Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
=\(\frac{\left(e^2-2e.m+m^2-e^2-2e.m-m^2\right).\left(y^2-2y+1-y^2-2y-1\right)}{a.16.n.h}.\frac{e}{u^{-1}}\)
= \(\frac{-4e.m.\left(-4y\right)}{a.16.n.h}.\frac{e}{u^{-1}}\)
=\(\frac{16e.m.y}{16a.n.h}.\frac{e}{\frac{1}{4}}\)
=\(\frac{e.m.y}{a.n.h}.e.u=\frac{e.m.y.e.u}{a.n.h}\)
ĐKXĐ: \(y\ge0;y\ne4;9\)
\(A=\left(\frac{8\sqrt{y}-4y+8y}{\left(2+\sqrt{y}\right)\left(2-\sqrt{y}\right)}\right):\left(\frac{\sqrt{y}-1}{\sqrt{y}\left(\sqrt{y}-2\right)}-\frac{2\left(\sqrt{y}-2\right)}{\sqrt{y}\left(\sqrt{y}-2\right)}\right)\)
\(=\left(\frac{4\sqrt{y}\left(2+\sqrt{y}\right)}{\left(2+\sqrt{y}\right)\left(2-\sqrt{y}\right)}\right):\left(\frac{-\sqrt{y}+3}{\sqrt{y}\left(\sqrt{y}-2\right)}\right)\)
\(=\left(\frac{4\sqrt{y}}{2-\sqrt{y}}\right):\left(\frac{\sqrt{y}-3}{\sqrt{y}\left(2-\sqrt{y}\right)}\right)\)
\(=\frac{4\sqrt{y}}{\left(2-\sqrt{y}\right)}.\frac{\sqrt{y}\left(2-\sqrt{y}\right)}{\left(\sqrt{y}-3\right)}=\frac{4y}{\sqrt{y}-3}\)
\(A=-2\Leftrightarrow\frac{4y}{\sqrt{y}-3}=-2\)
\(\Rightarrow2y=-\sqrt{y}+3\Rightarrow2y+\sqrt{y}-3=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{y}=1\\\sqrt{y}=-\frac{3}{2}< 0\left(l\right)\end{matrix}\right.\) \(\Rightarrow y=1\)
Câu 3:
\(\left\{{}\begin{matrix}mx+4y=9\\mx+m^2y=8m\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}mx+4y=9\\\left(m^2-4\right)y=8m-9\end{matrix}\right.\)
Để hpt đã cho có nghiệm \(\Leftrightarrow m\ne\pm2\)
Khi đó ta có: \(\left\{{}\begin{matrix}y=\frac{8m-9}{m^2-4}\\x=8-my=8-\frac{8m^2-9m}{m^2-4}=\frac{9m-32}{m^2-4}\end{matrix}\right.\)
\(2x+y+\frac{38}{m^2-4}=3\)
\(\Leftrightarrow\frac{18m-64}{m^2-4}+\frac{8m-9}{m^2-4}+\frac{38}{m^2-4}=3\)
\(\Leftrightarrow26m-35=3m^2-12\)
\(\Leftrightarrow3m^2-26m+23=0\Rightarrow\left[{}\begin{matrix}m=1\\m=\frac{23}{3}\end{matrix}\right.\)
Câu 4:
\(\left\{{}\begin{matrix}m^2x-my=2m^2\\4x-my=m+6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(m^2-4\right)x=2m^2-m-6\\4x-my=m+6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(m-2\right)\left(m+2\right)x=\left(m-2\right)\left(2m+3\right)\\4x-my=m+6\end{matrix}\right.\)
- Với \(m=-2\) hệ vô nghiệm
- Với \(m=2\) hệ có vô số nghiệm thỏa mãn \(2x-y=4\)
- Với \(m\ne\pm2\) hệ có nghiệm duy nhất:
\(\left\{{}\begin{matrix}x=\frac{2m+3}{m+2}\\y=mx-2m=\frac{2m^2+3m-2m^2-4m}{m+2}=\frac{-m}{m+2}\end{matrix}\right.\)
Câu 1: ĐKXĐ \(\left\{{}\begin{matrix}x\ne1\\y\ne-1\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\frac{1}{x-1}=u\\\frac{1}{y+1}=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2u+v=7\\5u-2v=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4u+2v=14\\5u-2v=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u=2\\v=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x-1}=2\\\frac{1}{y+1}=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-1=\frac{1}{2}\\y+1=\frac{1}{3}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{3}{2}\\y=-\frac{2}{3}\end{matrix}\right.\)
Câu 2:
Để hệ có nghiệm (x;y)=\(\left(2;-1\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}2m.2-\left(m+1\right).\left(-1\right)=m-n\\\left(m+2\right).2+3n\left(-1\right)=2m-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4m+n=-1\\3n=7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n=\frac{7}{3}\\m=\frac{5}{6}\end{matrix}\right.\)
a) ĐKXĐ : \(x,y\ge0;y\ne1;x+y\ne0\)
\(P=\frac{x}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)}-\frac{y}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+1\right)}-\frac{xy}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\)
\(=\frac{x\left(1+\sqrt{x}\right)-y\left(1-\sqrt{y}\right)-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{x}\right)\left(1-\sqrt{y}\right)}\)
\(=\frac{\left(x-y\right)+\left(x\sqrt{x}+y\sqrt{y}\right)-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{x}\right)\left(1-\sqrt{y}\right)}\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}+x-\sqrt{xy}+y-xy\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{x}\right)\left(1-\sqrt{y}\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)-\sqrt{y}\left(\sqrt{x}+1\right)+y\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}{\left(1+\sqrt{x}\right)\left(1+\sqrt{y}\right)}\)
\(=\frac{\sqrt{x}-\sqrt{y}+x-y\sqrt{x}}{1-\sqrt{y}}=\frac{\sqrt{x}\left(1-\sqrt{y}\right)\left(1+\sqrt{y}\right)-\sqrt{y}\left(1-\sqrt{y}\right)}{1-\sqrt{y}}\)
\(=\sqrt{x}+\sqrt{xy}+\sqrt{y}\)
Vậy P \(=\sqrt{x}+\sqrt{xy}+\sqrt{y}\)
b) ĐKXĐ : \(x,y\ge0;y\ne1;x+y\ne0\)
\(P=2\Leftrightarrow\) \(\sqrt{x}+\sqrt{xy}+\sqrt{y}=2\) ( * )
\(\Leftrightarrow\sqrt{x}\left(1+\sqrt{y}\right)-\left(\sqrt{y}+1\right)=1\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{y}+1\right)=1\)
Có : \(1+\sqrt{y}\ge1\Rightarrow\sqrt{x}-1\le1\Leftrightarrow0\le x\le4\Rightarrow x=0;1;2;3;4\)
Thay x = 0 ; 1 ; 2 ; 3 ;4 vào ( * )
Ta có các cặp giá trị : x =4 ; y = 0 và x = 2 ; y = 2 ( TM )
\(a,\frac{\sqrt{108x^3}}{\sqrt{12x}}=\frac{\sqrt{36.3.x^3}}{\sqrt{3.4.x}}=\frac{6\sqrt{3}.\sqrt{x}^3}{2\sqrt{3}.\sqrt{x}}=3\sqrt{x}^2=3x\)
\(b,\frac{\sqrt{13x^4y^6}}{\sqrt{208x^6y^6}}=\frac{\sqrt{13}.\sqrt{x^4}.\sqrt{y^6}}{\sqrt{16.13}.\sqrt{x^6}.\sqrt{y^6}}=\frac{\sqrt{13}.x^2y^3}{4\sqrt{13}x^3y^3}=\frac{1}{4x}\)
\(c,\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}+\sqrt{y}\right)^2\)
\(=\frac{\sqrt{x}^3+\sqrt{y}^3}{\sqrt{x}+\sqrt{y}}-\left(x+2\sqrt{xy}+y\right)\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x-2\sqrt{xy}-y\)
\(=x-\sqrt{xy}+y-x-2\sqrt{xy}-y=-3\sqrt{xy}\)
\(d,\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\frac{\sqrt{\left(\sqrt{x}-1\right)^2}}{\sqrt{\left(\sqrt{x}+1\right)^2}}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
Đk chỗ này là \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge\sqrt{1}\Rightarrow x\ge1\)nhé
\(e,\frac{x-1}{\sqrt{y}-1}.\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}-1}.\frac{y-2\sqrt{y}+1}{\left(x-1\right)^2}\)
\(=\frac{\left(x-1\right)\left(\sqrt{y}-1\right)^2}{\left(\sqrt{y}-1\right)\left(x-1\right)^2}=\frac{\sqrt{y}-1}{x-1}\)
\(\frac{\left[\left(e-m\right)^2-\left(e+m\right)\right]\left[\left(y-1\right)^2\left(y+1\right)^2\right]}{a.16.nh}.\frac{ê}{u^{-1}}\)
\(=\frac{\left[\left(e-m\right)^2\left(e+m\right)^2\right]\left[\left(y-1\right)^2\left(y+1\right)^2\right]}{16.anh}.êu\)
\(=\frac{\left(e^2-2em+m^2-e^2-2em.m^2\right)\left(y^2-2y+1-y^2-2y-1\right)}{16anh}.êu\)
\(=-\frac{4em\left(-4y\right)}{16anh}.êu\)
\(=\frac{emy}{anh}.êu\)
\(=\frac{em.yêu}{anh}\)
vào tìm trong câu hỏi tương tứ có đó