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2) x4 -16 =0 => x4 =16 => x4 = 44 hoặc (-4)4 => x = 4 hoặc -4
32010- ( 32009 + 32008 + ... + 3 + 1 )
Đặt A = 1 + 3 + ... + 32009
=> 3A = 3 + 32 + ... + 32010
=> 3A - A = 32010 - 1
Nên 32010 - ( 32010 - 1 ) = 1
đăng từng câu nhé bạn
chứ kiểu vậy thì ko có ai giải cho bạn đâu
\(a,\frac{(-10)^5}{3\cdot(-6)^4}=\frac{(-2\cdot5)^5}{3\cdot(-2\cdot3)^4}=\frac{(-2)^5\cdot5^5}{3\cdot(-2)^4\cdot3^4}=\frac{(-2)^5\cdot5^5}{(-2)^4\cdot3^5}=-2\cdot\frac{5^5}{3^5}=\frac{-6250}{243}\)
\(b,\frac{2^{15}\cdot9^4}{6^6\cdot8^3}=\frac{\left[2^3\right]^5\cdot\left[3^2\right]^4}{\left[3\cdot2\right]^6\cdot\left[2^3\right]^3}=\frac{2^{15}\cdot3^8}{3^6\cdot2^6\cdot2^9}=\frac{2^{15}\cdot3^8}{3^6\cdot2^{15}}=\frac{3^8}{3^6}=3^2=9\)
\(c,\left[1+\frac{2}{3}-\frac{1}{4}\right]\cdot\left[\frac{4}{5}-\frac{3}{4}\right]^2\)
\(=\left[\frac{12}{12}+\frac{8}{12}-\frac{3}{12}\right]\cdot\left[\frac{16}{20}-\frac{15}{20}\right]^2\)
\(=\frac{17}{12}\cdot\left[\frac{1}{20}\right]^2=\frac{17}{12}\cdot\frac{1^2}{20^2}=\frac{17}{12}\cdot\frac{1}{400}=\frac{17}{4800}\)
\(d,2^3+3\cdot\left[\frac{1}{2}\right]^0+\left[(-2)^2:\frac{1}{2}\right]\)
\(=8+3\cdot\frac{1^0}{2^0}+\left[4:\frac{1}{2}\right]\)
\(=8+3\cdot1+8=8+3+8=19\)
a) 169/196
b) 1/144
c) \(\frac{5^4.20^4}{25^5.4^5}=\frac{\left(5.20\right)^4}{\left(25.4\right)^5}=\frac{100^4}{100^5}=\frac{1}{100}\)
d) -2506/3
Lời giải:
$3^6-M=3^0+3^1+3^2+3^3+3^4+3^5$
$3(3^6-M)=3^1+3^2+3^3+3^4+3^5+3^6$
$\Rightarrow 3(3^6-M)-(3^6-M)=3^6-3^0$
$\Rightarrow 2(3^6-M)=3^6-1$
$\Rightarrow 2M = 2.3^6-(3^6-1)=3^6+1$
$\Rightarrow M=\frac{3^6+1}{2}$
M=36-(35+34+...+31+30)
Đặt A=35+34+...+31+30
3A=36+35+...+32+31
3A-A=36+35+...+32+31-35-34-...-31-30
2A=36-30=>A=\(\dfrac{3^6-3^0}{2}\)
Thay A vào M ta có:
M=36-\(\dfrac{3^6-3^0}{2}\)
M=\(\dfrac{2.3^6}{2}\)-\(\dfrac{3^6-3^0}{2}\)
M=\(\dfrac{3^6.\left(2-1\right)-1}{2}\)
M=\(\dfrac{3^6.1-1}{2}\)
M=\(\dfrac{3^6-1}{2}\)
M=364