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a) \(\frac{4}{3}-\frac{2}{5}\)
\(=\frac{20}{15}-\frac{6}{15}=\frac{14}{15}\)
b) \(\left|-\frac{1}{10}\right|-\left(-\frac{1}{3}\right)^2\div\frac{5}{9}\)
\(=\frac{1}{10}-\frac{1}{9}\cdot\frac{9}{5}\)
\(=\frac{1}{10}-\frac{1}{5}=\frac{1}{10}-\frac{2}{10}\)
\(=-\frac{1}{10}\)
c) Đề bài có vấn đề!!!
d) \(\left(-0,2\right)^2\cdot5-8^2\cdot\frac{9^4}{3^7}\cdot4^3\)
\(=0,04\cdot5-64\cdot\frac{\left(3^2\right)^4}{3^7}\cdot64\)
\(=0,2-4096\cdot\frac{3^8}{3^7}=0,2-4096\cdot3\)
\(=0,2-12288=-128878\)
3.
a) \(\left(x-1\right)^3=125\)
=> \(\left(x-1\right)^3=5^3\)
=> \(x-1=5\)
=> \(x=5+1\)
=> \(x=6\)
Vậy \(x=6.\)
b) \(2^{x+2}-2^x=96\)
=> \(2^x.\left(2^2-1\right)=96\)
=> \(2^x.3=96\)
=> \(2^x=96:3\)
=> \(2^x=32\)
=> \(2^x=2^5\)
=> \(x=5\)
Vậy \(x=5.\)
c) \(\left(2x+1\right)^3=343\)
=> \(\left(2x+1\right)^3=7^3\)
=> \(2x+1=7\)
=> \(2x=7-1\)
=> \(2x=6\)
=> \(x=6:2\)
=> \(x=3\)
Vậy \(x=3.\)
Chúc bạn học tốt!
a) \(\frac{75^3.3^7}{81^4.5^6}=\frac{5^3.3^3.5^3.3^7}{\left(3^4\right)^4.5^6}=\frac{5^6.3^3.3^7}{3^{16}.5^6}=\frac{3^{10}}{3^{16}}=\frac{1}{3^6}=\frac{1}{729}\)
b) \(\frac{6^6.4^2}{3^{12}.2^8}=\frac{2^6.3^6.\left(2^2\right)^2}{3^{12}.2^8}=\frac{2^6.3^6.2^4}{3^{12}.2^8}=\frac{2^{10}.3^6}{3^{12}.2^8}=\frac{2^2.1}{3^6}=\frac{4}{729}\)
c) \(\frac{34^5.2^5}{2^{14}.17^5}=\frac{2^5.17^5.2^5}{2^{14}.17^5}=\frac{2^{10}}{2^{14}}=\frac{1}{2^4}=\frac{1}{16}\)
a) a:b:c:d=2:3:4:5
<=>a/2=b/3=c/4=d/5=a+b+c+d/2+3+4+5=-42/14=-3
a/2=-3<=>a=-6
b/3=-3<=>b=-9
c/4=-3<=>c=-12
d/5=-3<=>d=-15
b)a/b=7/20<=>a/7=b/20
b/c=5/8 <=>b/5=c/8<=>b/20=c/32
<=>a/7=b/20=c/32=2a/14=5b/100=2c/64=2a+5b-2c/14+100+64=100/178=50/89
minh ko chac dung dau nha
Vũ Hồng Linh bạn check lại bài đầu dùm =_="
\(\left[-\frac{1}{3}\right]^3\cdot x=\frac{1}{81}\)
\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{3}\right]^3\)
\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{27}\right]\)
\(\Leftrightarrow x=\frac{1}{81}\cdot(-27)=-\frac{1}{3}\)
\(\left[x-\frac{1}{2}\right]^3=\frac{1}{27}\)
\(\Leftrightarrow\left[x-\frac{1}{2}\right]^3=\left[\frac{1}{3}\right]^3\)
=> Làm nốt
Mấy bài kia cũng làm tương tự
(- \(\dfrac{1}{3}\))3.\(x\) = \(\dfrac{1}{81}\)
\(x=\dfrac{1}{81}\) : (- \(\dfrac{1}{3}\))3
\(x\) = - (\(\dfrac{1}{3}\))4 :(\(\dfrac{1}{3}\))3
\(x=-\dfrac{1}{3}\)
Vậy \(x=-\dfrac{1}{3}\)
\(B=\left(1,6\cdot0,4\right)^2-\dfrac{57}{4}+\dfrac{7}{5}-\dfrac{7}{4}=\dfrac{256}{625}-16+\dfrac{7}{5}=-\dfrac{8869}{625}\)