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a) 74x.(3312+33332020+333333303030+3333333342424242)=32\frac{7}{4}x.\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)=3247​x.(1233​+20203333​+303030333333​+4242424233333333​)=32

74x.(3312+3320+3330+3342)=32\frac{7}{4}x.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)=3247​x.(1233​+2033​+3033​+4233​)=32

74x.(333.4+334.5+335.6+336.7)=32\frac{7}{4}x.\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)=3247​x.(3.433​+4.533​+5.633​+6.733​)=32

74x.33.(13−14+14−15+15−16+16−17)=32\frac{7}{4}x.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)=3247​x.33.(31​−41​+41​−51​+51​−61​+61​−71​)=32

74x.33.(13−17)=32\frac{7}{4}x.33.\left(\frac{1}{3}-\frac{1}{7}\right)=3247​x.33.(31​−71​)=32

74x.33⋅421=32\frac{7}{4}x.33\cdot\frac{4}{21}=3247​x.33⋅214​=32

b) 13+16+110+115+...+2x.(x−1)=20072009\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{2}{x.\left(x-1\right)}=\frac{2007}{2009}31​+61​+101​+151​+...+x.(x−1)2​=20092007​

26+212+220+230+...+2(x−1).x=20072009\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{\left(x-1\right).x}=\frac{2007}{2009}62​+122​+202​+302​+...+(x−1).x2​=20092007​

22.3+23.4+24.5+25.6+...+2(x−1).x=20072009\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{\left(x-1\right).x}=\frac{2007}{2009}2.32​+3.42​+4.52​+5.62​+...+(x−1).x2​=20092007​

2.(12−13+13−14+14−15+15−16+...+1x−1−1x)=200720092.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{x-1}-\frac{1}{x}\right)=\frac{2007}{2009}2.(21​−31​+31​−41​+41​−51​+51​−61​+...+x−11​−x1​)=20092007​

2.(12−1x)=200720092.\left(\frac{1}{2}-\frac{1}{x}\right)=\frac{2007}{2009}2.(21​−x1​)=20092007​

1−2x=200720091-\frac{2}{x}=\frac{2007}{2009}1−x2​=20092007​

2x=22009\frac{2}{x}=\frac{2}{2009}x2​=20092​

=> x = 2009

a, Ta có : \(3^{500}=\left(3^5\right)^{100}=243^{100}\)

\(7^{300}=\left(7^3\right)^{100}=343^{100}\)

Vì \(243< 343\)

\(\Rightarrow243^{100}< 343^{100}\)

Hay \(3^{500}< 7^{300}\)

a. 3⁵⁰⁰ = (3⁵)¹⁰⁰ = 243¹⁰⁰

    7³⁰⁰ = (7³)¹⁰⁰ = 343¹⁰⁰

    Vì 243¹⁰⁰ < 343¹⁰⁰ nên 3⁵⁰⁰ < 7³⁰⁰.

b.  2⁵⁰⁰ = (2⁵)¹⁰⁰ = 32¹⁰⁰

     5²⁰⁰ = (5²)¹⁰⁰ = 25¹⁰⁰

     Vì 32¹⁰⁰ > 25¹⁰⁰ nên 2⁵⁰⁰ > 5²⁰⁰.

c. 31¹¹ < 32¹¹ = (2⁵)¹¹ = 2⁵⁵

=>31¹¹ < 2⁵⁵

    17¹⁴ > 16¹⁴ = (2⁴)¹⁴ = 2⁵⁶

=>17¹⁴ > 2⁵⁶

=> 31¹¹ < 2⁵⁵ < 2⁵⁶ < 17¹⁴.

=> 31¹¹ < 17¹⁴.

d. 5³⁶ = (5³)¹² = 125¹²

    11²⁴ = (11²)¹² = 121¹²

Vì 125¹² > 121¹² nên 5³⁶ > 11²⁴

e. Vì 10⁹ < 10¹⁰ nên 10⁹ + 1 < 10¹⁰ + 1

    Vì 10¹⁰ < 10¹¹ nên 10¹⁰ + 1 < 10¹¹ + 1.

\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)

\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)

\(=1-\frac{1}{2!}+1-\frac{1}{3!}+\frac{1}{2!}-\frac{1}{4!}+...+\frac{1}{98!}-\frac{1}{100!}\)

\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)

Vậy \(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}< 2\left(đpcm\right)\)

\(\left(x^3+2x^5y^2-4\right)\left(-3xy\right)\)

\(=-3x^{3+1}y-3.2x^{5+1}y^{2+1}+\left(-4\right).-3xy\)

\(=3x^4y-6x^6y^3+12xy\)

gọi số tiền của 4 bạn lần lượt góp là a; b; c; d

có a + b + c + d = 120 000

có a = 1/2 (b + c + d)

suy ra a + 1/2a = 1/2(b + c + d)

3/2a = 1/2 * 120000 = 60000

suy ra a = 40000

dua theo phương pháp trên tính ra các bạn còn lại

40000 là đúng .