ĐKXĐ: \(\left\{{}\begin{matrix}x>=0;y>=0\\x^2+y^2\ne1^2+1^2=2\end{matrix}\right.\)

\(\dfrac{\sqrt{x}+1}{\sqrt{xy}+1}+\dfrac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{xy}+1}-\dfrac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}+1\)

\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)-\left(\sqrt{xy}+1\right)\left(\sqrt{xy}+\sqrt{x}\right)+xy-1}{xy-1}\)

\(=\dfrac{x\sqrt{y}-\sqrt{x}+\sqrt{xy}-1-xy-x\sqrt{y}-\sqrt{xy}-\sqrt{x}+xy-1}{xy-1}\)

\(=\dfrac{-2\sqrt{x}-2}{xy-1}\)

\(1-\dfrac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\dfrac{\sqrt{x}+1}{\sqrt[]{xy}+1}\)

\(=\dfrac{xy-1-\left(\sqrt{xy}+\sqrt{x}\right)\left(\sqrt{xy}+1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)}{xy-1}\)

\(=\dfrac{xy-1-xy-\sqrt{xy}-x\sqrt{y}-\sqrt{x}-x\sqrt{y}+\sqrt{x}-\sqrt{xy}+1}{xy-1}\)

\(=\dfrac{-2\sqrt{xy}-2x\sqrt{y}}{xy-1}\)

\(P=\left(\dfrac{\sqrt{x}+1}{\sqrt{xy}+1}+\dfrac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\right):\left(1-\dfrac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\dfrac{\sqrt{x}+1}{\sqrt[]{xy}+1}\right)\)

\(=\dfrac{-2\left(\sqrt{x}+1\right)}{xy-1}:\dfrac{-2\sqrt{xy}\left(\sqrt{x}+1\right)}{xy-1}\)

\(=\dfrac{-2\left(\sqrt{x}+1\right)}{xy-1}\cdot\dfrac{xy-1}{-2\sqrt{xy}\left(\sqrt{x}+1\right)}=\dfrac{1}{\sqrt{xy}}\)

1.7:

a: \(\left\{{}\begin{matrix}3x+2y=6\\2x-2y=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+2y+2x-2y=6+14\\x-y=7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5x=20\\x-y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=x-7=4-7=-3\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}0,3x+0,5y=3\\1,5x-2y=1,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1,2x+2y=12\\1,5x-2y=1,5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}1,2x+2y+1,5x-2y=12+1,5\\0,3x+0,5y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2,7x=13,5\\0,5y=3-0,3x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\0,5y=3-0,3\cdot5=1,5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=5\\y=3\end{matrix}\right.\)

c: \(\left\{{}\begin{matrix}-2x+6y=8\\3x-9y=-12\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-x+3y=4\\x-3y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0x=0\\x=3y-4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\in R\\3y=x+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in R\\y=\dfrac{x+4}{3}\end{matrix}\right.\)

1.6:

a: \(\left\{{}\begin{matrix}x-y=3\\3x-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+3\\3\left(y+3\right)-4y=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=y+3\\3y+9-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+3\\9-y=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=7\\x=7+3=10\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}7x-3y=13\\4x+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x-3y=13\\y=2-4x\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7x-3\left(2-4x\right)=13\\y=2-4x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19x-6=13\\y=2-4x\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}19x=19\\y=2-4x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2-4\cdot1=-2\end{matrix}\right.\)

c: \(\left\{{}\begin{matrix}0,5x-1,5y=1\\-x+3y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0,5x-1,5y=1\\x=3y-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}0,5\left(3y-2\right)-1,5y=1\\x=3y-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1,5y-1-1,5y=1\\x=3y-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-1=1\\x=3y-2\end{matrix}\right.\Leftrightarrow\left(x,y\right)\in\varnothing\)

Bạn ghi lại đề nhé

ĐỀ này bạn viết lại đi 

Với \(a>0;a\ne1\):

\(P=\left(\dfrac{a+3\sqrt{a}+2}{a+\sqrt{a}-2}-\dfrac{a+\sqrt{a}}{a-1}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{\sqrt{a}-1}\right)\\ =\left[\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}-\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right]:\left[\dfrac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right]\\ =\dfrac{\sqrt{a}+1-\sqrt{a}}{\sqrt{a}-1}:\dfrac{\sqrt{a}-1+\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\\ =\dfrac{1}{\sqrt{a}-1}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{2\sqrt{a}}=\dfrac{\sqrt{a}+1}{2\sqrt{a}}\)

Lời giải:

ĐKXĐ: $a\geq 0; a\neq 1$

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

\(a)\dfrac{7}{x+4}-\dfrac{3x}{x^2-16}\left(x\ne\pm4\right)\\ =\dfrac{7}{x+4}-\dfrac{3x}{\left(x+4\right)\left(x-4\right)}\\ =\dfrac{7\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}-\dfrac{3x}{\left(x+4\right)\left(x-4\right)}\\ =\dfrac{7x-28-3x}{\left(x+4\right)\left(x-4\right)}\\ =\dfrac{4x-28}{\left(x+4\right)\left(x-4\right)}\\ =\dfrac{4x-28}{x^2-16}\)

\(b)\dfrac{x^2-3}{\left(x-1\right)\left(x-2\right)}-\dfrac{x+1}{x-1}\left(x\ne1;x\ne2\right)\\ =\dfrac{x^2-3}{\left(x-1\right)\left(x-2\right)}-\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}\\ =\dfrac{x^2-3-\left(x^2-2x+x-2\right)}{\left(x-1\right)\left(x-2\right)}\\ =\dfrac{x^2-3-x^2+x+2}{\left(x-1\right)\left(x-2\right)}\\ =\dfrac{x-1}{\left(x-1\right)\left(x-2\right)}\\ =\dfrac{1}{x-2}\) 

\(c)\dfrac{x-3}{x^2-3x+2}+\dfrac{3}{x-2}\left(x\ne1;x\ne2\right)\\ =\dfrac{x-3}{\left(x-1\right)\left(x-2\right)}+\dfrac{3}{x-2}\\ =\dfrac{x-3}{\left(x-1\right)\left(x-2\right)}+\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}\\ =\dfrac{x-3+3x-3}{\left(x-1\right)\left(x-2\right)}\\ =\dfrac{4x-6}{\left(x-1\right)\left(x-2\right)}\)

a: ĐKXĐ: \(x\notin\left\{-4;4\right\}\)

\(\dfrac{7}{x+4}-\dfrac{3x}{x^2-16}\)

\(=\dfrac{7}{x+4}-\dfrac{3x}{\left(x-4\right)\left(x+4\right)}\)

\(=\dfrac{7\left(x-4\right)-3x}{\left(x+4\right)\left(x-4\right)}=\dfrac{4x-28}{\left(x+4\right)\left(x-4\right)}\)

b: ĐKXĐ: \(x\notin\left\{2;1\right\}\)

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

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

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

c: ĐKXĐ: \(x\notin\left\{1;2\right\}\)

\(\dfrac{x-3}{x^2-3x+2}+\dfrac{3}{x-2}\)

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

\(=\dfrac{x-3+3\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}=\dfrac{4x-6}{\left(x-1\right)\left(x-2\right)}\)

đề là rút gọn đk bn 

a,đk x khác 3 

 \(\dfrac{2}{x-3}-\dfrac{x-1}{x-3}=\dfrac{2-x+1}{x-3}=\dfrac{3-x}{x-3}=-1\)đ

b,đk x khác -1/2

 \(\dfrac{x-4}{2x+1}+\dfrac{3x-3}{2x+1}=\dfrac{x-4+3x-3}{2x+1}=\dfrac{4x-7}{2x+1}\)

c, đk x khác -4;4 

\(\dfrac{7}{x+4}-\dfrac{3x}{x^2-16}=\dfrac{7\left(x-4\right)-3x}{x^2-16}=\dfrac{7x-28-3x}{x^2-16}=\dfrac{4x-28}{x^2-16}\)

d, đk x khác -1 

\(\dfrac{3x-3}{2x+2}-\dfrac{6}{x+1}=\dfrac{3x-3-12}{2\left(x+1\right)}=\dfrac{3x-15}{2\left(x+1\right)}\)

\(\left\{{}\begin{matrix}3x+2y=6\\2x-2y=14\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x+2y=6\\3x+2x=14+6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x+2y=6\\5x=20\end{matrix}\right. \\ \Leftrightarrow\left\{{}\begin{matrix}3\cdot4+2y=6\\x=\dfrac{20}{5}=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2y=6-12=-6\\x=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{6}{2}=-3\\x=4\end{matrix}\right.\)

Vậy: ...

\(\left\{{}\begin{matrix}0,5x-1,5y=1\\-x+3y=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-3y=2\\-x+3y=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3y+2\\-\left(3y+2\right)+3y=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3y+2\\-3y-2+3y=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3y+2\\-2=2\end{matrix}\right.\)

=> Hpt vô nghiệm 

0,5x - 1,5y = 1 (1)

-x + 3y = 2 (2)

Từ (2) ta có:

x = 3y - 2 (3)

Thế (3) vào (1), ta có:

0,5(3y - 2) - 1,5y = 1

1,5y - 1 - 1,5y = 1

0y = 1 + 1

0y = 2 (vô lý)

Vậy 

ĐKXĐ: \(x\notin\left\{1;-1;2;-2\right\}\)

\(\dfrac{x+4}{x-1}+\dfrac{x-4}{x+1}=\dfrac{x+8}{x-2}+\dfrac{x-8}{x+2}+6\)

=>\(\dfrac{\left(x+4\right)\left(x+1\right)+\left(x-4\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x+8\right)\left(x+2\right)+\left(x-8\right)\left(x+2\right)+6\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

=>\(\dfrac{2x^2+8}{x^2-1}=\dfrac{x^2+10x+16+x^2-10x+16+6\left(x^2-4\right)}{x^2-4}\)

=>\(\dfrac{2x^2+8}{x^2-1}=\dfrac{2x^2+32+6x^2-24}{x^2-4}\)

=>\(\dfrac{2x^2+8}{x^2-1}=\dfrac{8x^2+8}{x^2-4}\)

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

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

=>\(4\left(x^4-1\right)=x^4-16\)

=>\(4x^4-4-x^4+16=0\)

=>\(3x^4+12=0\)(vô lý)

Vậy: Phương trình vô nghiệm