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\(\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)
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14 tháng 12 2021
\(a,=\dfrac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}:\dfrac{x-2+x+2}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{-8x}{\left(x-2\right)^2\left(x+2\right)^2}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{2x}=\dfrac{-4}{\left(x-2\right)\left(x+2\right)}\)
\(b,=\dfrac{5x^2+26xy+5y^2+5x^2-26xy+5y^2}{x\left(x-5y\right)\left(x+5y\right)}\cdot\dfrac{\left(x-5y\right)\left(x+5y\right)}{x^2+y^2}\\ =\dfrac{10\left(x^2+y^2\right)}{x\left(x^2+y^2\right)}=\dfrac{10}{x}\)
13 tháng 6 2022
a: \(=\dfrac{4x^2+4x+1-\left(4x^2-4x+1\right)}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{8x}{2x+1}\cdot\dfrac{5}{4x}=\dfrac{10}{2x+1}\)
c: \(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\left(\dfrac{x+1-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{2}{\left(x-1\right)}=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{x-1}{x^2+1}\)
ND
11 tháng 12 2017
\(a.\)
\(\left(x^2-25\right):\dfrac{2x+10}{3x-7}\)
\(=\left(x-5\right)\left(x+5\right).\dfrac{3x-7}{2\left(x+5\right)}\)
\(=\dfrac{\left(x-5\right)\left(x+5\right)\left(3x-7\right)}{2\left(x+5\right)}\)
\(=\dfrac{\left(x-5\right)\left(3x-7\right)}{2}\)
\(b.\)
\(\dfrac{x^2+x}{5x^2-10x+5}:\dfrac{3x+3}{5x-5}\)
\(=\dfrac{x\left(x+1\right)}{5\left(x^2-2x+1\right)}.\dfrac{5\left(x-1\right)}{3\left(x+3\right)}\)
\(=\dfrac{x\left(x+1\right)}{5\left(x-1\right)^2}.\dfrac{5\left(x-1\right)}{3\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right).5\left(x-1\right)}{5\left(x-1\right)^2.3\left(x+1\right)}\)
\(=\dfrac{x}{3\left(x-1\right)}\)
NH
11 tháng 12 2017
\(\dfrac{x^2+x}{5x^2-10x+5}:\dfrac{3x+3}{5x-5}=\dfrac{5x\left(x+1\right)\left(x-1\right)}{15\left(x-1\right)^2\left(x+1\right)}=\dfrac{x}{3\left(x-1\right)}\)\(\left(x^2-25\right):\dfrac{2x+10}{3x-7}=\dfrac{\left(x-5\right)\left(x+5\right)\left(3x-7\right)}{2\left(x+5\right)}=\dfrac{\left(x-5\right)\left(3x-7\right)}{2}\)
AN
10 tháng 11 2018
1.a)\(2.x-\dfrac{5}{4}=\dfrac{20}{15}\)
\(\Leftrightarrow2.x=\dfrac{20}{15}+\dfrac{5}{4}=\dfrac{4}{3}+\dfrac{5}{4}=\dfrac{16+15}{12}=\dfrac{31}{12}\)
\(\Leftrightarrow x=\dfrac{31}{12}:2=\dfrac{31}{12}.\dfrac{1}{2}=\dfrac{31}{24}\)
b)\(\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{8}\right)\)
\(\Leftrightarrow\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{2}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{3}=-\dfrac{1}{2}\)
\(\Leftrightarrow x=-\dfrac{1}{2}-\dfrac{1}{3}=-\dfrac{5}{6}\)
2.Theo đề bài, ta có: \(\dfrac{a}{2}=\dfrac{b}{3}\) và \(a+b=-15\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{a+b}{2+3}=\dfrac{-15}{5}=-3\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=-3\Rightarrow a=-6\\\dfrac{b}{3}=-3\Rightarrow b=-9\end{matrix}\right.\)
3.Ta xét từng trường hợp:
-TH1:\(\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>-1\\x< 2\end{matrix}\right.\)\(\Rightarrow x\in\left\{0;1\right\}\)
-TH2:\(\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\)\(\Rightarrow x\in\varnothing\)
Vậy \(x\in\left\{0;1\right\}\)
4.\(B=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^9=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^9=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^{18}=\left(\dfrac{3}{7}\right)^3=\dfrac{27}{343}\)
EC
2
13 tháng 5 2018
B = ( \(\dfrac{5}{7}.0,6:3\dfrac{1}{2}\)) . ( 40% - 1,4) . \(\left(-2\right)^3\)
= (\(\dfrac{5}{7}.\dfrac{3}{5}\): \(\dfrac{7}{2}\)) . ( \(\dfrac{2}{5}-\dfrac{7}{5}\)) . (-8)
= (\(\dfrac{15}{35}:\dfrac{7}{2}\)) . \(\dfrac{-5}{5}\) . ( -8)
= (\(\dfrac{3}{7}.\dfrac{2}{7}\)) . (-1) . (-8)
= \(\dfrac{6}{49}\) . (-1) . (-8)
= \(\dfrac{-6}{49}\) . (-8)
= \(\dfrac{48}{49}\)
Vậy: B =\(\dfrac{48}{49}\)
Nhớ tick nha
13 tháng 5 2018
B=(\(\dfrac{3}{7}.\dfrac{2}{7}\)).(\(\dfrac{4}{10}-\dfrac{14}{10}\)).(-8)
B=\(\dfrac{3}{7}.\left(-1\right)\left(-8\right)\)
B=\(\dfrac{24}{7}\)
22 tháng 12 2021
\(a,A=\dfrac{5-3}{5+2}=\dfrac{2}{7}\\ b,B=\dfrac{3x-9+2x+6-3x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ c,C=AB=\dfrac{x-3}{x+2}\cdot\dfrac{2}{x-3}=\dfrac{2}{x+2}\\ C=-\dfrac{1}{3}\Leftrightarrow x+2=-6\Leftrightarrow x=-8\left(tm\right)\)
Chúc bạn học tốt!
M
2
LH
9 tháng 12 2018
\(\dfrac{18x}{x^3-9x}-\dfrac{2-x}{x+3}+\dfrac{3}{3-x}\)
=\(\dfrac{18x}{x\left(x^2-9\right)}-\dfrac{\left(2-x\right)\left(x-3\right)x}{\left(x+3\right)\left(x-3\right)x}-\dfrac{3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)x}\)
=\(\dfrac{18x-\left(2-x\right)\left(x-3\right)x-3x\left(x+3\right)}{x\left(x+3\right)\left(x-3\right)}\)
=\(\dfrac{18x-\left(2x-6-x^2+3x\right)x-3x^2-9x}{x\left(x+3\right)\left(x-3\right)}\)
=\(\dfrac{18x-2x^2+6x+x^3-3x^2-3x^2-9x}{x\left(x+3\right)\left(x-3\right)}\)
=\(\dfrac{x^3-8x^2+15x}{x\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{x\left(x^2-8x+15\right)}{x\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{x^2-3x-5x+15}{x\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{x\left(x-3\right)-5\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{\left(x-3\right)\left(x-5\right)}{\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{\left(x-5\right)}{x+3}\)
\(\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)
\(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)
\(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x+1}+\dfrac{-\left(x+3\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}+\dfrac{-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{\left(3x+1\right)\left(x+1\right)-\left(x-1\right)^2-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{3x^2+4x+1-\left(x^2-2x+1\right)-\left(x^2+2x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+x+3x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right)+3\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x+3}{\left(x-1\right)^2}\)
\(\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)
\(=\dfrac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}+\dfrac{-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{3x^2+4x+1-x^2+2x-1-x^2-2x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+3x+x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{\left(x+3\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}=\dfrac{x+3}{\left(x-1\right)^2}\)