Rút gọn rồi tính:                                                                                                                                               \(a.\frac{16}{24}-\frac{1}{3}=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\)

\(b.\frac{4}{5}-\frac{12}{60}=\frac{4}{5}-\frac{1}{5}=\frac{3}{5}\)

Tính rồi rút gọn:

\(a.\frac{17}{6}-\frac{2}{6}=\frac{15}{6}\)

\(b.\frac{16}{15}-\frac{11}{15}=\frac{5}{15}=\frac{1}{3}\)

\(c.\frac{19}{12}-\frac{13}{12}=\frac{6}{12}=\frac{1}{2}\)

1a2/3-1/3=1/3            c4/5-1/5=3/5             tinh roi rut gon 17/6-2/6=15/6=5/2             16/15-11/15=5/15=1/3           19/12-13/12=6/12=1/2

chuc bn hoc tot

lớp 1 à cu

ii) (5x + 1)2 + (5x – 1)2 + 2(5x + 1)(5x – 1)

= [(5x + 1) + (5x - 1)]2

= (10x)2 = 100x2

ii) (4x + 1)2 + (4x – 1)2 – 2(4x + 1)(4x – 1)

= [(4x + 1) - (4x - 1)]2

= (4x + 1 - 4x + 1)2

= 22 = 4

\(a,\\ =\dfrac{5}{4}-\dfrac{3}{4}=\dfrac{2}{4}=\dfrac{1}{2}\\ b,\\ =\dfrac{2}{3}-\dfrac{2}{5}=\dfrac{10-6}{15}=\dfrac{4}{15}\\ c,\\ =\dfrac{5}{6}-\dfrac{1}{4}=\dfrac{20-6}{24}=\dfrac{14}{24}=\dfrac{7}{12}\\ d,\\ =\dfrac{4}{3}-\dfrac{1}{4}=\dfrac{12-3}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)

 cảm ơn chị ạ

Đáp án đúng : C

72/55

A=(15/10-4/10+1/10)/(18/12-8/12+1/12)

A=(6/5)/(11/12)

A=(6/5)*(12/11)

A= 72/55

minh chi biet cach tinh ra the thoi

chuc ban hoc gioi

Rút gọn rồi tính:

a) $\frac{16}{24}–\frac{1}{3}=\frac{2}{3}–\frac{1}{3}=\frac{2–1}{3}=\frac{1}{3}$     

b) $\frac{4}{5}–\frac{12}{60}=\frac{4}{5}–\frac{1}{5}=\frac{3}{5}$

Tính rồi rút gọn:

a) $\frac{17}{6}–\frac{2}{6}=\frac{17–2}{6}=\frac{15}{6}=\frac{5}{2}$             

b) $\frac{16}{15}–\frac{11}{15}=\frac{16–11}{15}=\frac{5}{15}=\frac{1}{3}$

c) $\frac{19}{12}–\frac{13}{12}=\frac{19–13}{12}=\frac{6}{12}=\frac{1}{2}$

\(A=\dfrac{2\sqrt{3}}{\sqrt{3}+1}+3\sqrt{\dfrac{1}{6}}\cdot\sqrt{\dfrac{1}{2}}-\sqrt{12}\)

\(A=\dfrac{2\sqrt{3}\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}+3\cdot\dfrac{1}{\sqrt{6}}\cdot\dfrac{1}{\sqrt{2}}-2\sqrt{3}\)

\(A=\dfrac{2\sqrt{3}\cdot\left(\sqrt{3}-1\right)}{2}+3\cdot\dfrac{1}{\sqrt{12}}-2\sqrt{3}\)

\(A=\sqrt{3}\cdot\left(\sqrt{3}-1\right)+3\cdot\dfrac{1}{2\sqrt{3}}-2\sqrt{3}\)

\(A=3-\sqrt{3}+\dfrac{3}{2\sqrt{3}}-2\sqrt{3}\)

\(A=3-3\sqrt{3}+\dfrac{\sqrt{3}}{2}\)

\(A=\dfrac{6+6\sqrt{3}+\sqrt{3}}{2}\)

\(A=\dfrac{6+7\sqrt{3}}{2}\)

1.

\(A=3log_{2^2}\sqrt{a}-log_{2^{-1}}a^2+2log_{a^{\dfrac{1}{2}}}a\)

\(=3.\dfrac{1}{2}.\dfrac{1}{2}log_2a-\left(-1\right).2.log_2a+2.2.log_2a\)

\(=\dfrac{27}{4}log_2a\)

2.

\(log_{12}36=\dfrac{log_236}{log_212}=\dfrac{log_2\left(3^2.2^2\right)}{log_2\left(3.2^2\right)}=\dfrac{log_23^2+log_22^2}{log_23+log_22^2}\)

\(=\dfrac{2.log_23+2}{log_23+2}=\dfrac{2a+2}{a+2}\)