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1) 1/x-1/y
=y/xy-x/xy
=y-x/xy
= - (x-y)/xy
= -1 (vì x-y=xy)
2)
(x- 1/2)*(y+1/3)*(z-2)=0
=> x-1/2 = 0 hoac y+1/3=0 hoac z-2=0
th1 :x-1/2=0 => x=1/2
x+2=y+3=z+4
mà x=1/2 => y= -1/2 ; z=-3/2
th2: y+1/3=0
th3 : z-2=0
(tự làm nha)
1) Với x,y khác 0, Ta có
\(\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}=-\left(\frac{x-y}{xy}\right)=-\left(\frac{xy}{xy}\right)=-1\)
Vậy \(\frac{1}{x}-\frac{1}{y}=-1\)
2) Ta có:
\(\left(x-\frac{1}{2}\right)\left(y+\frac{1}{3}\right)\left(z-2\right)=0\)
Trường hợp 1: x - 1/2 = 0 => x = 1/2 \(\Rightarrow\hept{\begin{cases}y=\frac{1}{2}+2-3=-\frac{1}{2}\\z=\frac{1}{2}+2-4=-\frac{3}{2}\end{cases}}\)
Trường hợp 2: y + 1/3 = 0 => y = -1/3 \(\Rightarrow\hept{\begin{cases}x=-\frac{1}{3}+3-2=\frac{2}{3}\\z=-\frac{1}{3}+3-4=-\frac{4}{3}\end{cases}}\)
Trường hợp 3: z - 2 = 0 => z = 2 \(\Rightarrow\hept{\begin{cases}x=2+4-2=4\\y=2+4-3=3\end{cases}}\)
Vậy......
a) \(\frac{1}{2}-\left(2x-\frac{3}{4}\right)=-\frac{5}{8}\)
\(\frac{1}{2}-2x+\frac{3}{4}=-\frac{5}{8}\)
\(\frac{1}{2}+\frac{3}{4}-2x=-\frac{5}{8}\)
\(1,25-2x=-0,625\)
\(2x=1,875\)
\(x=0,9375\)
b) \(-\frac{1}{2}-\frac{3}{2}.\left(x+\frac{5}{3}\right)=-\frac{1}{4}\)
\(-\frac{1}{2}-\frac{3}{2}.x-\frac{5}{2}=-\frac{1}{4}\)
\(-\frac{1}{2}-\frac{5}{2}-\frac{3}{2}.x=-\frac{1}{4}\)
\(-3-\frac{3}{2}.x=-\frac{1}{4}\)
...
đến đây thì b tự tính nha!
a) \(\left(\frac{3}{5}x-\frac{2}{3}x-x\right).\frac{1}{7}=\frac{-5}{21}\)
\(\Rightarrow\left(\frac{3}{5}-\frac{2}{3}-1\right).x=\frac{-5}{21}:\frac{1}{7}=\frac{-5}{3}\)
\(\Rightarrow\frac{-16}{15}.x=\frac{-5}{3}\Rightarrow x=\frac{-5}{3}:\frac{-16}{15}=\frac{25}{16}\)
b) \(\left(x-\frac{1}{4}\right)^2=\frac{1}{36}\)
\(\Rightarrow\left(x-\frac{1}{4}\right)^2=\left(±\frac{1}{6}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{4}=\frac{1}{6}\\x-\frac{1}{4}=\frac{-1}{6}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{12}\\x=\frac{1}{12}\end{cases}}\)
\(B=\left(\dfrac{1}{2^2}-1\right)\cdot\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{1000^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\cdot\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{1000}-1\right)\cdot\left(\dfrac{1}{2}+1\right)\cdot\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{1000}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-999}{1000}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot....\cdot\dfrac{1001}{1000}\)
\(=\dfrac{-1}{1000}\cdot\dfrac{1001}{2}=\dfrac{-1001}{2000}\)
Thanks