NHÁ ĂN CỨT

a) Vì \(-45< -16\) nên \(\left(-\dfrac{45}{17}\right)^{15}< \left(\dfrac{-16}{17}\right)^{15}\)

b) Vì \(21< 23\) nên \(\left(-\dfrac{8}{9}\right)^{21}< \left(-\dfrac{8}{9}\right)^{23}\)

c) \(27^{40}=3^{3^{40}}=3^{120}\)

\(64^{60}=8^{2^{60}}=8^{120}\)

Vì \(3< 8\) nên \(3^{120}< 8^{120}\) hay \(27^{40}< 64^{60}\)

con ai kooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo

a) \(4.8^6.2.8^3\)

\(=2^2.\left(2^3\right)^6.2.\left(2^3\right)^3\)

\(=2^2.2^{18}.2.2^9\)

\(=2^{2+18+1+9}\)

\(=2^{30}\)

______

b) \(12^2.2.12^3.6\)

\(=12^2.12^3.2.6\)

\(=12^2.12^3.12\)

\(=12^{2+3+1}\)

\(=12^6\)

c) \(6^3.2.6^4.3\)

\(=6^3.6^4.2.3\)

\(=6^3.6^4.6\)

\(=6^{3+4+1}\)

\(6^8\)

a) \(4\cdot8^6\cdot2\cdot8^3\)

\(=2^2\cdot\left(2^3\right)^6\cdot2\cdot\left(2^3\right)^3\)

\(=2^2\cdot2^{18}\cdot2\cdot2^9\)

\(=2^{30}\)

b) \(12^2\cdot2\cdot12^3\cdot6\)

\(=12^2\cdot12\cdot12^3\)

\(=12^6\)

c) \(6^3\cdot2\cdot6^4\cdot3\)

\(=6^3\cdot6\cdot6^4\)

\(=6^8\)

a) 25x² - 16

= (5x)² - 4²

= (5x - 4)(5x + 4)

b) 16a² - 9b²

= (4a)² - (3b)²

= (4a - 3b)(4a + 3b)

c) 8x³ + 1

= (2x)³ + 1³

= (2x + 1)(4x² - 2x + 1)

d) 125x³ + 27y³

= (5x)³ + (3y)³

= (5x + 3y)(25x² - 15xy + 9y²)

e) 8x³ - 125

= (2x)³ - 5³

= (2x - 5)(4x² + 10x + 25)

g) 27x³ - y³

= (3x)³ - y³

= (3x - y)(9x² + 3xy + y²)

a) \(25x^2-16=\left(5x-4\right)\left(5x+4\right)\)

b) \(16a^2-9b^2=\left(4a-3b\right)\left(4a+3b\right)\)

c) \(8x^3+1=\left(2x+1\right)\left(4x^2-2x+1\right)\)

d) \(125x^3+27y^3=\left(5x+3y\right)\left(25x^2-15xy+9y^2\right)\)

e) \(8x^3-125=\left(2x-5\right)\left(4x^2-10x+25\right)\)

g) \(27x^3-y^3=\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)

do 5^3=125 =>(5^3)^12=125^12 ban a

(5^3)^12=125^12 vì 5^3=125 giữ nguyên số mũ thì được lũy thừa 125^12

\(\left(x+6\right)^3=64\)

\(\Leftrightarrow\left(x+6\right)^3=4^3\)

\(\Leftrightarrow x+6=4\)

\(\Leftrightarrow x=-2\)

\(\left(x+6\right)^3=64\)

\(\sqrt[3]{\left(x+6\right)^3}=\sqrt[3]{64}\)

\(x+6=4\)

\(x=-2\)

a) \(a^2\cdot a^3\cdot a^7\cdot b^2\cdot b\)

\(=\left(a^2\cdot a^3\cdot a^7\right)\cdot\left(b^2\cdot b\right)\)

\(=a^{12}\cdot b^3\)

b) \(b^6\cdot b\cdot c^7\cdot c^8\)

\(=\left(b^6\cdot b\right)\cdot\left(c^7\cdot c^8\right)\)

\(=b^7\cdot c^{15}\)

c) \(a^8\cdot a^9\cdot a\cdot c\cdot c^{20}\)

\(=\left(a^8\cdot a^9\cdot a\right)\cdot\left(c\cdot c^{20}\right)\)

\(=a^{18}\cdot c^{21}\)

d) \(a^2\cdot a^3\cdot b^4\cdot c\cdot c^3\)

\(=\left(a^2\cdot a^3\right)\cdot b^4\cdot\left(c\cdot c^3\right)\)

\(=a^5\cdot b^4\cdot c^4\)

a) Kiểm tra lại nhé

b) \(b^6.b^7.c^8\)

\(=b^{6+7}.c^8=b^{13}.c^8\)

c) \(a^8.a^9.a.c.c^{20}\)

\(=a^{8+9+1}.c^{1+20}\)

\(=a^{18}.c^{21}\)

d) \(a^2.a^3.b^4.c.c^3\)

\(=a^{2+3}.b^4.c^{1+3}\)

\(=a^5.b^4.c^4\)

\(#WendyDang\)

a) \(2^3.3^2=8.9=72\)

b) \(5^{10}:5^7=5^2=25\)

c) \(2^6:2=2^5=32\)

d) \(7^4:7^4=7^0=1\)

e) \(9^5:9^5=9^0=1\)

a) $2^3.3^2=8.9=72$

b) $5^{10}:5^7=5^2=25$

c) ${}^5$

32

d) $1$

e) $1$