Rút gọn đa thức sau:
\(A=\left(5+3\right)\cdot\left(5^2+3^2\right)\cdot\left(5^3+3^3\right)\cdot......\cdot\left(5^{64}+3^{64}\right)-\frac{5^{128}+3^{128}}{2}\)
Những câu hỏi liên quan
Rút gọn biểu thức A= \(\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}\right)^2\cdot\left(-1\right)^{2017}}{\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}-\frac{71}{5}\)
26 tháng 8 2023 lúc 15:21
Rút gọn biểu thức
\(\left(1-\dfrac{1}{1+2}\right)\cdot\left(1-\dfrac{1}{1+2+3}\right)\cdot\cdot\cdot\cdot\left(1-\dfrac{1}{1+2+3+4+5+.....+2006}\right)\)
Giúp em với ạ
\(\left(1-\dfrac{1}{1+2}\right)\cdot\left(1-\dfrac{1}{1+2+3}\right)\cdot\left(\dfrac{1}{1+2+3+...+2006}\right)\)
\(=\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{6}\right)\cdot\left\{\dfrac{1}{\left(2006+1\right)\left[\left(2006-1\right):1+1\right]}\right\}\)
\(=\dfrac{2}{3}\cdot\dfrac{5}{6}\cdot\dfrac{1}{2007\cdot2006}\)
\(=\dfrac{10}{18}\cdot\dfrac{1}{4026042}\)
\(=\dfrac{5}{9}\cdot\dfrac{1}{4026042}\)
\(=\dfrac{5}{36234378}\)
Đúng 0
Bình luận (0)
\(\left[6\cdot\left(-\frac{1}{3}\right)^2-3\cdot\left(-\frac{1}{3}\right)+1\right]:\left(-\frac{1}{3}-1\right)\)
\(\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}\right)^2\cdot\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}\)
\(\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}^2\right)\cdot\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}\)
\(=\frac{\frac{8}{27}\cdot\frac{9}{16}\cdot\left(-1\right)}{\frac{4}{25}\cdot\left(-\frac{125}{1728}\right)}\)
\(=\frac{-\frac{1}{6}}{-\frac{5}{432}}=-\frac{1}{6}:\left(-\frac{5}{432}\right)=\frac{72}{5}\)
Đúng 0
Bình luận (0)
\(\left[6.\left(\frac{-1}{3}\right)^2-3.\left(\frac{-1}{3}\right)+1\right]:\left(\frac{-1}{3}-1\right)\)
\(=\left[6.\frac{1}{9}-\left(-1\right)+1\right]:\frac{-4}{3}\)
\(=\left[\frac{2}{3}-\left(-1\right)+1\right]:\frac{-4}{3}\)
\(=\frac{8}{3}:\frac{-4}{3}=\frac{-24}{12}=-2\)
~ Hok tốt ~
Đúng 0
Bình luận (0)
\(\left[6\cdot\left(-\frac{1}{3}\right)^2-3\cdot\left(-\frac{1}{3}\right)+1\right]:\left(-\frac{1}{3}-1\right)\)
\(=\left[6\cdot\left(-\frac{1}{9}\right)+1+1\right]:\left(-\frac{4}{3}\right)\)
\(=\left(-\frac{2}{3}+2\right):\left(-\frac{4}{3}\right)\)
\(=\frac{4}{3}:\left(-\frac{4}{3}\right)=-1\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
a)\(12\cdot\left(-\frac{2}{3}\right)^2+\frac{4}{3}\)
b)\(12,5\cdot\left(-\frac{5}{7}\right)+1,5\cdot\left(-\frac{5}{7}\right)\)
c)\(1:\left(\frac{2}{3}-\frac{3}{4}\right)^2\)
d)\(15\cdot\left(-\frac{2}{3}\right)^2-\frac{7}{3}\)
e)\(\frac{1}{2}\sqrt{64}-\sqrt{\frac{4}{25}}+\left(-1\right)^{2007}\)
a) 12. \(\frac{4}{9}\)+\(\frac{4}{3}\)=\(\frac{16}{3}\)+\(\frac{4}{3}\)=\(\frac{20}{3}\)
b) (\(\frac{-5}{7}\)) . (12,5+1,5)= (\(\frac{-5}{7}\)).14=-10
a) \(12.\left(-\frac{2}{3}\right)^2+\frac{4}{3}=12.\frac{4}{9}+\frac{4}{3}=\frac{16}{3}+\frac{4}{3}=\frac{20}{3}\)
b) \(12,5.\left(-\frac{5}{7}\right)+1,5.\left(-\frac{5}{7}\right)=-\frac{5}{7}.\left(12,5+1,5\right)=-\frac{5}{7}.14=-10\)
c) \(1:\left(\frac{2}{3}-\frac{3}{4}\right)^2=1:\left(-\frac{1}{12}\right)^2=1:\frac{1}{144}=1.144=144\)
d) \(15.\left(-\frac{2}{3}\right)^2-\frac{7}{3}=15.\frac{4}{9}-\frac{7}{3}=\frac{20}{3}-\frac{7}{3}=\frac{13}{3}\)
e) \(\frac{1}{2}\sqrt{64}-\sqrt{\frac{4}{25}}+\left(-1\right)^{2007}=\frac{1}{2}.8-\frac{2}{5}+\left(-1\right)=4-\frac{2}{5}-1=\frac{13}{5}\)
Rút gọn phân số sau: \(\frac{\left(2^{17}+5^{17}\right)\cdot\left(3^{14}-5^{12}\right)\cdot\left(2^4-4^2\right)}{15^2+5^3+67^7}\)
\(\frac{\left(2^{17}+5^{17}\right)\left(3^{14}-5^{12}\right)\left(2^4-4^2\right)}{15^2+5^3+67^7}=\frac{\left(2^{17}+5^{17}\right)\left(3^{14}-5^{12}\right).0}{15^2+5^3+67^7}=0\)
Đúng 0
Bình luận (0)
Phân tích thành nhân tử ;
1, left(x+2right)cdotleft(x+3right)cdotleft(x+4right)cdotleft(x+5right)-24
2, xcdotleft(x+4right)cdotleft(x+6right)cdotleft(x+10right)+128
3, left(x^2+5x+6right)cdotleft(x^2-15x+56right)-144
4, left(x-18right)cdotleft(x-7right)cdotleft(x+35right)cdotleft(x+90right)-67x^2
5, left(x-2right)cdotleft(x-3right)cdotleft(x-4right)cdotleft(x-6right)-72x^2
Đọc tiếp
Phân tích thành nhân tử ;
1, \(\left(x+2\right)\cdot\left(x+3\right)\cdot\left(x+4\right)\cdot\left(x+5\right)-24\)
2, \(x\cdot\left(x+4\right)\cdot\left(x+6\right)\cdot\left(x+10\right)+128\)
3, \(\left(x^2+5x+6\right)\cdot\left(x^2-15x+56\right)-144\)
4, \(\left(x-18\right)\cdot\left(x-7\right)\cdot\left(x+35\right)\cdot\left(x+90\right)-67x^2\)
5, \(\left(x-2\right)\cdot\left(x-3\right)\cdot\left(x-4\right)\cdot\left(x-6\right)-72x^2\)
1,(x+2)(x+5)(x+3)(x+4)-24=(x2+7x+10)(x2+7x+12)-24
Đặt x2+7x+10= t ta có t(t+2)-24=t2+2t-24=(t-4)(t+6)
hay (x2+7x+6)(x2+7x+16)
2,x(x+10)(x+4)(x+6)+128=(x2+10x)(x2+10x+24)+128
Đặt x2+10x=t ta có t(t+24)+128=t2+24t+128=(t+8)(t+16)
hay (x2+10x+8)(x2+10x+16)
3,(x+2)(x-7)(x+3)(x-8)-144=(x2-5x-14)(x2-5x-24)-144
Đặt x2-5x-14=t ta có t(t-10)-144=t2-10t-144=(t-18)(t+8)
Hay (x2-5x-32)(x2-5x-6)=(x2-5x-32)(x+1)(x-6)
Đúng 0
Bình luận (0)
Gái xinh review app chất cho cả nhà đây: https://www.facebook.com/watch/?v=485078328966618 Link tải app: https://www.facebook.com/watch/?v=485078328966618
Đúng 0
Bình luận (0)
Tính giá trị của biểu thức:a,(32)2-(-23)2-(-52)3b,left|frac{-1}{2}right|^2cdotleft(-32right)-left(-8right)+left|frac{1}{2}right|^3c,2^3+3cdotleft(frac{-5}{86}right)^0cdotleft(frac{1}{2}right)^2cdot4+left[left(-2right)^2:frac{1}{2}right]:8d,left|frac{5}{7}cdotleft(-14right)right|-left(frac{2}{3}right)^2cdotleft(-18right)+6^2cdotfrac{-1}{18}
Đọc tiếp
Tính giá trị của biểu thức:
a,(32)2-(-23)2-(-52)3
b,\(\left|\frac{-1}{2}\right|^2\cdot\left(-32\right)-\left(-8\right)+\left|\frac{1}{2}\right|^3\)
c,\(2^3+3\cdot\left(\frac{-5}{86}\right)^0\cdot\left(\frac{1}{2}\right)^2\cdot4+\left[\left(-2\right)^2:\frac{1}{2}\right]:8\)
d,\(\left|\frac{5}{7}\cdot\left(-14\right)\right|-\left(\frac{2}{3}\right)^2\cdot\left(-18\right)+6^2\cdot\frac{-1}{18}\)
So sánh hai số bằng cách vận dụng hằng đẳng thức :
A = \(4\cdot\left(3^2+1\right)\cdot\left(3^4+1\right).................\cdot\left(3^{64}+1\right)\)
và B = \(3^{128}-1\)
\(A=4\left(3^2+1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)
\(2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\)
\(2A=\left(3^{32}-1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\)
\(2A=\left(3^{64}-1\right)\left(3^{64}+1\right)\)
\(2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}< 3^{128}-1=B\)
Vậy \(A< B\)
Đúng 0
Bình luận (0)
Rút gọn:
\(\frac{1}{\left(x+y\right)^3}\cdot\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\frac{3}{\left(x+y\right)^4}\cdot\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+\frac{6}{\left(x+y\right)^5}\cdot\left(\frac{1}{x}+\frac{1}{y}\right)\)