rut go:
P=(1^4+4)(5^4+4)(9^4)...(21^4+4)/(3^4+4)(7^4+4)(11^4+4)...(23^4+4)
Những câu hỏi liên quan
Rút gọn biểu thức: P=[(1+4)(5^4+4)(9^4+4).......(21^4+4)] / [(3^4+4)(7^4+4)(11^4+4).......(23^4+4)]
Bài 1 Tính nhanh
a) -3/7 + 5/13 + 3/7
b) -5/21+-2/21+8/24
c) -5/11+(-6/11+2)
d) (-1/32+1/2)+15/32
e)5/17+ -6/13 + 3/4 + 7/-13+12/17
f) 7/23+-18/18+-4/9+16/23+-5/8
g)1/3+-3/4+3/5+-1/36+1/15+-2/9
h)-1/2+1/3+-1/4+-2/8+4/18+4/9
a)\(-\dfrac{3}{7}+\dfrac{5}{13}+\dfrac{3}{7}\)
=\(\left(-\dfrac{3}{7}+\dfrac{3}{7}\right)+\dfrac{5}{13}\)
=\(0+\dfrac{5}{13}\)
=\(\dfrac{5}{13}\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
Tính nhanh
A=2/1*5+2/5*9+2/9*13+...+2/97*101+2/101*105
B=4/2*5+4/5*2+4/8*11+...+4/44*47+4/47*50
C=3/1*3*5+3/3*5*7+3/5*7*9+...+3/19*21*23+3/21*23*25
A = 2/1*5 + 2/5*9 + ... + 2/101*105
= 1/2(4/1*5 + 4/5*9 + ... + 4/101*105)
= 1/2(1 - 1/5 + 1/5 - 1/9 + ... + 1/101 - 1/105)
= 1/2(1 - 1/105)
= 1/2 * 104/105 = 52/105
Sửa câu b. Phân số thứ 2 phải là 4/5*8
B = 4/2*5 + 4/5*8 + ... + 4/47*50
= 4/3(3/2*5 + 3/5*8 + ... + 3/47*50)
= 4/3(1/2 - 1/5 + 1/5 - 1/8 + ... + 1/47 - 1/50)
= 4/3(1/2 - 1/50)
= 4/3 * 24/50 = 16/25
Đúng 0
Bình luận (0)
Rút gọn phân thức P=\(\frac{\left(1^4+4\right)\left(5^4 +4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)
\(P=\frac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)\(=\frac{\left(1+4\right)\left(4^2+1\right)\left(6^2+1\right)\left(8^2+1\right)\left(10^2+1\right)...\left(20^2+1\right)\left(\cdot22^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)\left(6^2+1\right)\left(8^2+1\right)\left(10^2+1\right)\left(12^2+1\right)...\left(22^2+1\right)\left(24^2+1\right)}\)
\(=\frac{1+4}{\left(2^2+1\right)\left(24^2+1\right)}=\frac{5}{5\left(24^2+1\right)}=\frac{1}{24^2+1}=\frac{1}{577}\)
Đúng 1
Bình luận (0)
cái bước tách ra bn nhân lại là có kết quả y chang, VD:
\(\left(5^4+4\right)=\left(4^2+1\right)\left(6^2+1\right)=629\)
Đúng 1
Bình luận (0)
Rút gọn \(C=\dfrac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)
Tính
a) 4/3*7 + 4/7*11 + 4/11*15 + 4/15*19 + 4/19*23 + 4/23*27
b) 1/2*3 + 1/3*4 + 1/4*5 + 1/5*6 +1/6*7
c) 2/3*5 + 2/5*7 + 2/7*9 + 2/9*11 + 2/11*13 + 2/1*2 + 2/2*3 + 2/3*4 + 2/8*9 + 2/9*10
a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)
b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)
Đúng 0
Bình luận (0)
rút gọn biểu thức
P=\(\frac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)
\(P=\left(\frac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)....\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right).....\left(23^4+4\right)}\right)\). Rút gọn biểu thức
\(\frac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)....\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)....\left(23^4+4\right)}\) giúp mình nha:)
có \(a^4+4=a^4+4a^2+4-4a^2\)
=\(\left(a^2-2\right)^2-4a^2\)
=\(\left(a^2-2-2a\right)\left(a^2-2+2a\right)\)
=\(\left[\left(a-1\right)^2-1\right]\left[\left(a+1\right)^2-1\right]\)
thay vào biểu thức A ta có;
A=\(\frac{\left[\left(1-1\right)^2-1\right]\left[\left(1+1\right)^2-1\right]...\left[\left(21-1\right)^2-1\right]\left[\left(21+1\right)^2-1\right]}{\left[\left(3-1\right)^2-1\right]\left[\left(3+1\right)^2-1\right]...\left[\left(23-1\right)^2-1\right]\left[\left(23+1\right)^2-1\right]}\)
=\(\frac{-1.\left(2^2-1\right)...\left(20^2-1\right)\left(22^2-1\right)}{\left(2^2-1\right)\left(4^2-1\right)...\left(22^2-1\right)\left(24^2-1\right)}\)
=-1
kết bạn với mk nha !!!!!^-^
Đúng 0
Bình luận (0)