\(1+2+3+4x10=.......\)
\(1+2+3+4x10=.......\)
1+2+3+4x10
=1+2+3+40
=3+3+40
=6+40
=46
k mình nha
1 + 2 + 3 + 4 x 10
= 1 + 2 + 3 + 40
= 6 + 40
= 46
1+2+3+4 * 10
= 1+2+3+40
=3+3+40
=6+40
=46
k nha
so sánh 2/1x5+3/5x11+4/11x19+5/19x29+6/29x41 và 1/1x4+2/4x10+3/10x19+4/19x31
Đặt \(A=\dfrac{2}{1.5}+\dfrac{3}{5.11}+\dfrac{4}{11.19}+\dfrac{5}{19.29}+\dfrac{6}{29.41}\)
\(2A=\dfrac{4}{1.5}+\dfrac{6}{5.11}+\dfrac{8}{11.19}+\dfrac{10}{19.29}+\dfrac{12}{29.41}\)
\(2A=\dfrac{1}{1}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{29}+\dfrac{1}{29}-\dfrac{1}{41}\)
\(2A=\dfrac{1}{1}-\dfrac{1}{41}=\dfrac{40}{41}\)
\(A=\dfrac{40}{41}:2=\dfrac{20}{41}\)
Đặt \(B=\dfrac{1}{1.4}+\dfrac{2}{4.10}+\dfrac{3}{10.19}+\dfrac{4}{19.31}\)
Tương tự ta dễ dàng tình đc:
\(B=\dfrac{10}{31}=\dfrac{20}{62}\)
Ta có: \(\dfrac{20}{41}< \dfrac{20}{62}\Rightarrow A< B\)
Vậy ........................
Chúc bn học tốt
Đặt A=2/1.5+3/5.11+...+6/29.41
B=1/1.4+2/4.10+3/10.19+4/19.31
2A=4/1.5+6/5.11/+...+12/29.41
2A=1-1/5+1/5-1/11+...+1/29-1/41
2A=1-1/41
2A=40/41
A=20/41
3B=3/1.4+6/4.10+9/10.19+12/19.31
3B=1-1/4+1/4-1/10+1/10-1/19+1/19-1/31
3B=1-1/31
3B=30/31
B=10/31
Vì 10/31<20/41 nên B<A
Giải:
Đặt \(A=\dfrac{2}{1.5}+\dfrac{3}{5.11}+\dfrac{4}{11.19}+\dfrac{5}{19.29}+\dfrac{6}{29.41}\)
\(A=\left(\dfrac{4}{1.5}+\dfrac{6}{5.11}+\dfrac{8}{11.19}+\dfrac{10}{19.29}+\dfrac{12}{29.41}\right):2\)
\(A=\left(\dfrac{1}{1}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{29}+\dfrac{1}{29}-\dfrac{1}{41}\right):2\)
\(A=\left(\dfrac{1}{1}-\dfrac{1}{41}\right):2\)
\(A=\dfrac{40}{41}:2\)
\(A=\dfrac{20}{41}\)
Đặt \(B=\dfrac{1}{1.4}+\dfrac{2}{4.10}+\dfrac{3}{10.19}+\dfrac{4}{19.31}\)
\(B=\left(\dfrac{3}{1.4}+\dfrac{6}{4.10}+\dfrac{9}{10.19}+\dfrac{12}{19.31}\right):3\)
\(B=\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}\right):3\)
\(B=\left(\dfrac{1}{1}-\dfrac{1}{31}\right):3\)
\(B=\dfrac{30}{31}:3\)
\(B=\dfrac{10}{31}\)
Vì \(\dfrac{20}{41}>\dfrac{10}{31}\) nên A>B
2/5x10/7 x-3/4x10/7+5/2:2
Lời giải:
\(\frac{2}{5}\times \frac{10}{7}\times \frac{-3}{4}\times \frac{10}{7}+\frac{5}{2}:2=\frac{-30}{49}+\frac{5}{4}=\frac{125}{196}\)
Em nên gõ công thức trực quan để đề bài được rõ ràng nhé
9x10^4+4x10^2 = bao nhiêu và có chia hết cho 3 không?
Tìm x biết x2 -(9+2)2=4x10+23
x2-9-(-2)2=4x10+23
x2-9-4=40+8
x2 =40+8+4+9=61
x=\(\sqrt{61}\)
Tính nhanh
7^3x9+3^2x7^4-45x539
5x10^4+4x10^3+6x10^2+5x10+9
7x7x7x7x7x7x7x7x7x(40-4x10)
2x3=... 8-4=... 4x10=...
5x1=... 2x9=... 5x2=...
\(2\times3=6\)
\(8-4=4\)
\(4\times10=40\)
\(5\times1=5\)
\(2\times9=18\)
\(5\times2=10\)
6x10^7+5x10^5+4x10^3x2x10