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a) = (3.23)3.3-(2.3)4.25
= 33.(23)3.3-24.34.25
= 34.29-29.34
=0
1/ Để cho \(\left(n^2+3\right)⋮\left(n+1\right)\) thì
\(A=\frac{n^2+3}{n+1}\) là 1 số nguyên
Ta có: \(A=\frac{n^2+3}{n+1}=n-1+\frac{4}{n+1}\)
Để A nguyên thì (n + 1) phải là ước nguyên của 4 hay
\(\left(n+1\right)=\left(-4,-2,-1,1,2,4\right)\)
\(\Rightarrow x=\left(-5,-3,-2,0,1,3\right)\)
\(a,a^3\cdot a^9=a^{12}\)
\(b,\left(a^5\right)^7=a^{35}\)
\(c,\left(a^6\right)^4\cdot a^{12}=a^{24}\cdot a^{12}=a^{36}\)
\(d,4\cdot5^2-2\cdot3^2=2^2\cdot5-2\cdot3^2=2\cdot\left(2\cdot5+3^2\right)=2\cdot19=38\)
\(e,5^6:5^3+3^3\cdot3^2=5^3+3^5\)
a,a3.a9=a3+9=a12
b,(a5)7=a5.7=a35
Mấy câu tiếp theo bn lám tương tự!
a) \(=\frac{6}{7}+\frac{5}{8}:5-\frac{3}{16}.4\)
\(=\frac{6}{7}+\frac{1}{8}-\frac{3}{4}\)
\(=\frac{13}{56}\)
b) \(=\frac{3}{2}+\frac{1}{2}.\frac{7}{18}:\frac{7}{12}\)
\(=\frac{3}{2}+\frac{1}{3}\)
\(=\frac{11}{6}\)
\(1)A=a\frac{1}{3}+a\frac{1}{4}-a\frac{1}{6}=a\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{6}\right)=a\frac{5}{12}\)
Thay \(a=-\frac{3}{5}\) vào A,ta đc:
\(A=-\frac{3}{5}.\frac{5}{12}=-\frac{1}{4}\)
\(2)B=b\frac{5}{6}+b\frac{3}{4}-b\frac{1}{2}=b\left(\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)=b\frac{13}{12}\)
Thay \(b=\frac{12}{13}\) vào B, ta đc: \(B=b\frac{13}{12}=\frac{12}{13}.\frac{13}{12}=1\)
A=\(\frac{72^3.54^2}{108^4}=\frac{\left(2^3.3^2\right)^3.\left(2.3^3\right)^2}{\left(2^2.3^3\right)^4}=\frac{2^9.3^6.2^2.3^6}{2^8.3^{12}}=\frac{2^{11}.3^{12}}{2^8.3^{12}}=2^3=8\)
B= \(\frac{4^6.3^4.9^5}{6^{12}}=\frac{2^{12}.3^4.3^{10}}{2^{12}.3^{12}}=\frac{2^{12}.3^{14}}{2^{12}.3^{12}}=3^2=9\)
c) \(\frac{2^{13}+2^5}{2^{10}+2^2}=\frac{2^5\left(2^8+1\right)}{2^2\left(2^8+1\right)}=2^3=8\)
1.
\(\frac{72^3\times54^2}{108^4}=\frac{\left(8\times9\right)^3\times\left(27\times2\right)^2}{\left(27\times4\right)^4}=\frac{\left(2^3\times3^2\right)^3\times\left(3^3\times2\right)^2}{\left(3^3\times2^2\right)^4}=\frac{\left(2^3\right)^3\times\left(3^2\right)^3\times\left(3^3\right)^2\times2^2}{\left(3^3\right)^4\times\left(2^2\right)^4}=\frac{2^9\times3^6\times3^6\times2^2}{3^{12}\times2^8}=2^3=8\)
2.
\(\frac{4^6\times3^4\times9^5}{6^{12}}=\frac{\left(2^2\right)^6\times3^4\times\left(3^2\right)^5}{\left(2\times3\right)^{12}}=\frac{2^{12}\times3^4\times3^{10}}{2^{12}\times3^{12}}=3^2=9\)
3.
\(\frac{2^{13}+2^5}{2^{10}+2^2}=\frac{2^5\times\left(2^8+1\right)}{2^2\times\left(2^8+1\right)}=2^3=8\)
\(\left(2\right)K=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(K=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)
\(K=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(K=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(\left(3\right)L=\frac{5}{1\cdot4}+\frac{5}{4\cdot7}+\frac{5}{7\cdot10}+...+\frac{5}{100\cdot103}\)
\(L=\frac{5}{3}\cdot\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(L=\frac{5}{3}\cdot\left(\frac{1}{1}-\frac{1}{103}\right)=\frac{5}{3}\cdot\frac{102}{103}=\frac{510}{309}=\frac{170}{103}\)
Trả lời:
2,\(K=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(K=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(K=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(K=\frac{1}{2}-\frac{1}{100}\)
\(K=\frac{49}{100}\)
3,\(L=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)
\(L=\frac{5}{3}\times\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
\(L=\frac{5}{3}\times\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(L=\frac{5}{3}\times\left(\frac{1}{1}-\frac{1}{103}\right)\)
\(L=\frac{5}{3}\times\frac{102}{103}\)
\(L=\frac{170}{103}\)
Học tốt
\(160-\left(2^3\cdot5^2-6\cdot25\right)\)
= \(160-\left(8\cdot25-6\cdot25\right)\)
= \(160-25\left(8-6\right)\)
= \(160-25\cdot2=160-50=110\)
\(160-\left(2^3.5^2-6.25\right)\)
\(=160-\left(8.25-6.25\right)\)
\(=160-25.\left(8-6\right)\)
\(=160-25.2\)
\(=160-50\)
\(=110\)