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a) $\frac{2}{3} = \frac{{2 \times 7}}{{3 \times 7}} = \frac{{14}}{{21}}$ ; Giữ nguyên phân số $\frac{{16}}{{21}}$
$\frac{3}{7} = \frac{{3 \times 3}}{{7 \times 3}} = \frac{9}{{21}}$
Vì $\frac{9}{{21}} < \frac{{14}}{{21}} < \frac{{16}}{{21}}$ nên các phân số đã cho xếp theo thứ tự từ bé đến lớn là: $\frac{3}{7};\,\,\frac{2}{3};\,\,\,\frac{{16}}{{21}}$
b) $\frac{2}{9} = \frac{{2 \times 3}}{{9 \times 3}} = \frac{6}{{27}}$, Giữ nguyên phân số $\frac{4}{{27}}$
$\frac{1}{3} = \frac{{1 \times 9}}{{3 \times 9}} = \frac{9}{{27}}$
Vì $\frac{4}{{27}} < \frac{6}{{27}} < \frac{9}{{27}}$ nên $\frac{4}{{27}}$< $\frac{2}{9} < \frac{1}{3}$
Vậy các phân số đã cho xếp theo thứ tự từ bé đến lớn là $\frac{4}{{27}}$; $\frac{2}{9};\frac{1}{3}$
c) Giữ nguyên phân số $\frac{{11}}{{28}}$
$\frac{3}{4} = \frac{{3 \times 7}}{{4 \times 7}} = \frac{{21}}{{28}}$ ; $\frac{2}{7} = \frac{{2 \times 4}}{{7 \times 4}} = \frac{8}{{28}}$
Vì $\frac{8}{{28}} < \frac{{11}}{{28}} < \frac{{21}}{{28}}$ nên các phân số đã cho xếp theo thứ tự từ bé đến lớn là $\frac{2}{7};\frac{{11}}{{28}};\frac{3}{4}$
a) \(\dfrac{2}{5}-\dfrac{3}{15}\)
\(=\dfrac{2}{5}-\dfrac{3:3}{15:3}\)
\(=\dfrac{2}{5}-\dfrac{1}{5}\)
\(=\dfrac{1}{5}\)
b) \(\dfrac{9}{27}-\dfrac{2}{9}\)
\(=\dfrac{9:3}{27:3}-\dfrac{2}{9}\)
\(=\dfrac{3}{9}-\dfrac{2}{9}\)
\(=\dfrac{1}{9}\)
c) \(\dfrac{18}{24}-\dfrac{4}{8}\)
\(=\dfrac{18:6}{24:6}-\dfrac{4:2}{8:2}\)
\(=\dfrac{3}{4}-\dfrac{2}{4}\)
\(=\dfrac{1}{4}\)
d) \(\dfrac{6}{16}-\dfrac{10}{64}\)
\(=\dfrac{6\times2}{16\times2}-\dfrac{10:2}{64:2}\)
\(=\dfrac{12}{32}-\dfrac{5}{32}\)
\(=\dfrac{7}{32}\)
\(a,\dfrac{3}{5}+\dfrac{-5}{9}=\dfrac{27-25}{45}=\dfrac{2}{49}.\)
\(c,\dfrac{-27}{23}+\dfrac{5}{21}+\dfrac{4}{23}+\dfrac{16}{21}+\dfrac{1}{2}=\dfrac{-23}{23}+\dfrac{21}{21}+\dfrac{1}{2}=-1+1+\dfrac{1}{2}=\dfrac{1}{2}.\)
\(d,\dfrac{-8}{9}+\dfrac{1}{9}.\dfrac{2}{9}+\dfrac{1}{9}.\dfrac{7}{9}=\dfrac{-8}{9}+\dfrac{1}{9}.\left(\dfrac{2}{9}+\dfrac{7}{9}\right)=\dfrac{-8}{9}+\dfrac{1}{9}.1=\dfrac{-8+1}{9}=\dfrac{-7}{9}.\)
a)
=(x-2)3
b)\(\left(2-x\right)^3\)
c)\(\left(x+\dfrac{1}{3}\right)^3\)
d)\(\left(\dfrac{x}{2}+y\right)^3\)
e)
\(=\left(x-1\right)^2\left(x-1-15\right)+25\left[3\left(x-1\right)-5\right]\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-3-5\right)\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-8\right)\)
a: \(=\dfrac{-3}{4}\left(31+\dfrac{11}{23}+8+\dfrac{12}{23}\right)=\dfrac{-3}{4}\cdot40=-30\)
b: \(=\left(\dfrac{7}{3}+\dfrac{7}{2}\right):\left(-\dfrac{25}{6}+\dfrac{22}{7}\right)+\dfrac{15}{2}\)
\(=\dfrac{35}{6}:\dfrac{-175+132}{42}+\dfrac{15}{2}\)
\(=\dfrac{35}{6}\cdot\dfrac{42}{-43}+\dfrac{15}{2}\)
\(=\dfrac{35\cdot7}{-43}+\dfrac{15}{2}\)
\(=\dfrac{-70\cdot7+15\cdot43}{86}=\dfrac{155}{86}\)
c: \(=\dfrac{-7}{5}\left(4+\dfrac{5}{9}+5+\dfrac{4}{9}\right)=\dfrac{-7}{5}\cdot10=-14\)
d: \(=4+\dfrac{25}{16}+25\cdot\left(\dfrac{9}{16}\cdot\dfrac{64}{125}\cdot\dfrac{-8}{27}\right)\)
\(=\dfrac{89}{16}+25\cdot\dfrac{-32}{375}\)
\(=\dfrac{89}{16}-\dfrac{32}{15}=\dfrac{823}{240}\)
e: \(=\dfrac{2}{3}-4\cdot\left(\dfrac{2}{4}+\dfrac{3}{4}\right)=\dfrac{2}{3}-5=-\dfrac{13}{3}\)
a, 3.27.9
= 3.33.32
= 31+3+2
= 34+2
= 36
b; 25.5.125
= 52.51.53
= 52+1+3
= 53+3
= 56
c; 49.7.343
= 72.71.73
= 72+1+3
= 73+3
= 76
d; \(\dfrac{2}{3}\).\(\dfrac{4}{9}\).\(\dfrac{8}{27}\)
= \(\left(\dfrac{2}{3}\right)\)1.\(\left(\dfrac{2}{3}\right)\)2.\(\left(\dfrac{2}{3}\right)\)3
= \(\left(\dfrac{2}{3}\right)\)1+2+3
= \(\left(\dfrac{2}{3}\right)\)3+3
= \(\left(\dfrac{2}{3}\right)\)6
e; \(\dfrac{3}{4}\).\(\dfrac{9}{16}\).\(\dfrac{27}{64}\)
= \(\left(\dfrac{3}{4}\right)\)1.\(\left(\dfrac{3}{4}\right)\)2.\(\left(\dfrac{3}{4}\right)\)3
= \(\left(\dfrac{3}{4}\right)\)1+2+3
= \(\left(\dfrac{3}{4}\right)\)3+3
= \(\left(\dfrac{3}{4}\right)\)6
f; \(\dfrac{2}{3}\).\(\dfrac{8}{27}\).\(\dfrac{16}{81}\)
= \(\left(\dfrac{2}{3}\right)\)1.\(\left(\dfrac{2}{3}\right)\)3.\(\left(\dfrac{2}{3}\right)\)4
= \(\left(\dfrac{2}{3}\right)\)1+3+4
= \(\left(\dfrac{2}{3}\right)\)4+4
= \(\left(\dfrac{2}{3}\right)\)8