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Bài 1 :
a/ \(a^3.a^9=a^{3+9}=a^{12}\)
b/\(\left(a^5\right)^7=a^{5.7}=a^{35}\)
c/ \(\left(a^6\right).4.a^{12}=a^{24}.a^{12}.4=a^{24+12}.4=a^{36}.4\)
d/ \(\left(2^3\right)^5.\left(2^3\right)^3=2^{15}.2^9=2^{15+9}=2^{24}\)
e/ \(5^6:5^3+3^3.3^2\)
\(=5^3+3^5=125+243=368\)
i/ \(4.5^2-2.3^2\)
\(=2^2.5^2-2.3^2\)
\(=2^2.25-2^2.14\)
\(=2^2.\left(25-14\right)\)
\(=2^2.11\)
\(=4.11=44\)
a ,
\(x.x^2.x^3.x^4.x^5......x^{49}.x^{50}.x=x^{24.\left(1+49\right)+51}=x^{1251}\)
a) x . x2 . x3 . ... . x50
= x(1 + 2 + 3 + ... + 50)
= x1275
Bài 1:
\(\text{a) }x.x^2.x^3.x^4.x^5.....x^{49}.x^{50}\)
\(=x^{1+2+3+4+5+...+49+50}\)
\(=x^{\frac{51.50}{2}}\)
\(=x^{1275}\)
\(\text{b) Ta có:}\)
\(4^{15}=\left(2^2\right)^{15}=2^{2.15}=2^{30}\)
\(8^{11}=\left(2^3\right)^{11}=2^{3.11}=2^{33}\)
\(\text{Vì }2^{30}< 2^{33}\text{ nên }4^{15}< 8^{11}\)
Bài 2: Tìm x
\(\left(x-1\right)^4:3^2=3^6\)
\(\Rightarrow\left(x-1\right)^4=3^6\times3^2\)
\(\Rightarrow\left(x-1\right)^4=3^8\)
\(\Rightarrow\left(x-1\right)^4=3^{2.4}\)
\(\Rightarrow\left(x-1\right)^4=\left(3^2\right)^4\)
\(\Rightarrow x-1=9\)
\(\Rightarrow x=10\)
Bài 3 và bài 4 mk làm sau
Bài 1 : a) \(x.x^2.x^3.x^4.....x^{49}.x^{50}=x^{1+2+3+...+49+50}\) (Dễ rồi tự tính)
b) \(\hept{\begin{cases}4^{15}=\left(2^2\right)^{15}=2^{30}\\8^{11}=\left(2^3\right)^{11}=2^{33}\end{cases}}\)Rồi tự so sánh đi
Bài 2 :
\(\left(x-1\right)^4\div3^2=3^6\Leftrightarrow\left(x-1\right)^4=3^8=\left(3^2\right)^4=9^4\Leftrightarrow x-1=9\Leftrightarrow x=10\)
Bài 3 :
\(\hept{\begin{cases}27^{15}=\left(3^3\right)^{15}=3^{45}\\81^{11}=\left(3^4\right)^{11}=3^{44}\end{cases}}\) nt
a,\(=x^{1.2.3....49.50}\)
b,\(\Rightarrow\)2Q\(=2+2^2+2^3+...+2^{50}\)
2Q-Q\(=2+2^2+2^3+...+2^{50}-1-2-2^2-...-2^{49}\)
Q\(=2^{50}-1\)
Q+1=\(2^{50}\)
Mà Q+1=\(2^n\)
\(2^{50}=2^n\Rightarrow n=50\)
a) x1+2+3+...+50=x1275
b)Q=1+2+22+23+....+249
2Q=2+22+23+...+250
2Q-Q=250-1
Q+1=250 Mà Q+1=2n suy ra 250=2n
Vậy n=50
a 5.125.625=5.5^3.5^4=5^8
b 10.100.1000=10.10^2.10^3=10^6
c 8^4.16^5.32=2^3^4.2^4^5.2^5=2^12.2^20.2^5=2^37
a) = \(5^1\cdot5^3\cdot5^4=5^{1+3+4}=5^8\)
b) = \(10^1\cdot10^2\cdot10^3=10^{1+2+3}=10^6\)
c) = \(2^{12}\cdot2^{20}\cdot2^5=2^{12+20+5}=2^{37}\)
\(8-12x+6x^2-x^3\)
\(=\left(2-x\right)^3\)
\(125x^3-75x^2+15x-1\)
\(=\left(5x-1\right)^3\)
\(x^2-xz-9y^2+3yz\)
\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y-z\right)\)
\(x^3-x^2-5x+125\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
\(x^3+2x^2-6x-27\)
\(=x^3+5x^2+9x-3x^2-15x-27\)
\(=x\left(x^2+5x+9\right)-3\left(x^2+5x+9\right)\)
\(=\left(x-3\right)\left(x^2+5x+9\right)\)
\(12x^3+4x^2-27x-9\)
\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x+1\right)\left(4x^2-9\right)\)
\(=\left(3x+1\right)\left(2x-3\right)\left(2x+3\right)\)
\(4x^4+4x^3-x^2-x\)
\(=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=x\left(x+1\right)\left(4x^2-1\right)\)
\(=x\left(x+1\right)\left(2x-1\right)\left(2x+1\right)\)
a; 25 x 53 x \(\dfrac{1}{625}\) x 52
= 52 x 53 x \(\dfrac{1}{5^4}\) x 52
= 55 x \(\dfrac{1}{5^4}\) x 52
= 5 x 52
= 53
a)
\(25\cdot5^3\cdot\dfrac{1}{625}\cdot5^2\\ =\left(5^2\cdot5^3\cdot5^2\right)\cdot\dfrac{1}{625}\\ =5^7\cdot\dfrac{1}{5^4}\\ =5^3\)
b)
\(5^2\cdot3^5\cdot\left(\dfrac{3}{5}\right)^2\\ =5^2\cdot3^5\cdot\dfrac{3^2}{5^2}\\ =3^5\cdot3^2\\ =3^7\)
c)
\(\left(-\dfrac{1}{7}\right)^4\cdot49^2\\ =\dfrac{\left(-1\right)^4}{7^4}\cdot\left(7^2\right)^2\\ =\dfrac{1}{7^4}\cdot7^4\\ =1\)
d)
\(\left(\dfrac{1}{16}\right)^2:\left(\dfrac{1}{2}\right)^4\cdot\left(-\dfrac{1}{8}\right)^3\\ =\left[\left(\dfrac{1}{2}\right)^4\right]^2:\left(\dfrac{1}{2}\right)^4\cdot\left[\left(-\dfrac{1}{2}\right)^3\right]^3\\ =\left(\dfrac{1}{2}\right)^8:\left(\dfrac{1}{2}\right)^4\cdot\left(-\dfrac{1}{2}\right)^9\\ =\left(\dfrac{1}{2}\right)^4\cdot\left(-\dfrac{1}{2}\right)^9\\ =\left(\dfrac{1}{2}\right)^4\cdot-\left(\dfrac{1}{2}\right)^9\\ =-\left(\dfrac{1}{2}\right)^{13}\)