bài III: 

1: ĐKXĐ: \(\left\{{}\begin{matrix}x\ne1\\y\ne-2\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{3x}{x-1}-\dfrac{2}{y+2}=4\\\dfrac{2x}{x-1}+\dfrac{1}{y+2}=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{3x-3+3}{x-1}-\dfrac{2}{y+2}=4\\\dfrac{2x-2+2}{x-1}+\dfrac{1}{y+2}=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{3}{x-1}-\dfrac{2}{y+2}=4-3=1\\\dfrac{2}{x-1}+\dfrac{1}{y+2}=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{3}{x-1}-\dfrac{2}{y+2}=1\\\dfrac{4}{x-1}+\dfrac{2}{y+2}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{7}{x-1}=7\\\dfrac{2}{x-1}+\dfrac{1}{y+2}=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x-1=1\\y+2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\left(nhận\right)\)

2:

a: Phương trình hoành độ giao điểm là:

\(x^2=3x+m^2-1\)

=>\(x^2-3x-m^2+1=0\)

\(\text{Δ}=\left(-3\right)^2-4\cdot1\cdot\left(-m^2+1\right)=9+4m^2-4=4m^2+5>0\forall m\)

=>(P) luôn cắt (d) tại hai điểm phân biệt

b: Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=3\\x_1x_2=\dfrac{c}{a}=-m^2+1\end{matrix}\right.\)

\(\left(x_1+1\right)\left(x_2+1\right)=1\)

=>\(x_1x_2+\left(x_1+x_2\right)+1=1\)

=>\(-m^2+1+3=0\)

=>\(m^2=4\)

=>\(\left[{}\begin{matrix}m=2\\m=-2\end{matrix}\right.\)

a: Sửa đề: \(x^2+\dfrac{1}{x^2}-2\left(x+\dfrac{1}{x}\right)-6=0\)(1)

ĐKXĐ: x<>0

(1)=>\(\left(x+\dfrac{1}{x}\right)^2-2-2\left(x+\dfrac{1}{x}\right)-6=0\)

=>\(\left(x+\dfrac{1}{x}\right)^2-2\left(x+\dfrac{1}{x}\right)-8=0\)

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

=>\(\dfrac{x^2+1-4x}{x}\cdot\dfrac{x^2+1+2x}{x}=0\)

=>\(\left(x^2-4x+1\right)\left(x^2+2x+1\right)=0\)

=>\(\left[{}\begin{matrix}x^2-4x+1=0\\x^2+2x+1=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-1\left(nhận\right)\\x=2\pm\sqrt{3}\left(nhận\right)\end{matrix}\right.\)

b: \(x^2+2x=\sqrt{x^2+2x+1}+5\)

=>\(x^2+2x+1=\left|x+1\right|+6\)

=>\(\left(\left|x+1\right|\right)^2-\left|x+1\right|-6=0\)

=>\(\left(\left|x+1\right|-3\right)\left(\left|x+1\right|+2\right)=0\)

=>\(\left|x+1\right|-3=0\)

=>|x+1|=3

=>\(\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

a: \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)

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

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

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

b: \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)

\(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)

\(=\sqrt{5}-\sqrt{2}-\left(\sqrt{5}+\sqrt{2}\right)=-2\sqrt{2}\)

c: \(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)

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

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

d: \(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)

\(=\sqrt{\left(2\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\)

\(=2\sqrt{5}+2+\sqrt{5}-2=3\sqrt{5}\)

e: \(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)

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

\(=3-2\sqrt{2}+2\sqrt{2}+1=4\)
f: \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)

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

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

g: \(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)

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

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

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

h: \(\sqrt{21-12\sqrt{3}}-\sqrt{3}\)

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

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

i: \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)

\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)

j: \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)

\(=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)

\(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)

\(=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)

\(=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\)

\(=\sqrt{25+2\cdot5\cdot3\sqrt{2}+18}=\sqrt{\left(5+3\sqrt{2}\right)^2}\)

\(=5+3\sqrt{2}\)

k: \(\sqrt{5-\sqrt{13+4\sqrt{3}}}+\sqrt{3+\sqrt{13+4\sqrt{3}}}\)

\(=\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}+\sqrt{3+\sqrt{\left(2\sqrt{3}+1\right)^2}}\)

\(=\sqrt{5-2\sqrt{3}-1}+\sqrt{3+2\sqrt{3}+1}\)

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

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

l: \(\sqrt{1+\sqrt{3+\sqrt{13+4\sqrt{3}}}}+\sqrt{1-\sqrt{3-\sqrt{13-4\sqrt{3}}}}\)

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

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

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

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

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

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

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{3}+1+\sqrt{3}-1\right)=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

a: Xét tứ giác BCEF có \(\widehat{BFC}=\widehat{BEC}=90^0\)

nên BCEF là tứ giác nội tiếp

\(\sqrt[3]{26-15\sqrt[]{3}}=\sqrt[3]{8-3.4.\sqrt[]{3}+3.2.3-3\sqrt[]{3}}\)

\(=\sqrt[3]{2^3-3.2^2.\sqrt[]{3}+3.2.\left(\sqrt[]{3}\right)^2-\left(\sqrt[]{3}\right)^3}\)

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

\(\Rightarrow\left\{{}\begin{matrix}a=2\\b=-1\end{matrix}\right.\) \(\Rightarrow a^2-b^2=3\)

\(P=\dfrac{2x+3}{\sqrt{x}-2}=\dfrac{2\left(x-4\right)+11}{\sqrt{x}-2}=\dfrac{2\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+11}{\sqrt{x}-2}\)

\(=2\left(\sqrt{x}+2\right)+\dfrac{11}{\sqrt{x}-2}=2\left(\sqrt{x}-2\right)+\dfrac{11}{\sqrt{x}-2}+8\)

\(\ge2\sqrt{\dfrac{2\left(\sqrt{x}-2\right).11}{\sqrt{x}-2}}+8=2\sqrt{22}+8\)

ĐKXĐ: \(x\ge2\)

\(\sqrt{4x-8}-\sqrt{x-2}=2\)

\(\Leftrightarrow\sqrt{4\left(x-2\right)}-\sqrt{x-2}=2\)

\(\Leftrightarrow2\sqrt{x-2}-\sqrt{x-2}=2\)

\(\Leftrightarrow\sqrt{x-2}=2\)

\(\Leftrightarrow x-2=4\)

\(\Leftrightarrow x=6\) (thỏa mãn)

Chà, phải nói rằng pt này không "đẹp" (nghĩa là ý tưởng lời giải cho nó rất tối, đơn giản là dùng máy tính SOLVE nghiệm sau đó liên hợp và chứng minh nghiệm đó là duy nhất). Những phương trình thuần túy trâu bò trong lời giải thế này đi thi gần như ko bao giờ xuất hiện nên em cũng ko cần quan tâm nhiều đến nó đâu.

Dạ em cảm ơn ạ nhưng mà cho em hỏi xíu cách họ biến đổi chỗ này với