Tính nhanh:
\(A=\dfrac{1.5.6+2.10.12+3.5.18+4.20.24+5.25.30}{1.3.5+2.6.10+3.9.15+4.2.10+5.15.25}\)
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\(A=\frac{\text{1.5.6 + 2.10.12 + 3.15.18 + 4.20.24 + 5.25.30}}{\text{1.3.5 + 2.6.10 + 3.9.15 + 4.12.20 + 5.15.25 }}\)
\(=\frac{1.5.6+2.\left(1.5.6\right)+3.\left(1.5.6\right)+4.\left(1.5.6\right)+5.\left(1.5.6\right)}{1.3.5+2.\left(1.3.5\right)+3.\left(1.3.5\right)+4.\left(1.3.5\right)+5.\left(1.3.5\right)}\)
\(=\frac{30.\left(1+2+3+4+5\right)}{15.\left(1+2+3+4+5\right)}\)
\(=\frac{30}{15}=2\)
Vậy A=2.
\(=\frac{1.5.6+\left(1.5.6\right).2+\left(1.5.6\right).3+\left(1.5.6\right).4+\left(1.5.6\right).5}{1.3.5+\left(1.3.5\right).2+\left(1.3.5\right).3+\left(1.3.5\right).4+\left(1.3.5\right).5}\)
\(=\frac{\left(1.5.6\right).\left(1+2+3+4+5\right)}{\left(1.3.5\right).\left(1+2+3+4+5\right)}=\frac{1.5.6}{1.3.5}=\frac{1.1.2}{1.1.1}=2\)
A= \(\frac{1.5.3.2+2.10.2.6+2.15.9.2+4.20.12.2+5.25.15.2}{1.3.5+2.6.10+3.9.15+4.12.20+5.15.25}\)
A= \(\frac{2+2+2\cdot2+2+2}{0+0+3+0+0}\)
A= \(\frac{12}{3}\)
A= 4
Đầu tiên bạn tách ra, rút gọn rồi cộng lại,tính nha!
\(\frac{1\cdot5\cdot6+2\cdot10\cdot12+4\cdot20\cdot24+9\cdot45\cdot54}{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20+9\cdot27\cdot45}=\frac{1\cdot5\cdot6\cdot\left(1+2+4+9\right)}{1\cdot3\cdot5\cdot\left(1+2+4+9\right)}=2\)
\(A=\frac{1.5.6+2.10.12+4.20.24+9.45.54}{1.3.5+2.6.10+4.12.20+9.27.45}\)
\(A=\frac{1.5.6.\left(1^3+2^3+4^3+9^3\right)}{1.3.5.\left(1^3+2^3+4^3+9^3\right)}=2\)
\(\frac{1.5.6+2.1.5.2.6.2+4.1.5.6.4.4+9.1.5.6.9.9}{1.3.5+2.2.2.1.3.5+4.4.4.1.3.5+9.9.9.1.3.5}\) (chỗ này bạn đánh sai đề bài)
\(\frac{1.5.6\left(1+8+64+729\right)}{1.3.5\left(1+8+64+729\right)}\)
\(\frac{30}{15}=2\)
\(\frac{1.5.6+2.10.12+4.20.24+9.45.54}{1.3.5+2.6.10+4.12.20+9.27.45}\)
\(=\frac{1.3.5.2+2.6.10.2+4.12.20.2+9.27.45.2}{1.3.5+2.6.10+4.12.20+9.27.45}\)
\(=\frac{2\left(1.3.5+2.6.10+4.12.20+9.27.45\right)}{1.3.5+2.6.10+4.12.20+9.27.45}\)
\(=2\)
Bạn viết sai đề chỗ dấu chấm giữa số 20 và số 9 ở mẫu nên mình sửa lại thành dấu cộng rồi nha.
A=\(\dfrac{1.5.6+2.\left(1.5.6\right)+3.\left(1.5.6\right)+4.\left(1.5.6\right)+5.\left(1.5.6\right)}{1.3.5+2.\left(1.3.5\right)+3.\left(1.3.5\right)+4.\left(1.3.5\right)+5.\left(1.3.5\right)}\)
A=\(\dfrac{\left(1+2+3+4+5\right).\left(1.5.6\right)}{\left(1+2+3+4+5\right).\left(1.3.5\right)}\) = \(\dfrac{1.5.6}{1.3.5}\) = 2