\(\dfrac{6^3+3\times6^5+18^3}{-13\times6^3}\)
\(\dfrac{6^3+3\times6^3+3^3}{13}\)
Ta có : \(\dfrac{6^3+3.6^3+3^3}{13}=\dfrac{6^3.\left(1+3\right)+3^3}{13}=\dfrac{6^3.4+3^3}{13}\)
=> \(\dfrac{\left(3.2\right)^3.4+3^3}{13}=\dfrac{3^3.2^3.4+3^3}{13}=\dfrac{3^3.2^5+3^3}{13}\)
=> \(\dfrac{3^3.\left(2^5+1\right)}{13}=\dfrac{3^3.33}{13}=\dfrac{891}{13}\)
\(\dfrac{6^3+3\times6^3+3^3}{13}\)
= \(\dfrac{216+648+27}{13}\)
= \(\dfrac{891}{13}\)
@Trịnh Thị Thảo Nhi
\(\dfrac{4^2}{3\times5}\times\dfrac{5^2}{4\times6}\times\dfrac{6^2}{5\times7}\times\dfrac{7^2}{6\times8}\)
A =\(\dfrac{4^2}{3\times5}\) \(\times\)\(\dfrac{5^2}{4\times6}\) \(\times\) \(\dfrac{6^2}{5\times7}\) \(\times\) \(\dfrac{7^2}{6\times8}\)
A = \(\dfrac{4\times4\times5^2\times6^2\times7\times7}{3\times4\times5^2\times6^2\times7\times8}\)
A = \(\dfrac{4}{3}\) \(\times\) \(\dfrac{7}{8}\)
A = \(\dfrac{7}{6}\)
Tính\(\frac{^{6^3+3\times6^2+3^3}}{13}\)
63+3x62+33/13
=63+31x62+33/13
=62x31x(6+32)/13
=1620/13
Em nghĩ là thế !
bài 1 :
a) \(\dfrac{2}{5}+\dfrac{3}{8}=\) b)\(\dfrac{7}{6}-\dfrac{2}{3}=\) c)\(\dfrac{5}{9}\times6\) d)\(\dfrac{8}{5}:\dfrac{4}{7}=\)
bài 2:
a) \(\dfrac{4}{5}+\) x =\(\dfrac{5}{6}\) b)x : \(\dfrac{7}{10}=5\)
bài 3 : hai xe ô tô chở được tất cả 16 tấn 8 tạ hàng . Xe ô tô thứ nhất chở được nhiều hơn xe ô tô thứ hai 2 tấn 6 tạ hàng . Hỏi mỗi xe chở được bao nhiêu tạ hàng
bài 4 :
145 \(\times\) 69 + 22 x 145 +145 x 8 + 145 =
bài 1
a)\(=\dfrac{16}{40}+\dfrac{15}{40}=\dfrac{31}{40}\)
b)\(=\dfrac{7}{6}-\dfrac{4}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)
c)\(=\dfrac{30}{9}=\dfrac{10}{3}\)
d)\(=\dfrac{8}{5}\times\dfrac{7}{4}=\dfrac{56}{20}=\dfrac{14}{5}\)
bài 2
a)\(x=\dfrac{5}{6}-\dfrac{4}{5}=\dfrac{25}{30}-\dfrac{24}{30}=\dfrac{1}{30}\)
b)\(x=5\times\dfrac{10}{7}=\dfrac{50}{7}\)
bài 4 :
145 ×× 69 + 22 x 145 +145 x 8 + 145
\(=145\times\left(69+22+8+1\right)=145\times100=14500\)
bài 1:
a, \(\dfrac{2}{5}+\dfrac{3}{8}=\dfrac{16}{40}+\dfrac{15}{40}=\dfrac{31}{40}\)
b,\(\dfrac{7}{6}-\dfrac{2}{3}=\dfrac{7}{6}-\dfrac{4}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)
c,\(\dfrac{5}{9}x6=\dfrac{5}{9}x\dfrac{6}{1}=\dfrac{30}{9}\)
d,\(\dfrac{8}{5}:\dfrac{4}{7}=\dfrac{8}{5}x\dfrac{7}{4}=\dfrac{14}{5}\)
bài 2 :
\(a,\dfrac{4}{5}+x=\dfrac{5}{6}\)
\(x=\dfrac{5}{6}-\dfrac{4}{5}\)
\(x=\dfrac{1}{30}\)
b, \(x:\dfrac{7}{10}=5\)
\(x\) \(=5x\dfrac{7}{10}\)
\(x\) \(=\dfrac{35}{10}\)
bài 3 :
đổi :16 tấn 8 tạ = 168 tạ
2 tấn 6 tạ = 26 tạ
xe ô tô thứ nhất chở số tạ hàng là:
( 168 + 26 ) : 2= 97 ( tạ)
xe ô tô thứ hai chở số tạ hàng là:
97 - 26 = 71 ( tạ)
đáp số :xe ô tô thứ nhất : 97 tạ thóc
xe ô tô thứ hai : 71 tạ thóc
Tính nhanh:\(\frac{1\times2\times3+2\times4\times6+3\times6\times9+4\times8\times12+5\times10\times15}{1\times3\times5+2\times6\times10+3\times9\times15+4\times12\times20+5\times15\times25}-\frac{1+2+3+2+4+6+3+6+9+4+8+12+5+10+15}{1+3+5+2+6+10+3+9+15+4+12+20+5+15+25}\)
Tính = cách nhanh nhất
\(\frac{6^3+3\times6^2+3^3}{-13}\)
= (2.3)^3+ 3. ( 2.3)^2 + 3^3 / -13
= 2^3 .3^3 + 3. 3^2 . 2^2 + 3^3 / -13
= 3^3. ( 2^3 + 2^2 + 1) /-13
= 27.13/-13
= -27
\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{6^2\left(6+3\right)+3^2\cdot3}{-13}=\frac{36\cdot9+9\cdot3}{-13}=\frac{39\cdot9}{-13}=-3\cdot9=-27\)
Tính :\(\frac{5^2\times6^{11}\times6^2\times12^6\times15^2}{2\times6^{12}\times10^4-81^2\times960^3}\)
CẦN GẤP !! TRÌNH BÀY ĐẦY ĐỦ !!!
rút gọn:
\(\frac{5^2\times6^{11}\times16^2+6^2\times12^6\times15^2}{2\times6^{12}\times10^4-81^2\times960^3}\)
\(\frac{5^2\times6^{11}\times16^2+6^2\times12^6\times15^2}{2\times6^{12}\times10^4-81^2\times960^3}\)
\(=\frac{5^2\times\left(2\times3\right)^{11}\times\left(2^4\right)^2+\left(2\times3\right)^2\times\left(2^2\times3\right)^6\times\left(3\times5\right)^2}{2\times\left(2\times3\right)^{12}\times\left(2\times5\right)^4-\left(3^4\right)^2\times\left(2^6\times3\times5\right)^3}\)
\(=\frac{5^2\times2^{19}\times3^{11}+2^{14}\times3^{10}\times5^3}{2^{17}\times5^4\times3^{12}-3^{11}\times2^{18}\times5^3}\)
\(=\frac{5^2\times3^{10}\times2^{14}\times\left(2^5\times3+5\right)}{2^{17}\times5^3\times3^{11}\times\left(5\times3-2\right)}\)
\(=\frac{2^5\times3+5}{2^3\times5\times3\times12}\)
\(=\frac{32\times3+5}{8\times15\times12}=\frac{96+5}{120\times12}=\frac{101}{1440}\)
Tính
a, 4\(\times\left(-\dfrac{1}{2}\right)^3-2\times\left(-\dfrac{1}{2}\right)^2+3\times\left(-\dfrac{1}{2}\right)+1\)
b, \(\dfrac{4^6\times9^5+6^9\times120}{-8^4\times3^{12}+6^{11}}\)
c, \(\dfrac{155-\dfrac{10}{7}-\dfrac{5}{11}+\dfrac{5}{23}}{403-\dfrac{26}{7}-\dfrac{13}{11}+\dfrac{12}{23}}+\dfrac{\dfrac{3}{5}+\dfrac{3}{13}-0,9}{\dfrac{7}{91}+0,2-\dfrac{3}{10}}\)
d,\(\dfrac{30\times4^7\times3^{29}-5\times14^5\times2^{12}}{54\times6^{14}\times9^7-12\times8^5\times7^5}\)
a, \(4\times\left(-\dfrac{1}{2}\right)^3-2\times\left(-\dfrac{1}{2}\right)^2+3\times\left(-\dfrac{1}{2}\right)+1\)
\(=\left(-\dfrac{1}{2}\right)\left[\left(4\times-\dfrac{1}{2}\right)-\left(2\times-\dfrac{1}{2}\right)+3\right]+1\)
\(=\left(-\dfrac{1}{2}\right)\left(-2+1+3\right)+1\)
\(=\left(-\dfrac{1}{2}\right)2+1\)
\(=-1+1\)
\(=0\)
@Trịnh Thị Thảo Nhi
a, 4×(−12)3−2×(−12)2+3×(−12)+14×(−12)3−2×(−12)2+3×(−12)+1
=(−12)[(4×−12)−(2×−12)+3]+1=(−12)[(4×−12)−(2×−12)+3]+1
=(−12)(−2+1+3)+1=(−12)(−2+1+3)+1
=(−12)2+1=(−12)2+1
=−1+1=−1+1
=0=0