\(\sqrt{2}\cdot\left(\sqrt{2-\sqrt{3}}+2\right)-\sqrt{3}\)

\(=\sqrt{4-2\sqrt{3}}+\sqrt{4}-\sqrt{3}\\ =\sqrt{3-2\sqrt{3}+1}+2-\sqrt{3}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}+2-\sqrt{3}\\ \)

\(=\left|\sqrt{3}-1\right|+2-\sqrt{3}\) 

\(=\sqrt{3}-1+2-\sqrt{3}\) ( vì \(\sqrt{3}-1>0\) )

`=1`

a)

`A=2^2 *5^2 -3^2 +30`

`=4*25-9+30`

`=100-9+30`

`=91+30`

`=121`

`=11^2`

b)

`B=5*3^2 +4*3^2`

`=5*9+4*9`

`=45+36`

`=81`

`=9^2`

c)

`C=5*4^3 +2^4 *5+41`

`=5*64+16*5+41`

`=320+80+41`

`=441`

`=21^2`

d)

`D=5^3 +6^3 +59`

`=125+216+59`

`=400`

`=20^2`

a)

\(\dfrac{8\cdot5-8\cdot2}{16}=\dfrac{8\left(5-2\right)}{16}=\dfrac{3}{2}\)

b)

\(\dfrac{17\cdot5-17}{3-20}=\dfrac{17\left(5-1\right)}{-17}=\dfrac{4}{-1}=-4\)

\(< =>\dfrac{4x+6}{x+3}\le0\\ \)

\(\\ < =>\left[{}\begin{matrix}\left\{{}\begin{matrix}4x+6\le0\\x+3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}4x+6\ge0\\x+3\le0\end{matrix}\right.\end{matrix}\right.\\\)

\(< =>\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le-\dfrac{3}{2}\\x\ge-3\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-\dfrac{3}{2}\\x\ge-3\end{matrix}\right.\end{matrix}\right.\)

`<=>-3/2 <= x <= 3`

sao nó cứ lỗi công thức thế nhỉ?

Phần còn lại nó ở đây nhé

\(< =>\dfrac{4x+6}{x+3}\le0\\ < =>\left[{}\begin{matrix}\left\{{}\begin{matrix}4x+6\le0\\x+3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}4x+6\ge0\\x+3\le0\end{matrix}\right.\end{matrix}\right.\\ < =>\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le-\dfrac{3}{2}\\x\ge-3\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge-\dfrac{3}{2}\\x\le-3\end{matrix}\right.\left(voli\right)}\end{matrix}\right.\\ < =>-3\le x\le-\dfrac{3}{2}\)

a)

\(B=\left(\dfrac{3}{x^2+6x+9}-\dfrac{1}{x+3}\right):\left(\dfrac{3}{x^2-9}+\dfrac{1}{3-x}\right)\\ =\left(\dfrac{3}{\left(x+3\right)^2}-\dfrac{x+3}{\left(x+3\right)^2}\right):\left(\dfrac{3}{\left(x-3\right)\left(x+3\right)}-\dfrac{1}{x-3}\right)\)

\(=\dfrac{3-x-3}{\left(x+3\right)^2}:\left(\dfrac{3}{\left(x-3\right)\left(x+3\right)}-\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\right)\)

\(=\dfrac{-x}{\left(x+3\right)^2}:\dfrac{3-x-3}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{-x}{\left(x+3\right)^2}\cdot\dfrac{\left(x-3\right)\left(x+3\right)}{-x}\\ =\dfrac{x-3}{x+3}\)

b)

`x^2 -2x=0` 

`<=>x(x-2)=0`

`<=>x=0` hoặc `x-2=0`

`<=>x=0(ktm)` hoặc `x=2(tm)`

Với x=2 thì

\(\dfrac{2-3}{2+3}=\dfrac{-1}{5}\)

c)

Dể `B<=-3`

`=>(x-3)/(x+3) <=-3`

`<=>(x-3)/(x+3)+3 <=0`

`<=>(x-3)/(x+3)+(3(x+3))/(x+3) <=0`

`<=>(x-3+3x+9)/(x+3) <=0`

các số mà ta có thể viết được: `99;92;97;93;29;22;27;23;79;72;77;73;39;32;37;33`

`=>` Ta viết được 16 chữ số

hinh 3