\(=\left(-3+3\sqrt{6}+4+2\sqrt{6}-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

=(căn 6-11)(căn 6+11)

=6-121=-115

\(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}\right)^2-1^2}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}\right)^2-2^2}-\dfrac{12\left(3+\sqrt{6}\right)}{3^2-\left(\sqrt{6}\right)^2}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right)\left(\sqrt{6}+11\right)\)

\(=\left[3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)

\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)

\(=\left(\sqrt{6}\right)^2-11^2\)

\(=6-121\)

\(=-115\)

a) \(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)

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

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

\(=\dfrac{\left|3-\sqrt{5}\right|}{\sqrt{9-5}}\)

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

b) \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)

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

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

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

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

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

a: \(=\sqrt{\dfrac{\left(3-\sqrt{5}\right)\left(3-\sqrt{5}\right)}{4}}=\dfrac{3-\sqrt{5}}{2}\)

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

d: \(=\left(-3+3\sqrt{6}+4+2\sqrt{6}-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

=(căn 6-11)(căn 6+11)

=6-121=-115

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

\(\dfrac{24}{54}\)

các phân số bằng \(\dfrac{4}{9}\)là

\(\dfrac{24}{54}\)

\(\dfrac{1212}{1515}+\dfrac{1212}{3535}+\dfrac{1212}{6363}+\dfrac{1212}{9999}\)

=\(\dfrac{12}{15}+\dfrac{12}{35}+\dfrac{12}{63}+\dfrac{12}{99}\)

=\(\dfrac{16}{11}\)

Giải:

\(\dfrac{1212}{1515}+\dfrac{1212}{3535}+\dfrac{1212}{6363}+\dfrac{1212}{9999}\) 

\(=\dfrac{12}{15}+\dfrac{12}{35}+\dfrac{12}{63}+\dfrac{12}{99}\) 

\(=12.\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}\right)\) 

\(=12.\dfrac{1}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\right)\) 

\(=6.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)\) 

\(=6.\left(\dfrac{1}{3}-\dfrac{1}{11}\right)\) 

\(=6.\dfrac{8}{33}\) 

\(=\dfrac{16}{11}\)

\(\dfrac{-4}{50}\text{=}\dfrac{-2}{25}\)

\(\dfrac{6}{25}\text{=}\dfrac{6}{25}\)

\(\dfrac{-27}{54}\text{=}\dfrac{-1}{2}\)

\(\dfrac{-18}{-75}\text{=}\dfrac{6}{25}\)

\(\dfrac{28}{-56}\text{=}-\dfrac{1}{2}\)

suy ra chỉ có phân số \(\dfrac{-4}{50}\) là không bằng phân số nào trong các phân số còn lại

-1/2

\(\dfrac{6}{25}\)

\(-\dfrac{4}{50}=\dfrac{-2}{25}\)

\(-\dfrac{27}{54}=-\dfrac{1}{2}\)

\(-\dfrac{18}{75}=-\dfrac{6}{25}\)

\(\dfrac{28}{-56}=-\dfrac{1}{2}\)

\(TC:\)

\(-\dfrac{27}{54}=\dfrac{28}{-56}\)

A+B>0

=>2*căn x-1>0

=>x>1/4

a: ĐKXĐ: x<>0; y<>0

\(\begin{cases}\frac{1}{x}+\frac{1}{y}=\frac12\\ \frac{3}{x}-\frac{4}{y}=-1\end{cases}\Rightarrow\begin{cases}\frac{1}{y}=\frac12-\frac{1}{x}\\ \frac{3}{x}-4\left(\frac12-\frac{1}{x}\right)=-1\end{cases}\)

=>\(\begin{cases}\frac{1}{y}=\frac12-\frac{1}{x}\\ \frac{3}{x}-2+\frac{4}{x}=-1\end{cases}\Rightarrow\begin{cases}\frac{7}{x}=1\\ \frac{1}{y}=\frac12-\frac{1}{x}\end{cases}\Rightarrow\begin{cases}x=7\\ \frac{1}{y}=\frac12-\frac17=\frac{5}{14}\end{cases}\)

=>x=7(nhận ) và y=14/5(nhận)

b: ĐKXĐ: 2x-y<>0 và x+y<>0

\(\begin{cases}\frac{3}{2x-y}-\frac{6}{x+y}=-1\\ \frac{1}{2x-y}-\frac{1}{x+y}=0\end{cases}\Rightarrow\begin{cases}\frac{1}{2x-y}=\frac{1}{x+y}\\ \frac{3}{x+y}-\frac{6}{x+y}=1\end{cases}\)

=>\(\begin{cases}2x-y=x+y\\ \frac{-3}{x+y}=1\end{cases}\Rightarrow\begin{cases}2x-y=x+y\\ x+y=-3\end{cases}\)

=>\(\begin{cases}2x-y=-3\\ x+y=-3\end{cases}\Rightarrow\begin{cases}2x-y+x+y=-3-3\\ x+y=-3\end{cases}\Rightarrow\begin{cases}3x=-6\\ y=-3-x\end{cases}\)

=>x=-2(nhận) và y=-3-(-2)=-3+2=-1(nhận)

c: ĐKXĐ: x<>-1 và y<>3

\(\begin{cases}\frac{5x}{x+1}+\frac{y}{y-3}=27\\ \frac{2x}{x+1}-\frac{3y}{y-3}=4\end{cases}\Rightarrow\begin{cases}\frac{y}{y-3}=27-\frac{5x}{x+1}\\ \frac{2x}{x+1}-3\left(27-\frac{5x}{x+1}\right)=4\end{cases}\)

=>\(\begin{cases}\frac{2x}{x+1}-81+\frac{15x}{x+1}=4\\ \frac{y}{y-3}=27-\frac{5x}{x+1}\end{cases}\Rightarrow\begin{cases}\frac{17x}{x+1}=85\\ \frac{y}{y-3}=27-\frac{5x}{x+1}\end{cases}\)

=>\(\begin{cases}\frac{x}{x+1}=5\\ \frac{y}{y-3}=27-5\cdot5=2\end{cases}\Rightarrow\begin{cases}5x+5=x\\ 2y-6=y\end{cases}\)

=>4x=-5 và y=6

=>x=-5/4(nhận) và y=6(nhận)

d: ĐKXĐ: x<>-2; y<>0

\(\begin{cases}\frac{7}{x+2}+\frac{3}{y}=2\\ \frac{4}{x+2}-\frac{1}{y}=\frac52\end{cases}\Rightarrow\begin{cases}\frac{1}{y}=\frac{4}{x+2}-\frac52\\ \frac{7}{x+2}+3\left(\frac{4}{x+2}-\frac52\right)=2\end{cases}\)

=>\(\begin{cases}\frac{1}{y}=\frac{4}{x+2}-\frac52\\ \frac{19}{x+2}-\frac{15}{2}=2\end{cases}\Rightarrow\begin{cases}\frac{19}{x+2}=\frac{19}{2}\\ \frac{1}{y}=\frac{4}{x+2}-\frac52\end{cases}\)

=>\(\begin{cases}x+2=2\\ \frac{1}{y}=\frac42-\frac52=\frac{-1}{2}\end{cases}\Rightarrow\begin{cases}x=0\\ y=-2\end{cases}\) (nhận)