?

ĐKXĐ : x \(\ge-1\)

\(x^3+\left(3x^2-4x-4\right)\sqrt{x+1}=0\)

<=> \(x^3+3x^2\sqrt{x+1}-4\left(x+1\right)\sqrt{x+1}=0\)

<=> \(x^3+3x^2\sqrt{x+1}-4\left(\sqrt{x+1}\right)^3=0\)

<=> \(\left(x^3-x^2\sqrt{x+1}\right)+4\left[x^2\sqrt{x+1}-\left(\sqrt{x+1}\right)^3\right]=0\)

\(\Leftrightarrow x^2\left(x-\sqrt{x+1}\right)+4\sqrt{x+1}\left[x^2-\left(\sqrt{x+1}\right)^2\right]=0\)

<=> \(x^2\left(x-\sqrt{x+1}\right)+4\sqrt{x+1}\left(x-\sqrt{x+1}\right)\left(x+\sqrt{x+1}\right)=0\)

<=> \(\left(x-\sqrt{x+1}\right)\left(x^2+4x\sqrt{x+1}+4x+4\right)=0\)

<=> \(\left(x-\sqrt{x+1}\right)\left(x+2\sqrt{x+1}\right)^2=0\)

<=> \(\left[{}\begin{matrix}x=\sqrt{x+1}\left(1\right)\\x=-2\sqrt{x+1}\left(2\right)\end{matrix}\right.\)

Giải (1) ta có \(x=\sqrt{x+1}\Leftrightarrow\left\{{}\begin{matrix}x^2=x+1\\x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{\sqrt{5}+1}{2}\\x=\dfrac{1-\sqrt{5}}{2}\left(\text{loại}\right)\end{matrix}\right.\\x\ge0\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{\sqrt{5}+1}{2}\)

Giải (2) ta có : \(x=-2\sqrt{x+1}\Leftrightarrow\left\{{}\begin{matrix}x^2-4x-4=0\\x\ge-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\pm\sqrt{8}+2\\x\ge-1\end{matrix}\right.\Leftrightarrow x=\sqrt{8}+2\)

\(x^3+\left(3x^2-4x-4\right)\sqrt{x+1}=0\left(đk:x\ge-1\right)\)

\(\Leftrightarrow x^3+3x^2\sqrt{x+1}-4\left(x+1\right)\sqrt{x+1}=0\)

\(\Leftrightarrow x^3+3x^2\sqrt{x+1}-4\sqrt{x+1}^3=0\left(1\right)\)

\(TH:x=-1\Rightarrow\left(1\right)\Leftrightarrow-1=0\left(ktm\right)\)

\(TH:x>-1\Rightarrow\left(1\right)\Leftrightarrow\left(\dfrac{x}{\sqrt{x+1}}\right)^3+3\left(\dfrac{x}{\sqrt{x+1}}\right)^2-4=0\)

\(đặt:\dfrac{x}{\sqrt{x+1}}=a\Rightarrow a^3+3a^2-4=0\Leftrightarrow\left(a+2\right)^2\left(a-1\right)=0\Leftrightarrow\left[{}\begin{matrix}a=1=\dfrac{x}{\sqrt{x+1}}\Leftrightarrow\sqrt{x+1}=x\left(2\right)\\a=-2=\dfrac{x}{\sqrt{x+1}}\Leftrightarrow2\sqrt{x+1}=-x\left(3\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2=x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=\dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow x=\dfrac{1+\sqrt{5}}{2}\)

\(\left(3\right)\Leftrightarrow\left\{{}\begin{matrix}-1< x\le0\\4\left(x+1\right)=x^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-1< x\le0\\\left[{}\begin{matrix}x=2+2\sqrt{2}\\x=2-2\sqrt{2}\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow x=2-2\sqrt{2}\)

Gọi số có 2 chữ số ban đầu là \(\overline{ab}\left(a\ne0\right)\)

Ta có \(a+b=9\)

Khi đổi chỗ 2 chữ số ta được số mới là \(\overline{ba}\)

Ta có: \(\overline{ab}-\overline{ba}=27\Rightarrow\left(10a+b\right)-\left(10b+a\right)=27\)

\(\Rightarrow9a-9b=27\Rightarrow a-b=3\)

Ta có hệ phương trình: \(\left\{{}\begin{matrix}a+b=9\\a-b=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=6\\b=3\end{matrix}\right.\)

Vậy số cần tìm là 63.

a,A=\(\left(2+\dfrac{2+\sqrt{3}}{\sqrt{3}+1}\right)\left(2-\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)\)

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

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

=\(\dfrac{3\sqrt{3}+4}{\sqrt{3}+1}\times\dfrac{3\sqrt{3}-5}{\sqrt{3}-1}\)

=\(\dfrac{\left(3\sqrt{3}+4\right)\left(3\sqrt{3}-5\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

=\(\dfrac{27-15\sqrt{3}+12\sqrt{3}-20}{3-1}\)

=\(\dfrac{7-3\sqrt{3}}{2}\)

b,B=\(\left(\dfrac{\sqrt{a}}{a-\sqrt{ab}}-\dfrac{\sqrt{a}}{\sqrt{ab}-b}\right)\left(a\sqrt{a}-b\sqrt{a}\right)\)

=\(\left(\dfrac{\sqrt{a}.\sqrt{b}-\sqrt{a}.\sqrt{a}}{\sqrt{ab}.\left(\sqrt{a}-\sqrt{b}\right)}\right).\sqrt{a}.\left(a-b\right)\)

=\(\left(\dfrac{\sqrt{ab}-a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\right).\sqrt{a}\left(a-b\right)\)

=\(\left(\dfrac{-\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\right).\sqrt{a}\left(a-b\right)\)

=\(\dfrac{-1}{\sqrt{b}}.\sqrt{a}\left(a-b\right)\)

=\(\dfrac{-\sqrt{a}\left(a-b\right)}{\sqrt{b}}\)

11. C góc đối cộng  lại = 180 độ

12. A

A + C = 180
2C + C = 180 => C = 60 => A = 120

Ta có \(A=\dfrac{2}{xy}+\dfrac{3}{x^2+y^2}=\dfrac{4}{2xy}+\dfrac{3}{x^2+y^2}=3\left(\dfrac{1}{2xy}+\dfrac{1}{x^2+y^2}\right)+\dfrac{1}{2xy}\)Lại có \(\dfrac{1}{2xy}+\dfrac{1}{x^2+y^2}\ge\dfrac{4}{x^2+2xy+y^2}=\dfrac{4}{\left(x+y\right)^2}=4\) (vì \(x+y=1\))

Và \(xy\le\dfrac{\left(x+y\right)^2}{4}=\dfrac{1}{4}\Leftrightarrow2xy\le\dfrac{1}{2}\Leftrightarrow\dfrac{1}{2xy}\ge2\)

Do đó \(A\ge3.4+2=14\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)

Vậy GTNN của A là 14 khi \(x=y=\dfrac{1}{2}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x+3\ge0\\3-2x\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-3\\x\le\dfrac{3}{2}\end{matrix}\right.\)

PT\(\Leftrightarrow x-1+4\left(\sqrt{x+3}-2\right)+2\left(\sqrt{3-2x}-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)+4.\dfrac{x+3-4}{\sqrt{x+3}+2}+2.\dfrac{3-2x-1}{\sqrt{3-2x}+1}=0\)

\(\Leftrightarrow\left(x-1\right)+4.\dfrac{x-1}{\sqrt{x+3}+2}-2.\dfrac{2\left(x-1\right)}{\sqrt{3-2x}+1}=0\)

\(\Leftrightarrow\left(x-1\right)\left[1+\dfrac{4}{\sqrt{x+3}+2}-\dfrac{4}{\sqrt{3-2x}+1}\right]=0\)

TH1:\(x-1=0\Leftrightarrow x=1\left(tmđk\right)\)

TH2:\(1+\dfrac{4}{\sqrt{x+3}+2}-\dfrac{4}{\sqrt{3-2x}+1}=0\)(VN)