\(g)8^4\cdot2^3\cdot16^2=2^{12}\cdot2^3\cdot2^8=2^{23}\)

\(h)2^3\cdot2^2\cdot8^3=2^3\cdot2^2\cdot2^9=2^{14}\)

\(i)y\cdot y^7=y^8\)

\(a)3^{x}\cdot3=243\)

\(\Rightarrow3^{x+1}=3^5\)

\(\Rightarrow x+1=5\Rightarrow x=4\)

\(b)64\cdot4^{x}=168\)

\(4^3\cdot4^{x}=168\)

\(4^{x+3}=168\)

vì 168 không phải là luỹ thừa của 4 nên x thuộc rỗng

\(c)2^{x}\cdot162=1024\)

\(2^{x}\cdot2\cdot3^4=2^{10}\)

\(2^{x+1}\cdot81=2^{10}\)

vì 81 không phải là luỹ thừa của 2 nên x thuộc rỗng

\(d)2^{x}=16\)

\(2^{x}=4^2\)

\(2^{x}=\left(2^2\right)^2\)

\(2^{x}=2^4\)

\(\Rightarrow x=4\)

\(a)4^9:4^4=4^5\)

\(b)17^8:17^5=17^3\)

\(c)2^{10}:8^2=2^{10}:2^6=2^4\)

\(d)18^{10}:3^{10}=\frac{3^{10}\cdot6^{10}}{3^{10}}=6^{10}\)

\(e)27^5:81^3=\frac{3^5\cdot9^5}{9^6}=\frac{3^5}{9}=\frac{3^5}{3^2}=3^3\)

\(f)10^6:100=10^6:10^2=10^4\)

\(g)5^9:25^3=5^9:5^6=5^3\)

\(h)4^{10}:64^3=2^{20}:2^{18}=2^2\)

\(i)2^{25}:32^4=2^{25}:2^{20}=2^5\)

\(g)18^4:9^4=\frac{2^4\cdot9^4}{9^4}=2^4\)

câu a)

\(2^2=4;2^3=8;2^4=16;2^5=32;2^6=64\)

\(2^7=128;2^8=256;2^9=512;2^{10}=1024\)

câu b)

\(3^2=9;3^3=27;3^4=81;3^5=243\)

câu c)

\(4^2=16;4^3=64;4^4=256\)

câu d)

\(5^2=25;5^3=125;5^4=625\)

\(a)1^3+2^3=\left(1+2\right)^2=3^2=9\)

\(b)1^3+2^3+3^3=\left(1+2+3\right)^2=6^2=36\)

\(c)1^3+2^3+3^3+4^3=\left(1+2+3+4\right)^2=10^2=100\)

số kg cá 10 chú gấu con trong đàn đó ăn hết là:

10 x 4,5 = 45(kg)

đáp số: 45kg

\(a)3\left(5x-1\right)-x\left(x-2\right)+x^2-13x=7\)

\(15x-3-x^2+2x+x^2-13x=7\)

\(4x=10\Rightarrow x=\frac{10}{4}=\frac52\)

\(b)4\left(x+2\right)-7\left(2x-1\right)+9\left(3x-4\right)=30\)

\(4x+8-14x+7+27x-36=30\)

\(17x=51\Rightarrow x=3\)

\(a)\left(3-\left|-3\right|\right)^2-6^2\)

\(=\left(3-3\right)^2-6^2\)

\(=0-6^2=-36\)

\(b)\left(\left|2-5\right|\right)^2-\left|-9\right|\)

\(=3^2-9=9-9=0\)

ㅤ

đkxđ: x khác 1; x khác 2

\(\frac{4}{x-1}-\frac{5}{x-2}=-3\)

\(\frac{4\cdot\left(x-2\right)-5\cdot\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}=-3\)

\(\frac{4x-8-5x+5}{x^2-3x+2}=-3\)

\(\Rightarrow-x-3=-3x^2+9x-6\)

\(\Rightarrow3x^2-9x+6-x-3=0\)

\(3x^2-10x+3=0\)

\(\triangle=\left(-10\right)^2-4\cdot3\cdot3=64\)

\(x_1=\frac{10-8}{2\cdot3}=\frac13\)

\(x_2=\frac{10+8}{2\cdot3}=3\)

Vậy phương trình có 2 nghiệm là \(x_1=\frac13;x_2=3\)