightarrow x=y=z=\frac{2}{3}\)

a) Có \(P=1.\sqrt{2x+yz}+1.\sqrt{2y+xz}+1.\sqrt{2z+xy}\)

\(\le\sqrt{\left(1^2+1^2+1^2\right)\left(2x+yz+2y+xz+2z+xy\right)}\)(BĐT Bunyakovsky) 

\(=\sqrt{3.\left[2\left(x+y+z\right)+xy+yz+zx\right]}\)

\(\le\sqrt{3\left[4+\frac{\left(x+y+z\right)^2}{3}\right]}=\sqrt{3\left(4+\frac{4}{3}\right)}=4\)

Dấu "=" xảy ra <=> x = y = z = 2/3