\(tan45^o=\dfrac{CD-1,8}{10}\) (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 

Đề thiếu. Bạn xem lại đề.

2b:

A=[m+1;m+2]; B=(3-2m;5-2m)

Để A giao B =rỗng thì \(\left[\begin{array}{l}m+2\le3-2m\\ m+1\ge5-2m\end{array}\right.\Rightarrow\left[\begin{array}{l}3m\le1\\ 3m\ge4\end{array}\right.\Rightarrow\left[\begin{array}{l}m<=\frac13\\ m\ge\frac43\end{array}\right.\)

=>Để A giao B khác rỗng thì \(\frac13

Em là thần đồng cờ vua nhưng bài này thì chịu

❤

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