BÀI 1 \(\left(\frac{a+\sqrt{a}}{\sqrt{a}+1}+1\right).\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}-1\right)\)ĐKXĐ(a>0,a khác 1)
RÚT GỌN HỘ MK NHA!!!Giair chi tiết giúp mk vs
BÀI 2 Cho hpt\(\hept{\begin{cases}x+my=1\\x+2y=3\end{cases}}\)Tìm m để hpt có nghiệm (x,y) thỏa mãn x-y=1
\(\left(\frac{a+\sqrt{a}}{\sqrt{a}+1}+\frac{\sqrt{a}+1}{\sqrt{a}+1}\right)\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}-1}\right)\)
\(\left(\frac{a+\sqrt{a}+\sqrt{a}+1}{\sqrt{a}+1}\right)\left(\frac{a-\sqrt{a}-\sqrt{a}+1}{\sqrt{a}-1}\right)\)
\(\left(\frac{a+2\sqrt{a}+1}{\sqrt{a}+1}\right)\left(\frac{a-2\sqrt{a}+1}{\sqrt{a}-1}\right)\)
\(\frac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}+1}\times\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}-1}\)
\(\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)\)
\(a-1\)
\(\hept{\begin{cases}x+my=1\left(1\right)\\x+2y=3\left(2\right)\end{cases}}\)
Từ (1) ta có :
\(x+my=1\)
\(x=1-my\)
Từ (2) ta có :
\(x+2y=3\)
\(1-my+2y=3\)
\(2y-my=2\)
\(y\left(2-m\right)=2\)
\(y=\frac{2}{2-m}\)
Mà \(x-y=1\)
\(1-my-\frac{2}{2-m}=1\)
\(1-\frac{2m}{2-m}-\frac{2}{2-m}=1\)
\(\frac{2m}{2-m}+\frac{2}{2-m}-1=-1\)
\(\frac{2m+2}{2-m}=0\)
\(2m+2=0\)
\(m=-1\)