Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(\frac{64}{2^n}=16\Leftrightarrow\frac{64}{2^n}=\frac{64}{4}\Leftrightarrow2^n=4\Leftrightarrow n=2\)
b, \(\left(\frac{1}{3}\right)^{2n-1}=\left(\frac{1}{3}\right)^3\Leftrightarrow2n-1=3\Leftrightarrow n=2\)
a)\(\frac{64}{2^n}=16\Leftrightarrow2^n.16=64\Leftrightarrow2^n=4\Leftrightarrow2^n=2^2\Leftrightarrow n=2\)
b)\(\left(\frac{1}{3}\right)^{2n-1}=\frac{1}{27}\)
\(\Leftrightarrow\left(\frac{1}{3}\right)^{2n-1}=\left(\frac{1}{3}\right)^3\)
\(\Leftrightarrow2n-1=3\Leftrightarrow2n=4\Leftrightarrow n=2\)
a, \(6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)=6\frac{5}{7}-1\frac{3}{4}-2\frac{5}{7}=6\frac{5}{7}-2\frac{5}{7}-\frac{7}{4}=4-\frac{7}{4}=\frac{16}{4}-\frac{7}{4}=\frac{9}{4}\)
b, (-4/5+4//3)+(-5/4+14/5)-7/3 = -4/5+4/3+-5/4+14/5-7/3=(-4/5+14/5)+(4/3-7/3)+-5/4=2+(-1)+-5/4=1+-5/4=-1/4
Phần b dấu ''/'' là dấu phần nha , mk viết ra phân số thì nó dài quá nên mk viết vậy , t.i.c.k mk hennnn ^^ thanks @@
Nếu thích kb qua face vs mk nick : Akiko Ayano nhé . Kb và ib nch vs nhau nè
Bài 1:
a) b) c) sẽ có bạn giải cho em thôi vì nó dễ tính tay cũng đc
d) \(\frac{4}{2.5}+\frac{4}{5.8}+...+\frac{4}{23.26}\)
\(=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{23.26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{23}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\frac{6}{13}\)
\(=\frac{8}{13}\)
Bài 2:
a) b) c)
d)\(|\frac{5}{8}x+\frac{6}{7}|-\frac{4}{7}=\frac{10}{7}\)
\(\Leftrightarrow|\frac{5}{8}x+\frac{6}{7}|=2\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x+\frac{6}{7}=2\\\frac{5}{8}x+\frac{6}{7}=-2\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x=\frac{8}{7}\\\frac{5}{8}x=\frac{-20}{7}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{64}{35}\\x=\frac{-32}{7}\end{cases}}}\)
Vậy \(x\in\left\{\frac{64}{35};\frac{-32}{7}\right\}\)
Bài 1 :
a) \(\left(\frac{2}{5}-\frac{5}{8}\right):\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-9}{40}:\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-27}{44}+\frac{1}{8}\)
\(=\frac{-43}{88}\)
a) Ta có:
\(x-\left\{\left[-x-\left(x+3\right)\right]-\left[\left(x+2018\right)-\left(x+2019\right)\right]+21\right\}\)
\(=x-\left\{\left[-x-x-3\right]-\left[x+2018-x-2019\right]+21\right\}\)
\(=x-\left\{\left[-2x-3\right]-\left[2018-2019\right]+21\right\}\)
\(=x+2x+-3+1-21\)
\(=3x-23\)
=> \(3x-23=2020\)
\(3x=2020+23=2043\)
=> \(x=2043:3=681\)
Nhầm
\(=x-\left\{-2x-3+1+21\right\}\\ =x+2x+3-1-21\)
\(=3x-17\\ =>3x-17=2020\\ 3x=2020+17=2037\\ x=2037:3=679\)
a,-3/5.2/7+-3/7.3/5+-3/7
=-3/7.2/5+(-3/7).3/5+(-3/7)
=-3/7(2/5+3/5+1)
=-3/7.2
=-6/7
Ta có :
\(B=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+...+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)
\(=\frac{12}{4.16}+\frac{20}{16.36}+...+\frac{388}{9216.9604}+\frac{396}{9604.10000}\)
\(=\frac{1}{4}-\frac{1}{16}+\frac{1}{16}-\frac{1}{36}+...+\frac{1}{9604}-\frac{1}{10000}\)
\(=\frac{1}{4}-\frac{1}{10000}< \frac{1}{4}\)
\(\Leftrightarrow B< \frac{1}{4}\)
B=\(\frac{12}{4.16}\)+\(\frac{20}{16.36}\)+...+\(\frac{396}{9604.10000}\)
Ta có:\(\frac{12}{4.16}\)=\(\frac{1}{4}\)-\(\frac{1}{16}\)
\(\frac{20}{16.36}\)=\(\frac{1}{16}\)-\(\frac{1}{36}\)
...
Khi đó:B=\(\frac{1}{4}\)-\(\frac{1}{16}\)+\(\frac{1}{16}\)-\(\frac{1}{36}\)+...+\(\frac{1}{9604}\)-\(\frac{1}{10000}\)=\(\frac{1}{4}\)-\(\frac{1}{10000}\)<\(\frac{1}{4}\)
Vậy: B<\(\frac{1}{4}\)
Đặt : \(ƯCLN\left(a,b\right)=d\)
\(\Rightarrow a=d.m\)\(;\)\(b=d.n\)\(\left(m,n\in N;\left(a,b\right)=1;m>n\right)\)
\(\Rightarrow BCNN\left(a,b\right)=d.m.n\)
Ta có : \(\frac{ƯCLN\left(a,b\right)}{BCNN\left(a,b\right)}=\frac{1}{6}\)
\(\Rightarrow\frac{d}{d.m.n}=\frac{1}{6}\)
\(\Rightarrow m.n=6\)
\(\Rightarrow a-b=d\left(m-n\right)=5\)
Ta lại có : \(\left(m,n\right)=1\)\(;\)\(m.n=6\)\(;\)\(m>n\)
\(\Rightarrow\left(m,n\right)\in\left\{\left(6;1\right);\left(3;2\right)\right\}\)
Xét từng TH :
+) TH1 : \(m=6\)\(;\)\(n=1\)
\(\Rightarrow d\left(m-n\right)=5\)
\(\Rightarrow d\left(6-1\right)=5\)
\(\Rightarrow d.5=5\)
\(\Rightarrow d=1\)
\(\Rightarrow a=d.m=1.6=6\)
\(\Rightarrow b=d.n=1.1=1\)
+) TH2 : \(m=3\)\(;\)\(n=2\)
\(\Rightarrow d\left(m-n\right)=5\)
\(\Rightarrow d\left(3-2\right)=5\)
\(\Rightarrow d.1=5\)
\(\Rightarrow d=5\)
\(\Rightarrow a=d.m=5.3=15\)
\(\Rightarrow b=d.n=5.2=10\)
Vậy \(\left(a,b\right)\in\left\{\left(6;1\right);\left(15;10\right)\right\}\)
Cho mk hỏi
BCNN(a,b)=a.b=d.n.d.m
Thì sao có thể =d.n.m được
Chúc bn học tốt
Thanks bn nhiều