2.Thuc hien phep tinh
\(\dfrac{1}{\dfrac{3}{\dfrac{4}{3}}+1}\) la \(\dfrac{3}{4}\) nhe cau
thuc hien phep tinh
\(\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)
\(\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)
\(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)
\(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x+1}+\dfrac{-\left(x+3\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}+\dfrac{-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{\left(3x+1\right)\left(x+1\right)-\left(x-1\right)^2-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{3x^2+4x+1-\left(x^2-2x+1\right)-\left(x^2+2x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+x+3x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right)+3\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x+3}{\left(x-1\right)^2}\)
\(\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)
\(=\dfrac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}+\dfrac{-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{3x^2+4x+1-x^2+2x-1-x^2-2x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+3x+x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{\left(x+3\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}=\dfrac{x+3}{\left(x-1\right)^2}\)
Thuc hien phep tinh :\(\dfrac{18x}{x^3-9x}-\dfrac{2-x}{x+3}+\dfrac{3}{3-x}\)
\(\dfrac{18x}{x^3-9x}-\dfrac{2-x}{x+3}+\dfrac{3}{3-x}\)
=\(\dfrac{18x}{x\left(x^2-9\right)}-\dfrac{\left(2-x\right)\left(x-3\right)x}{\left(x+3\right)\left(x-3\right)x}-\dfrac{3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)x}\)
=\(\dfrac{18x-\left(2-x\right)\left(x-3\right)x-3x\left(x+3\right)}{x\left(x+3\right)\left(x-3\right)}\)
=\(\dfrac{18x-\left(2x-6-x^2+3x\right)x-3x^2-9x}{x\left(x+3\right)\left(x-3\right)}\)
=\(\dfrac{18x-2x^2+6x+x^3-3x^2-3x^2-9x}{x\left(x+3\right)\left(x-3\right)}\)
=\(\dfrac{x^3-8x^2+15x}{x\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{x\left(x^2-8x+15\right)}{x\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{x^2-3x-5x+15}{x\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{x\left(x-3\right)-5\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{\left(x-3\right)\left(x-5\right)}{\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{\left(x-5\right)}{x+3}\)
thuc hien phep tinh:
a,2xyz+4xyz-\(\dfrac{1}{2}\)xyz
b,\(\dfrac{x^2}{2}\)+\(\dfrac{x^2}{3}\)+\(\dfrac{x^2}{4}\)
1) thuc hien phep tinh sau:
a)B=\(\left(\dfrac{5}{7}.0,6:3\dfrac{1}{2}\right)\) .(40%-1,4).(-2)^3
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
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=\(\dfrac{24}{7}\)
thuc hien phep tinh:
a,2xyz+4xyz-\(\dfrac{1}{2}\)xyz
b,\(\dfrac{x^2}{2}\)+\(\dfrac{x^2}{3}\)+\(\dfrac{x^2}{4}\)
2xyz+4xyz-\(\frac{1}{2}\) xyz
=(2+4-\(\frac{1}{2}\) )(xxxyyyzzz)
=3.5x3 y3z3
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)\(\left(\dfrac{5x+y}{x^2-5xy}+\dfrac{5x-y}{x^2+5xy}\right)\cdot\dfrac{x^2-25y^2}{x^2+y^2}\)
\(a,=\dfrac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}:\dfrac{x-2+x+2}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{-8x}{\left(x-2\right)^2\left(x+2\right)^2}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{2x}=\dfrac{-4}{\left(x-2\right)\left(x+2\right)}\)
\(b,=\dfrac{5x^2+26xy+5y^2+5x^2-26xy+5y^2}{x\left(x-5y\right)\left(x+5y\right)}\cdot\dfrac{\left(x-5y\right)\left(x+5y\right)}{x^2+y^2}\\ =\dfrac{10\left(x^2+y^2\right)}{x\left(x^2+y^2\right)}=\dfrac{10}{x}\)
thuc hien phep tinh:
a, 1\(\dfrac{1}{2}\).\(2\dfrac{1}{3}\)+1\(\dfrac{1}{3}\).\(\dfrac{1}{2}\)
b, \(\dfrac{1}{9}\).\(\dfrac{2}{145}\)-4\(\dfrac{1}{3}\).2\(\dfrac{2}{145}\)+\(\dfrac{2}{145}\)
Giải:
a) \(1\dfrac{1}{2}.2\dfrac{1}{3}+1\dfrac{1}{3}.\dfrac{1}{2}\)
\(=\dfrac{3}{2}.\dfrac{7}{3}+\dfrac{4}{3}.\dfrac{1}{2}\)
\(=\dfrac{21}{6}+\dfrac{4}{6}\)
\(=\dfrac{1}{6}\left(21+4\right)\)
\(=\dfrac{25}{6}\)
b) \(\dfrac{1}{9}.\dfrac{2}{145}-4\dfrac{1}{3}.2\dfrac{2}{145}+\dfrac{2}{145}\)
\(=\dfrac{1}{9}.\dfrac{2}{145}-\dfrac{13}{3}.\dfrac{292}{145}+\dfrac{2}{145}\)
\(=\dfrac{2}{145}\left(\dfrac{1}{9}-\dfrac{13}{3}.146+1\right)\)
\(=\dfrac{2}{145}\left(-\dfrac{5684}{9}\right)\)
\(=-\dfrac{392}{45}\)
Vậy ...
Thuc hien phep tinh;
a/ \(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)
b/ \(\left(\dfrac{1}{25}+\dfrac{1}{5}+1\right):\left(\dfrac{1}{25}-\dfrac{1}{5}-1\right)\)
c/ \(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}\)+ \(\dfrac{0,6-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-0,16-\dfrac{4}{125}-\dfrac{4}{625}}\)
d/ \(\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
a: \(=\left(1+\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}\)
\(=1+1+\dfrac{1}{2}=2+\dfrac{1}{2}=\dfrac{5}{2}\)
b: \(=\left(\dfrac{1}{25}+\dfrac{5}{25}+\dfrac{25}{25}\right):\left(\dfrac{1}{25}-\dfrac{5}{25}-\dfrac{25}{25}\right)\)
\(=\dfrac{31}{25}:\dfrac{-29}{25}=\dfrac{-31}{29}\)
c: \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{\dfrac{3}{5}-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-\dfrac{4}{25}-\dfrac{4}{125}-\dfrac{4}{625}}\)
=1/4+3/4
=1
thuc hien phep tinh:
a,\(1\dfrac{1}{2}\). \(2\dfrac{1}{3}\)+ \(1\dfrac{1}{3}\). \(\dfrac{1}{2}\)
b,\(\dfrac{1}{9}\).\(\dfrac{2}{145}\)\(-4\dfrac{1}{3}\). \(\dfrac{2}{145}\)+\(\dfrac{2}{145}\)
Giải:
a) \(1\dfrac{1}{2}.2\dfrac{1}{3}+1\dfrac{1}{3}.\dfrac{1}{2}\)
\(=\dfrac{3}{2}.\dfrac{7}{3}+\dfrac{4}{3}.\dfrac{1}{2}\)
\(=\dfrac{21}{6}+\dfrac{4}{6}\)
\(=\dfrac{1}{6}\left(21+4\right)\)
\(=\dfrac{1}{6}.25=\dfrac{25}{6}\)
b) \(\dfrac{1}{9}.\dfrac{2}{145}-4\dfrac{1}{3}.\dfrac{2}{145}+\dfrac{2}{145}\)
\(=\dfrac{1}{9}.\dfrac{2}{145}-\dfrac{13}{3}.\dfrac{2}{145}+\dfrac{2}{145}\)
\(=\dfrac{2}{145}\left(\dfrac{1}{9}-\dfrac{13}{3}+1\right)\)
\(=\dfrac{2}{145}\left(-\dfrac{29}{9}\right)\)
\(=-\dfrac{2}{45}\)
Vậy ...