bai 1 thuc hien phep tinh a)\(\dfrac{\left(\dfrac{1}{x^2+4x+4}-\dfrac{1}{x^2-4x+4}\right)}{\left(\dfrac{1}{x+2}+\dfrac{1}{x-2}\right)}\)
b)\(\left(\dfrac{5x+y}{x^2-5xy}+\dfrac{5x-y}{x^2+5xy}\right)\cdot\dfrac{x^2-25y^2}{x^2+y^2}\)
\(a,=\dfrac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}:\dfrac{x-2+x+2}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{-8x}{\left(x-2\right)^2\left(x+2\right)^2}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{2x}=\dfrac{-4}{\left(x-2\right)\left(x+2\right)}\)
\(b,=\dfrac{5x^2+26xy+5y^2+5x^2-26xy+5y^2}{x\left(x-5y\right)\left(x+5y\right)}\cdot\dfrac{\left(x-5y\right)\left(x+5y\right)}{x^2+y^2}\\ =\dfrac{10\left(x^2+y^2\right)}{x\left(x^2+y^2\right)}=\dfrac{10}{x}\)
Bài 1. (2 điểm) Thực hiện các phép tính
a) $\left(6 x^{3} y^{2}-27 x^{3} y\right) \, : \, 3 x y$.
b) $\left(\dfrac{2}{3^{2}} x^{4}\right) \cdot\left(3 y x^{5}\right)$.
c) $\dfrac{x^{2}}{x^{2}-4}+\dfrac{1}{x-2}+\dfrac{1}{x+2}$.
d) $\dfrac{2}{x-y}-\left(\dfrac{x}{x-1}-\dfrac{2}{y-x}\right)-\left(\dfrac{-2}{x+y}-\dfrac{x}{x-1}\right)$
cho 3 số x,y,z đôi 1 khác nhau và chứng minh rằng :
\(\dfrac{y-z}{\left(x-y\right)\cdot\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\cdot\left(y-x\right)}+\dfrac{y-x}{\left(z-x\right)\cdot\left(z-y\right)}=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\)
Ta có: \(\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}=\dfrac{y-x+x-z}{\left(x-y\right)\left(x-z\right)}\)\(=\dfrac{y-x}{\left(x-y\right)\left(x-z\right)}+\dfrac{x-z}{\left(x-y\right)\left(x-z\right)}\) \(=\dfrac{1}{z-x}+\dfrac{1}{x-y}\)
Tương tự:
\(\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}=\dfrac{1}{x-y}+\dfrac{1}{y-z}\)
\(\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}=\dfrac{1}{y-z}+\dfrac{1}{z-x}\)
\(\Rightarrow\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}+\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}\) \(=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\) \(\left(đpcm\right)\)
Tìm $x$, biết :
a) $\left(\dfrac{1}{2}+1,5\right) \cdot x=\dfrac{1}{5}$
b) $\left(-1 \dfrac{3}{5}+x\right): \dfrac{12}{13}=2 \dfrac{1}{6}$
c) $\left(x: 2 \dfrac{1}{3}\right) \cdot \dfrac{1}{7}=\dfrac{-3}{8}$
d) $\dfrac{-4}{7} \cdot x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1 \dfrac{2}{3}\right)$
\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)
\(\Rightarrow2x=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{10}\)
\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=2\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)
\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)
\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)
\(\Leftrightarrow x=-\dfrac{49}{8}\)
\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)
\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)
\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)
\(\Leftrightarrow x=\dfrac{413}{160}\)
Cho 2 biểu thức
\(M=\left[\dfrac{x^2-y^2}{x^2+2xy+y^2}+\dfrac{2}{xy}:\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\right]\cdot\dfrac{1}{x-y}\)
\(N=\dfrac{1}{x+y}+\dfrac{2xy}{\left(x^2-y^2\right)\cdot\left(x+y\right)}+\dfrac{3}{x^2-2x+2}\)
a/ Rút gọn M, N
b/ Với giá trị nào của x, y thì M - N có GTNN ? Tìm GTNN đó.
Rút gọn:
\(\left[\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{2}{x+y}\cdot\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\right]:\dfrac{x^3+y^3}{x^2y^2}-\dfrac{x+y}{x^2-xy+y^2}\)
\(\left[\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{2}{x+y}.\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\right]:\dfrac{x^3+y^3}{x^2y^2}-\dfrac{x+y}{x^2-xy+y^2}\)
\(=\left[\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{2}{x+y}.\dfrac{x+y}{xy}\right].\dfrac{x^2y^2}{x^3+y^3}-\dfrac{x+y}{x^2-xy+y^2}\)
\(=\left[\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{2}{xy}\right].\dfrac{x^2y^2}{\left(x+y\right)\left(x^2-xy+y^2\right)}-\dfrac{x+y}{x^2-xy+y^2}\)
\(=\dfrac{y^2+x^2+2xy}{x^2y^2}.\dfrac{x^2y^2}{\left(x+y\right)\left(x^2-xy+y^2\right)}-\dfrac{x+y}{x^2-xy+y^2}\)
\(=\dfrac{\left(x+y\right)^2}{\left(x+y\right)\left(x^2-xy+y^2\right)}-\dfrac{x+y}{x^2-xy+y^2}\)
=\(=\dfrac{x+y}{x^2-xy+y^2}-\dfrac{x+y}{x^2-xy+y^2}=0\)
Mina ơi~~~Ai giải giùm em vài bài này với a~~Em làm rùi nhưng cứ thấy hoang mang quá nên hỏi mina cho chắc a~Em cảm ơn mina nhiều a~
Bài 1:
b,\(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)
c,\(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
d,\(\dfrac{x+1}{x+2}:\left(\dfrac{x+2}{x+3}:\dfrac{x+3}{x+1}\right)\)
c,\(\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)
d,\(\dfrac{xy}{ab}+\dfrac{\left(x-a\right)\left(y-a\right)}{a\left(a-b\right)}-\dfrac{\left(x-b\right)\left(y-b\right)}{b\left(a-b\right)}\)
e,\(\dfrac{x^3}{x-1}-\dfrac{x^2}{x+1}-\dfrac{1}{x-1}+\dfrac{1}{x+1}\)
g,\(\left(\dfrac{x-y}{x+y}+\dfrac{x+y}{x-y}\right).\left(\dfrac{x^2+y^2}{2xy}+1\right).\dfrac{xy}{x^2+y^2}\)
h,\(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
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Giải phương trình
\(1,\dfrac{x^2-2x-3}{x-1}+\dfrac{x^2-8x+20}{x-4}=\dfrac{x^2-4x+6}{x-2}+\dfrac{x^2-6x+12}{x-3}\)
\(2,\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\cdot[1+\dfrac{1}{x\cdot\left(x+2\right)}]=\dfrac{31}{16}\left(x\in N\right)\)
Tính:
\(D=\dfrac{4x^2-1}{\left(x-y\right)\cdot\left(x+y\right)}+\dfrac{4y^2-1}{\left(y-z\right)\cdot\left(y-x\right)}+\dfrac{4z^2-1}{\left(z-x\right)\cdot\left(z-y\right)}\)
