A = \(\left\{16780\right\}\)

$\left\{16780\right\}$

\(tan45^o=\dfrac{CD-1,8}{10}\) (CD là chiều cao cây bàng)

\(\Rightarrow CD-1,8=10.tan45^o\)

\(\Rightarrow CD=10.1+1,8=11,8\left(m\right)\)

20m

đề là j vậy bạn ?

THẰNG ĐIÊN 

1: \(\sqrt6-\sqrt5-\sqrt3+\sqrt{10}\)

\(=\left(\sqrt6-\sqrt3\right)+\left(\sqrt{10}-\sqrt5\right)\)

\(=\sqrt3\left(\sqrt2-1\right)+\sqrt5\left(\sqrt2-1\right)=\left(\sqrt2-1\right)\left(\sqrt3+\sqrt5\right)\)

2: \(x-2\sqrt{x}+\sqrt{xy}-2\sqrt{y}\)

\(=\sqrt{x}\cdot\left(\sqrt{x}-2\right)+\sqrt{y}\left(\sqrt{x}-2\right)\)

\(=\left(\sqrt{x}-2\right)\left(\sqrt{x}+\sqrt{y}\right)\)

3: \(x-y-2\sqrt{x}+2\sqrt{y}\)

\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)-2\left(\sqrt{x}-\sqrt{y}\right)\)

\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}-2\right)\)

4: \(\sqrt{a^3b}+\sqrt{ab^3}+\sqrt{\left(a+b\right)^2}\)

\(=a\cdot\sqrt{ab}+b\cdot\sqrt{ab}+a+b\)

\(=\sqrt{ab}\left(a+b\right)+\left(a+b\right)=\left(a+b\right)\left(\sqrt{ab}+1\right)\)

5: \(\sqrt{xm}-\sqrt{yn}-\sqrt{xn}+\sqrt{ym}\)

\(=\left(\sqrt{xm}-\sqrt{xn}\right)+\left(\sqrt{ym}-\sqrt{yn}\right)\)

\(=\sqrt{x}\left(\sqrt{m}-\sqrt{n}\right)+\sqrt{y}\left(\sqrt{m}-\sqrt{n}\right)=\left(\sqrt{m}-\sqrt{n}\right)\left(\sqrt{x}+\sqrt{y}\right)\)

6: \(\sqrt{ab}+\sqrt{a}+\sqrt{b}+1\)

\(=\sqrt{a}\left(\sqrt{b}+1\right)+\left(\sqrt{b}+1\right)\)

\(=\left(\sqrt{b}+1\right)\left(\sqrt{a}+1\right)\)

7: \(ab+b\sqrt{a}+\sqrt{a}+1\)

\(=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)\)

\(=\left(\sqrt{a}+1\right)\cdot\left(b\sqrt{a}+1\right)\)

Áp dụng BĐT Cauchy cho cặp số dương \(\dfrac{1}{\left(z+x\right)};\dfrac{1}{\left(z+y\right)}\)

\(\dfrac{1}{\left(z+x\right)}+\dfrac{1}{\left(z+y\right)}\ge\dfrac{1}{2}.\dfrac{1}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\)

\(\Rightarrow\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\left(1\right)\)

Tương tự ta được

\(\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}\le\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}\left(2\right)\)

\(\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}\left(3\right)\)

\(\left(1\right)+\left(2\right)+\left(3\right)\) ta được :

\(P=\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}+\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}+\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}+\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}+\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\)

\(\Rightarrow P\le2\left(x+y+z\right)=2.3=6\)

\(\Rightarrow GTLN\left(P\right)=6\left(tạix=y=z=1\right)\)

Công thức Heron được áp dụng cho tất cả tam giác nên nó cũng được áp dụng cho tam giác tù hoặc vuông.

Nhớ tích cho mk nha 

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