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) \(6^3\cdot2\cdot6^4\cdot3\)

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

\(=6^8\)

5^{2x - 3} = 125

<=> 5^{2x - 3} = 5³

<=> 2x - 3 = 3

<=> x = 3

có chắc đúng ko vậy bạn

Lời giải :

1. \(\left(\frac{1}{2}a+b\right)^3+\left(\frac{1}{2}a-b\right)^3\)

\(=\frac{a^3}{8}+\frac{3a^2b}{4}+\frac{3ab^2}{2}+b^3+\frac{a^3}{8}-\frac{3a^2b}{4}+\frac{3ab^2}{2}-b^3\)

\(=\frac{a^3}{4}+3ab^2\)

Lời giải :

2. \(x^3-3x^2+3x-1=0\)

\(\Leftrightarrow\left(x-1\right)^3=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vậy...

1) \(\left(\frac{1}{2}a+b\right)^3+\left(\frac{1}{2}a-b\right)^3\)

\(=\left(\frac{a}{2}+b\right)^2+\left(\frac{a}{2}-b\right)^2\)

\(=\left(\frac{a}{2}+b\right)\left[\left(\frac{a}{2}\right)^2+2.\frac{a}{b}b+b^2\right]+\left(\frac{a}{2}-b\right)\left[\left(\frac{a}{2}\right)^2-2.\frac{a}{2}b+b^2\right]\)

\(=\frac{a}{2}\left[\left(\frac{a}{2}\right)^2+2.\frac{a}{2}b+b^2\right]+b\left[\left(\frac{a}{2}\right)^2+2.\frac{a}{2}b+b^2\right]+\frac{a}{2}\left[\left(\frac{a}{2}\right)^2-2.\frac{a}{2}b+b^2\right]\)\(-b\left[\left(\frac{a}{2}\right)^2-2.\frac{a}{2}b+b^2\right]\)

\(=\frac{a^3}{8}+\frac{a^2b}{2}+\frac{ab^2}{2}+\frac{ba^2}{4}+b^2a+b^3+\frac{a^3}{8}-\frac{a^2b}{2}+\frac{ab^2}{2}-\frac{ba^2}{4}+b^2a-b^3\)

\(=\frac{a^3}{4}+3ab^2\)

2) \(x^3-3x^2+3x-1=0\)

\(\Leftrightarrow x^3-3x^2.1+3.x.1^2-1^3=0\)

\(\Leftrightarrow\left(x+1\right)^3=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=0-1\)

\(\Rightarrow x=-1\)

3) \(A=\left(4x-1\right)^3-\left(4x-3\right)\left(16x^2+3\right)\)

\(A=64x^3-32x^2+4x-16x^2+8x-1-64x^3-12x+48x^2+9\)

\(A=8\)

Vậy: biểu thức không phụ thuộc vào biến

1) \(\left(x+5\right)^3-x^3-125\)

\(=\left(x+5\right)\left(x^2+2x.5+5^2\right)-x^3-125\)

\(=x\left(x^2+2x.5+5^2\right)+5\left(x^2+2x.5+5^2\right)-x^3-125\)

\(=x^3+10x^2+25x+5x^2+50x+125-x^3-125\)

\(=15x^2+75x\)

2) \(\left(x-2\right)^3+6\left(x+1\right)^2-x^3+12=0\)

\(\Leftrightarrow x^3-4x^2+4x-2x^2+8x-8+6x^2+12x+6-x^3+12=0\)

\(\Leftrightarrow24x+10=0\)

\(\Leftrightarrow24x=0-10\)

\(\Leftrightarrow24x=-10\)

\(\Leftrightarrow x=-\frac{10}{24}=-\frac{5}{12}\)

\(\Rightarrow x=-\frac{5}{12}\)

3) \(\left(x-1\right)^3-x^3+3x^2-3x+1\)

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

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

\(=x^3-2x^2+x-x^2+2x-1-x^3-3x^2-3x+1\)

\(=0\)

Vậy: biểu thức không phụ thuộc vào biến

S = 1 + 3 + 3² + ... + 3¹⁰⁰⁰

⇒ 3S = 3 + 3² + 3³ + ... + 3¹⁰⁰¹

⇒ 2S = 3S - S

= (3 + 3² + 3³ + ... + 3¹⁰⁰¹) - (1 + 3 + 3² + ... + 3¹⁰⁰⁰)

= 3¹⁰⁰¹ - 1

⇒ S = (3¹⁰⁰¹ - 1) : 2

3S=3+3²+3³+...+3¹⁰⁰¹

3S-S=(3+3²+3³+...+3¹⁰⁰¹)-(1+3+3²+...+3¹⁰⁰⁰)

2S= 3¹⁰⁰¹-1

S=(3¹⁰⁰¹-1):2

(-1)^3+(-2)^3+...+(-99)^3 <1