Anh chép nhầm đề bài ý cuối và ý bên trên rồi ạ.

\(\dfrac{1}{\sqrt{5}+\sqrt{3}}-\dfrac{1}{\sqrt{5}-\sqrt{3}}\\ =\dfrac{\sqrt{5}-\sqrt{3}}{5-3}-\dfrac{\sqrt{5}+\sqrt{3}}{5-3}\\ =\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}}{2}\\ =\dfrac{-2\sqrt{3}}{2}=-\sqrt{3}\\ ---\\ \dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}\\ =\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{\left|\sqrt{3}-1\right|}{\sqrt{2}\left(\sqrt{3}-1\right)}\\ =\dfrac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)

\(---\\ \sqrt{127-48\sqrt{7}}-\sqrt{127+48\sqrt{7}}\\ =\sqrt{\left(3\sqrt{7}\right)^2-2.3\sqrt{7}.8+8^2}-\sqrt{\left(3\sqrt{7}\right)^2+2.3\sqrt{7}.8+8^2}\\ =\sqrt{\left(3\sqrt{7}-8\right)^2}-\sqrt{\left(3\sqrt{7}+8\right)^2}\\ =\left|3\sqrt{7}-8\right|-\left|3\sqrt{7}+8\right|\\ =8-3\sqrt{7}-3\sqrt{7}-8\\ =-6\sqrt{7}\\ ---\\ \dfrac{\sqrt{10}+\sqrt{6}}{2\sqrt{5}+\sqrt{12}}=\dfrac{\sqrt{10}+\sqrt{6}}{2\sqrt{5}+2\sqrt{3}}\\ =\dfrac{\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)}{2\left(\sqrt{5}+\sqrt{3}\right)}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)

#$\mathtt{Toru}$

ĐKXĐ: \(x\in\mathbb{R}\)

\(\dfrac{1}{\left(x^2+x+1\right)^2}+\dfrac{1}{\left(x^2+x+2\right)^2}=\dfrac{13}{36}\) (1)

Đặt \(x^2+x+1=t;\left(t>0\right)\), thì (1) trở thành:

\(\dfrac{1}{t^2}+\dfrac{1}{\left(t+1\right)^2}=\dfrac{13}{36}\\ \Leftrightarrow\dfrac{36\left(t+1\right)^2}{36t^2\left(t+1\right)^2}+\dfrac{36t^2}{36t^2\left(t+1\right)^2}=\dfrac{13t^2\left(t+1\right)^2}{36t^2\left(t+1\right)^2}\\ \Rightarrow36\left(t^2+2t+1\right)+36t^2=13t^2\left(t^2+2t+1\right)\\ \Leftrightarrow13t^4+26t^3+13t^2=72t^2+72t+36\\ \Leftrightarrow13t^4+26t^3-59t^2-72t-36=0\\ \Leftrightarrow13t^4-26t^3+52t^3-104t^2+45t^2-90t+18t-36=0\\ \Leftrightarrow13t^3\left(t-2\right)+52t^2\left(t-2\right)+45t\left(t-2\right)+18\left(t-2\right)\\=0\\ \Leftrightarrow\left(t-2\right)\left(13t^3+52t^2+45t+18\right)=0\)

\(\Leftrightarrow\left(t-2\right)\left(13t^3+39t^2+13t^2+39t+6t+18\right)=0\\ \Leftrightarrow\left(t-2\right)\left[13t^2\left(t+3\right)+13t\left(t+3\right)+6\left(t+3\right)\right]=0\\ \Leftrightarrow\left(t-2\right)\left(t+3\right)\left(13t^2+13t+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t-2=0\\t+3=0\\13t^2+13t+6=0\end{matrix}\right.\)

+, Với `t-2=0=>x^2+x+1-2=0`

`<=>x^2+x-1=0`

`<=>[x^2+2.x. 1/2+(1/2)^2]-1/4-1=0`

`<=>(x+1/2)^2-5/4=0`

`<=>(x+{1-\sqrt5}/{2})(x+{1+\sqrt5}/{2})=0`

`<=>[(x={-1+\sqrt5}/{2}),(x={-1-\sqrt5}/{2}):}`

+, Với `t+3=0<=>t=-3` (ktm)

+, Với `13t^2+13t+6=0`

`<=>13(t^2+t)+6=0`

`<=>13[t^2+2.t. 1/2+(1/2)^2]-13/4+6=0`

`<=>13(t+1/2)^2+21/4=0` (vô lí)

Vậy phương trình đã cho có tập nghiệm là: $S=\{\dfrac{-1+\sqrt5}{2};\dfrac{-1-\sqrt5}{2}\}$.

#$\mathtt{Toru}$

\(\left|1-\left|x+2\right|\right|=2x-1\) (1)

+, Với \(x\ge-2\Rightarrow x+2\ge0\), (1) trở thành:

\(\left|1-\left(x+2\right)\right|=2x-1\\ \Rightarrow\left|-x-1\right|=2x-1\\ \Rightarrow\left|x+1\right|=2x-1\\ \Rightarrow\left[{}\begin{matrix}x+1=2x-1\left(dk:x\ge-1\right)\\-x-1=2x-1\left(dk:-2\le x< -1\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-x=1+1\\x+2x=-1+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=0\left(ktm\right)\end{matrix}\right.\)

+, Với \(x< -2\Rightarrow x+2< 0\), (1) trở thành:

\(\left|1-\left(-x-2\right)\right|=2x-1\\ \Rightarrow\left|x+3\right|=2x-1\\ \Rightarrow\left[{}\begin{matrix}x+3=2x-1\left(dk:-3\le x< -2\right)\\-x-3=2x-1\left(dk:x< -3\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-x=3+1\\x+2x=-3+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\left(ktm\right)\\x=\dfrac{-2}{3}\left(ktm\right)\end{matrix}\right.\)

Vậy \(x=2\) là giá trị cần tìm.

#$\mathtt{Toru}$

\(M=-x^2+6x-11\\ =-\left(x^2-6x+9\right)+9-11\\ =-\left(x-3\right)^2-2\)

Ta thấy: \(\left(x-3\right)^2\ge0\forall x\)

\(\Rightarrow-\left(x-3\right)^2\le0\forall x\\ \Rightarrow-\left(x-3\right)^2-2\le-2\forall x\\ \Rightarrow M\le-2\forall x\)

Dấu "=" xảy ra khi: \(x-3=0\Leftrightarrow x=3\)

Vậy \(M_{max}=-2\Leftrightarrow x=3\).

Bạn xem lại bài từ dòng thứ 2 nhé, đâu thể suy ra được như vậy?

Bạn xem lại dòng thứ 4 nhé, vì 1-2x+1=2-2x.