\(\hept{\begin{cases}x^2+y^2+xy-x=5\\xy-y=1\end{cases}}\)

\(\left(x+y\right)^2-\left(x+y\right)=6\)

đặt t = x+y 

\(t^2-t=6\)

\(t^2-t-6=0\)

\(t^2-3t+2t-6=0\)

\(t\left(t-3\right)+2\left(t-3\right)=0\)

\(\orbr{\begin{cases}t-3=0\\t+2=0\end{cases}\orbr{\begin{cases}t=3\\t=-2\end{cases}}}\)

\(TH1:x+y=3< =>x=3-y\)

\(y\left(3-y-1\right)=1\)

\(y\left(2-y\right)=1\)

\(2y-y^2=1\)

\(y^2-2y+1=0\)

\(\left(y-1\right)^2=0\)

\(y=1< =>x=3-1=2\)

\(TH2:x+y=-2,=>x=-2-y\)

\(y\left(-2-y-1\right)=1\)

\(y\left(-3-y\right)=1\)

\(-3y-y^2=1\)

\(y^2+3y+1=0\)

\(\sqrt{\Delta}=\sqrt{5}\)

\(x_1=\frac{-3+\sqrt{5}}{2}< =>x=-\frac{1+\sqrt{5}}{2}\)

\(x_2=\frac{-3-\sqrt{5}}{2}< =>x=\frac{-1+\sqrt{5}}{2}\)

vậy hệ pt có 3 tập nghiệm là ..........

Trong các hàm số trên, các hàm số bậc nhất là: 

\(y=25\left(x+5\right),y=\frac{10x+7}{9}\).

Giải nhanh giùm mik vs , mik cần gấp

P = \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)

P = \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)

P = \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)

P = \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)

P = \(\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}=\sqrt{4+\sqrt{25}}=\sqrt{9}=3\)

Q = \(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)

Q = \(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{2+2\sqrt{3}+\sqrt{\left(16-8\sqrt{2}+2\right)}}}}\)

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

Q = \(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}}\)

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

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

Q = \(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)

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

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

D = \(\sqrt{2+\sqrt{3}+\sqrt{4-2\sqrt{3}-\sqrt{21-12\sqrt{3}}}}\)

D = \(\sqrt{2+\sqrt{3}+\sqrt{4-2\sqrt{3}-\sqrt{\left(2\sqrt{3}-3\right)^2}}}\)

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

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

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

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

E² = \(2-\sqrt{2\sqrt{5}-2}+2+\sqrt{2\sqrt{5}-2}-2\sqrt{\left(2-\sqrt{2\sqrt{5}-2}\right)\left(2+\sqrt{2\sqrt{5}-2}\right)}\)

E² = \(4-2\sqrt{\left(4-2\sqrt{5}+2\right)}=4-2\sqrt{\left(\sqrt{5}-1\right)}^2=4-2\sqrt{5}+2=\left(\sqrt{5}-1\right)^2\)

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

\(7:a,\sqrt{2-x}=3\)

\(\left|2-x\right|=3^2=9\)

\(\orbr{\begin{cases}2-x=9\\2-x=-9\end{cases}\orbr{\begin{cases}x=-7\left(KTM\right)\\x=11\left(TM\right)\end{cases}}}\)

\(b,\sqrt{4-4x+x^2}=3\)

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

\(\left|2-x\right|=3\)

\(\orbr{\begin{cases}2-x=3\\2-x=-3\end{cases}\orbr{\begin{cases}x=-1\left(TM\right)\\x=5\left(TM\right)\end{cases}}}\)

\(c,\sqrt{4+x^2}+x=3\)

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

\(4+x^2=\left(3-x\right)^2\)

\(4+x^2=9-6x+x^2\)

\(x=\frac{5}{6}\left(TM\right)\)

\(d,\frac{1}{2}\sqrt{16x-32}-2\sqrt{4x-8}+\sqrt{9x-18}=5\)

\(2\sqrt{x-2}-4\sqrt{x-2}+3\sqrt{x-2}=5\)

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

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

\(\left|x-2\right|=25\)

\(\orbr{\begin{cases}x-2=25\\x-2=-25\end{cases}\orbr{\begin{cases}x=27\left(TM\right)\\x=-23\left(KTM\right)\end{cases}}}\)

thank

\(\sqrt{1-8b+16b^2}\)

\(\sqrt{\left(1-4b\right)^2}\)

vậy với mọi giá trị thực của x thì biều thức đxđ

\(\sqrt{\frac{3a-4}{-5}}\sqrt{2x^2}\sqrt{2x^2+1}\)

\(\hept{\begin{cases}2x^2+1\ge0\\2x^2\ge0\\3x-4\le0\end{cases}}\)

\(\hept{\begin{cases}x\\x\ge0\\x\le\frac{4}{3}\end{cases}}\)đoạn x là tất cả các số thực

\(< =>0\le x\le\frac{4}{3}\)

\(\sqrt{\frac{-5}{x^2+1}}\)

\(x^2+1\ne0;x^2+1\le0< =>x^2+1< 0\left(1\right)\)

\(x^2+1\ge1\left(2\right)\)

\(\left(1\right);\left(2\right)< =>\)ko có giá trị x thỏa mãn để pt đc xác định