ĐK : \(x\ne0\)

Ta có \(x^4+2x^3y+x^2.y^2=7x+9\)

\(\Leftrightarrow x^2.\left(x+y\right)^2=7x+9\)

\(\Rightarrow x\left(x+y\right)=\sqrt{7x+9}\left(x\ge-\dfrac{9}{7}\right)\)(1)

Lại có \(x.\left(y-x+1\right)=3\Leftrightarrow x.\left(x+y\right)=2x^2-x+3\) (2) 

Thay (2) vào (1) ta được \(2x^2-x+3=\sqrt{7x+9}\)

\(\Leftrightarrow2x^2-x-1=\sqrt{7x+9}-4\)

\(\Leftrightarrow\left(x-1\right).\left(2x+1\right)=\dfrac{7.\left(x-1\right)}{\sqrt{7x+9}+4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\2x+1=\dfrac{7}{\sqrt{7x+9}+4}\end{matrix}\right.\)

Với \(2x+1=\dfrac{7}{\sqrt{7x+9}+4}\) (*)

\(\Leftrightarrow2x=\dfrac{3-\sqrt{7x+9}}{\sqrt{7x+9}+4}\)

\(\Leftrightarrow2x+\dfrac{7x}{\left(\sqrt{7x+9}+4\right).\left(\sqrt{7x+9}+3\right)}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(\text{loại}\right)\\2+\dfrac{7}{\left(\sqrt{7x+9}+4\right).\left(\sqrt{7x+9}+3\right)}=0\left(3\right)\end{matrix}\right.\)

Dễ thấy (3) vô nghiệm nên phương trình (*) vô nghiệm

Với x = 1 => y = 3 

Tập nghiệm (x;y) = (1;3)

√4−√7−√4+√7+√7=√2(√4−√7−√4+√7+√7)√2=√8−2√7−√8+2√7+√14√2=√7−2√7+1−√7+2√7+1+√14√2=√(√7−1)2−√(√7+1)2+√14√2=∣∣√7−1∣∣−∣∣√7+1∣∣+√14√2=√7−1−√7−1+√14√2=√14−2√2=√2(√7−√2)√2=√7−√2

Lời giải:
\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}=\sqrt{\frac{8-2\sqrt{7}}{2}}-\sqrt{\frac{8+2\sqrt{7}}{2}}=\sqrt{\frac{(\sqrt{7}-1)^2}{2}}-\sqrt{\frac{(\sqrt{7}+1)^2}{2}}\)

\(=\frac{|\sqrt{7}-1|}{\sqrt{2}}-\frac{|\sqrt{7}+1|}{\sqrt{2}}=\frac{\sqrt{7}-1-(\sqrt{7}+1)}{\sqrt{2}}=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)