a, \(M=\sqrt{2}-\sqrt{6-2\sqrt{5}}=\sqrt{2}-\left(\sqrt{5}-1\right)=\sqrt{2}-\sqrt{5}+1\)

b, \(N=-\sqrt{2}-\sqrt{5}\left(\sqrt{5}-\sqrt{2}\right)=-\sqrt{2}-5+\sqrt{10}\)

c, \(P=\sqrt{\left(2\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{3}\right)^2}\)

\(=2\sqrt{2}+\sqrt{3}+2\sqrt{2}-\sqrt{3}=4\sqrt{2}\)

- Đặt \(\left\{{}\begin{matrix}\sqrt{4x^2-20x+28}=a>0\\3x^2-15x+20=b>0\end{matrix}\right.\)

- Khi đó, ta có hệ:

\(\left\{{}\begin{matrix}a=b\\a^2-\dfrac{4}{3}b=\dfrac{4}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b\\a^2-\dfrac{4}{3}a=\dfrac{4}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b\\a^2-\dfrac{4}{3}a-\dfrac{4}{3}=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b\\\left(a^2-\dfrac{4}{3}a+\dfrac{4}{9}\right)-\dfrac{16}{9}=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b\\\left(a-\dfrac{2}{3}\right)^2-\dfrac{16}{9}=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b\\\left(a-\dfrac{2}{3}-\dfrac{4}{3}\right)\left(a-\dfrac{2}{3}+\dfrac{4}{3}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b\\\left[{}\begin{matrix}a=2\\a=-\dfrac{2}{3}\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a=b=2\left(nhận\right)\\a=b=-\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{4x^2-20x+28}=2\\3x^2-15x+20=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x^2-20x+24=0\\3x^2-15x+18=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4\left(x-2\right)\left(x-3\right)=0\\3\left(x-2\right)\left(x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

- Vậy \(S=\left\{2;3\right\}\)

\(2\sqrt{x^2-5x+7}=3x^2-15x+20\)

đk x^2 - 5x + 7 > 0 

Đặt \(\sqrt{x^2-5x+7}\) = t 

\(2t=3t^2-1\Leftrightarrow3t^2-2t-1=0\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-\dfrac{1}{3}\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow x^2-5x+6=0\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)(tm) 

áp dụng bđt Cosi cho các số a, b, c không âm, ta có:

\(\left\{{}\begin{matrix}a+b\ge2\sqrt{ab}\left(1\right)\\a+c\ge2\sqrt{ac}\left(2\right)\\b+c\ge2\sqrt{bc}\left(3\right)\end{matrix}\right.\)

lấy (1) + (2) + (3) vế theo vế, ta được:

\(2\left(a+b+c\right)\ge2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)\)

\(\Leftrightarrow a+b+c\ge\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\) (đpcm)

\(\sqrt{\dfrac{9}{7-4\sqrt{3}}}-\sqrt{\dfrac{4}{7+4\sqrt{3}}}\)

\(=\sqrt{\dfrac{9\left(7+4\sqrt{3}\right)}{7^2-\left(4\sqrt{3}\right)^2}}-\sqrt{\dfrac{4\left(7-4\sqrt{3}\right)}{7^2-\left(4\sqrt{3}\right)^2}}\)

\(=\dfrac{3\sqrt{7+4\sqrt{3}}-2\sqrt{7-4\sqrt{3}}}{\sqrt{49-48}}\)

\(=3\sqrt{4+2.2.\sqrt{3}+3}-2\sqrt{4-2.2.\sqrt{3}+3}\)

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

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

\(=6+3\sqrt{3}-2+2\sqrt{3}\)

\(=4+5\sqrt{3}\)

Điều kiện : \(x\le3\)

Đặt \(3-x=t\left(t\ge0\right)\Rightarrow5-x=2+t\)

Khi này ta có phương trình :

\(\sqrt{t}+\sqrt{t+2}< 2\)

\(\Leftrightarrow t+2\sqrt{t\left(t+2\right)}+t+2< 2\)

\(\Leftrightarrow2t+2\sqrt{t\left(t+2\right)}< 0\)

\(\Leftrightarrow t+\sqrt{t\left(t+2\right)}< 0\) ( vô lí do \(t\ge0\) )

Vậy bất phương trình đã cho vô nghiệm

Lời giải:

Thay vì dấu < thì dấu $\leq$ đúng hơn 

CMR: $(ax+by)^2\leq (a^2+b^2)(x^2+y^2)$

$\Leftrightarrow (a^2+b^2)(x^2+y^2)-(ax+by)^2\geq 0$

$\Leftrightarrow (a^2x^2+a^2y^2+b^2x^2+b^2y^2)-(a^2x^2+b^2y^2+2axby)\geq 0$

$\Leftrightarrow a^2y^2+b^2x^2-2axby\geq 0$

$\Leftrightarrow (ay-bx)^2\geq 0$ (luôn đúng) 

Vậy ta có đpcm.

Dấu "=" xảy ra khi $ay=bx$

bạn thấy đó:\(A^2-B=0\)

nên : Nên theo cái dấu mà bạn viết thì VP=0

bạn thấy vế trái có thể =0 ko?