9/14:6/7 +3/12:2*1/2

 =3/4+1/8:2*1/2

=3/4+1/16*1/2

=3/4+1/32

=25/32

\(\frac{9}{14}\): \(\frac{6}{7}\) + \(\frac{3}{12}\): \(2\)x\(\frac{1}{2}\)

= \(\frac{3}{4}\)\(+\)\(\frac{1}{8}\)x\(\frac{1}{2}\)

=\(\frac{3}{4}\)\(+\)\(\frac{1}{16}\)

=\(\frac{9}{16}\)\(+\)\(\frac{1}{16}\)

= \(\frac{10}{16}\)

=\(\frac{5}{8}\)

nhé

\(B=\frac{9^{12}\cdot4^8}{27^7\cdot2^{17}}\)

\(B=\frac{\left(3^2\right)^{12}\cdot\left(2^2\right)^8}{\left(3^3\right)^7\cdot2^{17}}\)

\(B=\frac{3^{24}\cdot2^{16}}{3^{21}\cdot2^{17}}\)

\(B=\frac{3^3}{2}\)

\(B=\frac{27}{2}\)

\(B=\frac{9^{12}.4^8}{27^7.2^{17}}\)

\(\Rightarrow B=\frac{\left(3^2\right)^{12}.\left(2^2\right)^8}{\left(3^3\right)^7.2^{17}}\)

\(\Rightarrow B=\frac{3^{24}.2^{16}}{3^{21}.2^{17}}\)

\(\Rightarrow B=\frac{3^3}{2}=\frac{27}{2}\)

\(B=\dfrac{3}{2}-\dfrac{2}{21}-\left\{\dfrac{7}{12}-\left[\dfrac{15}{21}-\left(\dfrac{1}{3}-\dfrac{5}{4}\right)-\left(\dfrac{2}{7}+\dfrac{1}{3}\right)\right]\right\}\)

\(B=\dfrac{3}{2}-\dfrac{2}{21}-\left\{\dfrac{7}{12}-\left[\dfrac{15}{21}-\left(-\dfrac{11}{12}\right)-\dfrac{13}{21}\right]\right\}\)

\(B=\dfrac{3}{2}-\dfrac{2}{21}-\left\{\dfrac{7}{12}-\dfrac{85}{84}\right\}\)

\(B=\dfrac{3}{2}-\dfrac{2}{21}-\left(-\dfrac{3}{7}\right)\)

\(B=\dfrac{11}{6}\)

\(=\dfrac{3}{2}-\dfrac{2}{21}-\dfrac{7}{12}+\left[\dfrac{15}{21}-\dfrac{1}{3}+\dfrac{5}{4}-\dfrac{2}{7}-\dfrac{1}{3}\right]\)

=11/12-2/21+5/7-2/3+5/4-2/7

=11/12-2/3+5/4-2/21+3/7

=11/12-8/12+15/12-2/21+9/21

=18/12+7/21

=3/2+1/3

=9/6+2/6=11/6

\(\frac{x-12}{3}=\frac{x+1}{4}\)

=>(x-12).4=(x+1)*3

    4x-48=3x+3

    4x-3x=48+3

    x=51

(x-12)/3=(x+1)/4

(x-12)*4=(x+1)*3

x*4-12*4=x*3+1*3

4x-48=3x+3

4x-3x=3+48

x=51

Câu hỏi của Lê Tiến Cường - Toán lớp 6 - Học toán với OnlineMath

\(A=\frac{3}{2}+\frac{7}{6}+\frac{13}{12}+...+\frac{10101}{10100}=\frac{2+1}{2}+\frac{6+1}{6}+\frac{12+1}{12}+...+\frac{10100+1}{10100}\)

\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{6}\right)+\left(1+\frac{1}{12}\right)+....+\left(1+\frac{1}{10100}\right)\)

\(A=\left(1+\frac{1}{1\times2}\right)+\left(1+\frac{1}{2\times3}\right)+\left(1+\frac{1}{3\times4}\right)+...+\left(1+\frac{1}{100\times101}\right)\)

\(A=\left(1+1+1+....+1\right)+\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{100\times101}\right)\)

\(A=100+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{100}-\frac{1}{101}\right)\)

\(A=100+1-\frac{1}{101}=101-\frac{1}{101}< 101=B\)

\(\Rightarrow A< B\)

So easy

a, 12- (-4)= 16
b, 12- (-14)= 26
c, (-13)-(-5)=-8
d, (-2)-(-10)= 8