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Bài 8:
a: \(\left(\dfrac{2}{5}+\dfrac{3}{4}\right)^2=\left(\dfrac{8+15}{20}\right)^2=\left(\dfrac{23}{20}\right)^2=\dfrac{529}{400}\)
b: \(\left(\dfrac{5}{4}-\dfrac{1}{6}\right)^2=\left(\dfrac{15}{12}-\dfrac{2}{12}\right)^2=\left(\dfrac{13}{12}\right)^2=\dfrac{169}{144}\)
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!
\(\frac{1}{9}.3^4.3^x=3^7\)
\(\Leftrightarrow3^x=3^7:\frac{1}{9}:3^4=243\)
\(\Leftrightarrow3^x=3^5\)
\(\Leftrightarrow x=5\)
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) \(\frac{34^5.2^5}{2^{14}.17^5}=\frac{2^5.17^5.2^5}{2^{14}.17^5}=\frac{2^{10}}{2^{14}}=\frac{1}{2^4}=\frac{1}{16}\)
Thêm nữa câu a) Tính: M(x) + N(x)+ P(x)
B) Tính M(x) - N (x) - P(x)
ok rồi giúp mình với nha
a,\(\left(\dfrac{3}{7}+\dfrac{1}{2}\right)^2\)
\(=\left(\dfrac{13}{14}\right)^2\)
\(=\dfrac{169}{196}\)
b,\(\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)
\(=\left(\dfrac{-1}{12}\right)^2\)
\(=\dfrac{1}{144}\)
c,\(\dfrac{5^4.20^4}{25^5.4^5}\)
\(=\dfrac{100^4}{100^5}\)
\(=\dfrac{1}{100}\)
d,\(\left(\dfrac{-10}{3}\right)^5.\left(\dfrac{-6}{5}\right)^4\)
\(=\left(\dfrac{-10}{3}\right)^4.\left(\dfrac{-6}{5}\right)^4.\left(\dfrac{-10}{3}\right)\)
\(=\left(\dfrac{\left(-10\right)}{3}.\dfrac{\left(-6\right)}{5}\right)^4.\left(\dfrac{-10}{3}\right)\)
\(=4^4.\left(\dfrac{-10}{3}\right)\)
\(=256.\left(\dfrac{-10}{3}\right)\)
\(=\dfrac{-2560}{3}\)
\(4^3=64\) \(4^2=16\)
\(4^3+4^2=64+16=80\)
\(4^3+4^2=4^{3+2}=4^5\)
Hc tốt
@nth