a) công thức . \(\frac{đáy.chiềucao}{2}\)

b) Áp dụng định lý pitago ta có

\(BC^2=AB^2+AC^2\)

=> AC^2=\(BC^2-AB^2=^{10^2}-6^2=64\)

=>\(AC=8\)

A)Xét tam giác ABC vuông tại A(gt),có:

SABC=(AB.AC)/2

B)Xét tam giác ABC vuông tại A(gt),có:

AB^2+AC^2=BC^2(ĐL Pytago)

Thay số:36+AC^×=100

<=>AC=căn64=8cm

Ta có:SABC=(AB.AC)/2

Thay số:SABC=24cm^2

Mà SABC=(AH.BC)/2

=>(AH.BC)/2=24

Thay số:AH=24.2:10=4,8cm

SABC=24CM^2(cmt)

\(A=2x^2+2xy+y^2-2x+2y+2\)

\(\Rightarrow2A=4x^2+4xy+2y^2-4x+4y+4\)

              \(=\left(4x^2+4xy+y^2\right)-2\left(2x+y\right).1+1+y^2+6y+9-6\)

               \(=\left(2x+y\right)^2-2\left(2x+y\right)+1+\left(y+3\right)^2-6\)

                \(=\left(2x+y-1\right)^2+\left(y+3\right)^2-6\)

vì \(\left(2x+y-1\right)^2\ge0\forall x,y;\left(y+3\right)^2\ge0\forall y\)nên

\(2A=\left(2x+y-1\right)+\left(y+3\right)-6\ge-6\forall x,y\)

hay \(2A\ge-6\Rightarrow A\ge-3\Rightarrow minA=-3\Leftrightarrow\hept{\begin{cases}2x+y-1=0\\y+3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=-3\end{cases}}}\)

lấy cả 2 vế trừ đi 3

\(\frac{x-2010-2011}{2009}+\frac{x-2009-2011}{2010}+\frac{x-2009-2010}{2011}=3\)

\(\Leftrightarrow\left(\frac{x-2010-2011}{2009}-1\right)+\left(\frac{x-2009-2011}{2010}-1\right)+\left(\frac{x-2009-2010}{2011}-1\right)=0\)

\(\Leftrightarrow\frac{x-6030}{2009}+\frac{x-6030}{2010}+\frac{x-6030}{2011}=0\)

\(\Leftrightarrow\left(x-6030\right)\left(\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}\right)\)

\(\Leftrightarrow x-6030=0\)(vì \(\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}>0\))

\(\Leftrightarrow x=6030\)

Vậy ................

\(B=-x^2-y^2+xy+2x+2y\)

\(\Rightarrow-2B=2x^2+2y^2-2xy-2x-4y\)

                    \(=\left(x^2-2xy+y^2\right)+\left(x^2-4x+4\right)+\left(y^2-4y+4\right)-8\)

                     \(=\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2-8\)

vì \(\left(x-y\right)^2\ge0\forall x,y;\left(x-2\right)^2\ge0\forall x;\left(y-2\right)\ge0\forall y\)nên

\(-2B=\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2-8\ge8\)

hay \(-2B\ge-8\Rightarrow B\le4\)

\(\Rightarrow maxB=4\Leftrightarrow\hept{\begin{cases}x-y=0\\x-2=0\\y-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=y\\x=2\\y=2\end{cases}}}\)