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f) \(\frac{3^3.\left(0,5\right)^5}{\left(1,5\right)^4}=\frac{3^3.\left(0,5\right)^5}{\left[3.\left(0,5\right)\right]^4}=\frac{3^3.\left(0,5\right)^5}{3^4.\left(0,5\right)^4}=\frac{0,5}{3}=\frac{1}{6}\)
b) \(\frac{2^3+3.2^6-4^3}{2^3+3^2}=\frac{2^3.\left(1+3.2^3-2^3\right)}{2^3+3^2}=\frac{2^3.17}{17}=2^3=8\)
Các phần còn lại tương tự, bạn tự làm nhé !
(*) Lưu ý ở những bài rút gọn có chứa lũy thừa thì bạn đưa số đó về số nguyên tố rồi thực hiện như bình thường .
VD : \(4^3=\left(2^2\right)^3=2^6\) ( đưa về số nguyên tố là 2 )
\(6^3=\left(2.3\right)^3=2^3.3^3\) ( đưa về tích hai số nguyên tố )
a) \(\frac{5^5.20^3-5^4.20^3+5^7.4^5}{\left(20+5\right)^3.4^5}\)
= \(\frac{5^5.\left(2^2.5\right)^3-5^4.\left(2^2.5\right)^3+5^7.\left(2^2\right)^5}{\left(5^2\right)^3.\left(2^2\right)^5}\)
= \(\frac{5^5.2^6.5^3-5^4.2^6.5^3+5^7.2^{10}}{5^6.2^{10}}\)
= \(\frac{5^8.2^6-5^7.2^6+5^7.2^{10}}{5^6.2^{10}}\)
= \(\frac{5^7.2^6.\left(5-1+2^4\right)}{5^6.2^{10}}\)
= \(\frac{5.20}{2^4}=\frac{25}{4}\)
\(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)
\(=\frac{2^{12}\cdot3^{10}+2^3\cdot3\cdot5\cdot2^9\cdot3^9}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)
\(=\frac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)
\(=\frac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\left(6-1\right)}\)
\(=\frac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot5}=\frac{2^2}{5}=\frac{4}{5}\)
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
a, 205.510/1005
=205.55.55/1005
=1005.55/1005
=55
=3125
b, (0,9)5/(0,3)6
=(0,3.3)5/0,36
=0,55.35/0,36
=35/0,3
=810
c, 63+3.62+33/-13
=(2.3)3+3.(3.2)2+33/-13
=23.33+3.32.22+33/-13
=33.23+33.22+33/-13
=33(23+22+1)/-13
=27.13/-13
=-27
d, 46.95+69.120/84.312-611
=(22)6.(32)5+(2.3)9.3.23.5/(23)4.312-(2.3)11
=212.310+29.39.3.23.5/212.312-211.311
=212.310+212.310.5/211.311.2.3-211.311
=212.310.(1+5)/211.311(6-1)
=212.310.6/211.311.5
=2.6/3.5
=12/15
=4/5
Tính:
2) \(\left(\frac{2}{3}\right)^3-\left(\frac{3}{4}\right)^2.\left(-1\right)^5\)
\(=\frac{8}{27}-\frac{9}{16}.\left(-1\right)\)
\(=\frac{8}{27}-\left(-\frac{9}{16}\right)\)
\(=\frac{371}{432}.\)
Xin lỗi, anh chỉ làm câu này thôi em.
Chúc em học tốt!
a)\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{3^3\cdot2^3+3^3\cdot2^2+3^3}{-13}=\frac{3^3\left(2^3+2^2+1\right)}{-13}=-3^3=-27\)
b) \(\frac{2^3+3\cdot2^6-4^3}{2^3+3^2}=\frac{8+3\cdot64-64}{8+9}=\frac{8+192-64}{17}=\frac{136}{17}=8\)
c) \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{11}\cdot3^{10}\left(2+2\cdot5\right)}{2^{11}\cdot3^{10}\cdot\left(2\cdot3^2-3\right)}=\frac{12}{18-3}=\frac{12}{15}\)
d) \(\frac{5^5\cdot20^3-5^4\cdot20^3+5^7\cdot4^5}{\left(20+5\right)^3\cdot4^5}=\frac{5^5\cdot20^3-5^4\cdot20^3+20^3\cdot20^2\cdot5^2}{5^6\cdot4^5}=\frac{20^3\left(5^5-5^4+5^4\cdot4^2\right)}{20^5\cdot5}\)\(=\frac{5^4\left(5-1+16\right)}{20^2\cdot5}=\frac{5^4\cdot20}{20^2\cdot5}=\frac{5^3}{20}=\frac{5^3}{5\cdot4}=\frac{25}{4}\)
Bài giải
a)\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{3^3\cdot2^3+3^3\cdot2^2+3^3}{-13}=\frac{3^3\left(2^3+2^2+1\right)}{-13}=-3^3=-27\)
b) \(\frac{2^3+3\cdot2^6-4^3}{2^3+3^2}=\frac{8+3\cdot64-64}{8+9}=\frac{8+192-64}{17}=\frac{136}{17}=8\)
c) \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{11}\cdot3^{10}\left(2+2\cdot5\right)}{2^{11}\cdot3^{10}\cdot\left(2\cdot3^2-3\right)}=\frac{12}{18-3}=\frac{12}{15}\)
d) \(\frac{5^5\cdot20^3-5^4\cdot20^3+5^7\cdot4^5}{\left(20+5\right)^3\cdot4^5}=\frac{5^5\cdot20^3-5^4\cdot20^3+20^3\cdot20^2\cdot5^2}{5^6\cdot4^5}=\frac{20^3\left(5^5-5^4+5^4\cdot4^2\right)}{20^5\cdot5}\)\(=\frac{5^4\left(5-1+16\right)}{20^2\cdot5}=\frac{5^4\cdot20}{20^2\cdot5}=\frac{5^3}{20}=\frac{5^3}{5\cdot4}=\frac{25}{4}\)