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

❤

?

\(\left\{{}\begin{matrix}A\subset X\\X\subset B\end{matrix}\right.\) 

\(\Rightarrow\left\{{}\begin{matrix}X=\left\{1;2;3;4\right\}\\X=\left\{1;2;3;4;5\right\}\\X=\left\{1;2;3;4;5;6\right\}\\X=\left\{1;2;3;4;5;6;7\right\}\end{matrix}\right.\)

x ϵ {1;2;3;4}

x ϵ {1;2;3;4;5}

x ϵ {1;2;3;4;5;6}

x ϵ {1;2;3;4;5;6;7}