Ta có $\sqrt{x^2-7x+19}-\sqrt{x^2-7x+15}=2$

$=>2M=(\sqrt{x^2-7x+19}-\sqrt{x^2-7x+15})(\sqrt{x^2-7x+19}+\sqrt{x^2-7x+15})$

$=>2M=\sqrt{x^2-7x+19}^2-\sqrt{x^2-7x+15}^2$

$=>2M=(x^2-7x+19)-(x^2-7x+15)=4$

$=>M=2$

\(2.M=\left(x^2-7x+19\right)-\left(x^2-7x+15\right)=4\Rightarrow M=2\)

thảo có ghi lộn đề k

ta có:\(\frac{x-2\sqrt{x}+1}{x-\sqrt{x}+1}=\frac{1}{2}\)

\(\Rightarrow x-3\sqrt{x}+1=0\)

\(\Rightarrow\hept{\begin{cases}x+1=3\sqrt{x}\\x-3\sqrt{x}=-1\end{cases}}\)

lại có \(B=\frac{3x\sqrt{x}+10x+19}{x^2+7x+15}\)

\(=\frac{3x\sqrt{x}-9x+19x+19}{x^2-9x+16x+15}\)

\(=\frac{3\sqrt{x}\left(x-3\sqrt{x}\right)+19\left(x+1\right)}{\left(x+3\sqrt{x}\right)\left(x-3\sqrt{x}\right)+16x+15}\)

\(=\frac{-3\sqrt{x}+19\times3\sqrt{x}}{-1\times\left(x+3\sqrt{x}\right)+16x+15}\)

\(=\frac{57\sqrt{x}-3\sqrt{x}}{15x+15-3\sqrt{x}}\)

\(=\frac{54\sqrt{x}}{15\left(x+1\right)-3\sqrt{x}}\)

\(=\frac{54\sqrt{x}}{45\sqrt{x}-3\sqrt{x}}\)

\(=\frac{54\sqrt{x}}{42\sqrt{x}}=\frac{27}{21}\)

Ta có:

\(\sqrt{x^2-x+19}+\sqrt{7x^2+8x+13}+\sqrt{13x^2+17x+7}\)

\(=\sqrt{\frac{1}{4}\left(2x-1\right)^2+\frac{75}{4}}+\sqrt{\left(2x-1\right)^2+3\left(x+2\right)^2}+\sqrt{\frac{1}{4}\left(2x-1\right)^2+\frac{3}{4}\left(4x+3\right)^2}\)

\(\ge\sqrt{\frac{75}{4}}+\sqrt{3\left(x+2\right)^2}+\sqrt{\frac{3}{4}\left(4x+3\right)^2}\)

\(=\frac{5\sqrt{3}}{2}+\sqrt{3}\left(x+2\right)+\frac{\sqrt{3}\left(4x+3\right)}{2}=3\sqrt{3}\left(x+2\right)\)

Dấu = xảy ra khi ....

ĐK:\(3x^2-6x-6\ge0;\left(2-x\right)^5\ge0;x\le2\)

\(\Leftrightarrow\sqrt{3x^2-6x-6}-\sqrt{3}=3\sqrt{\left(2-x\right)^5}-27\sqrt{3}+\left(7x-19\right)\sqrt{2-x}+26\sqrt{3}\)

\(\Leftrightarrow\frac{3x^2-6x-9}{\sqrt{3x^2-6x-6}+\sqrt{3}}=3\left(\frac{\left(2-x\right)^5-243}{\sqrt{\left(2-x\right)^5}+9\sqrt{3}}\right)+\frac{\left(7x-19\right)^2\left(2-x\right)-2028}{\left(7x-19\right)\sqrt{2-x}-26\sqrt{3}}\)

\(\Leftrightarrow\left(x+1\right)\left[...\right]=0\)

Ta c/m đc [...] khác 0.

Vậy x=-1(TM)

ĐKXĐ: \(x\le1-\sqrt{3}\)

\(\Leftrightarrow\sqrt{3x^2-6x-6}=3\left(2-x\right)^2\sqrt{2-x}+\left(7x-19\right)\sqrt{2-x}\)

\(\Leftrightarrow\sqrt{3x^2-6x-6}=\left(3x^2-5x-7\right)\sqrt{2-x}\)

\(\Leftrightarrow\sqrt{3x^2-6x-6}=\left(3x^2-6x-6-\left(2-x\right)+1\right)\sqrt{2-x}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{3x^2-6x-6}=a\ge0\\\sqrt{2-x}=b\ge0\end{matrix}\right.\) ta được:

\(a=\left(a^2-b^2+1\right)b\Leftrightarrow b\left(a^2-b^2\right)+b-a=0\)

\(\Leftrightarrow b\left(a+b\right)\left(a-b\right)-\left(a-b\right)=0\)

\(\Leftrightarrow\left(ab+b^2-1\right)\left(a-b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\ab=1-b^2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x^2-6x-6}=\sqrt{2-x}\\\sqrt{\left(3x^2-6x-6\right)\left(2-x\right)}=1-\left(2-x\right)\end{matrix}\right.\)

TH1: \(3x^2-5x-8=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=\frac{8}{3}>1-\sqrt{3}\left(l\right)\end{matrix}\right.\)

TH2: \(\sqrt{\left(3x^2-6x-6\right)\left(2-x\right)}=x-1\)

Do \(x\le1-\sqrt{3}\Rightarrow x-1\le-\sqrt{3}\Rightarrow VP< 0\) và \(VT\ge0\Rightarrow ptvn\)

Vậy pt có nghiệm duy nhất \(x=-1\)