Do (2x-5)²⁰⁰⁰>0

(3y+4)²⁰⁰²>0

Mà (2x-5)²⁰⁰⁰+(3y+4)²⁰⁰²<0

=>(2x-5)²⁰⁰⁰=0 (3y+4)²⁰⁰²=0

<=>x=2,5 y=4/3

2) a) 2^x.(2² - 1) = 96 => 2^x. 3 = 96 => 2^x = 96 : 3 = 32 = 2⁵ => x = 5

b) 7^x. (7² + 2. 7^-1) = 345 => 7^x. \(\frac{345}{7}\) = 345 => 7^x = 7 => x = 1

c)  3^x-1. (1 + 5) = 162 => 3^x-1 . 6 = 162 => 3^x-1 = 162 : 6 = 27 = 3³ => x - 1 = 3 => x = 3 + 1 = 4

1) a)  (3³)ⁿ = 9.3ⁿ => 3³ⁿ = 3².3ⁿ = 3²⁺ⁿ => 3n = 2 + n => 3n - n = 2 => 2n = 2 => n = 1

b) 3^-2+4+n = 3⁷ => 2 + n = 7 => n = 7 - 2 = 5

c) 2ⁿ.(2^-1 + 4) = 9.2⁵ => 2ⁿ.\(\frac{9}{2}\) = 9.2⁵ => 2^n-1 = 2⁵ => n - 1 = 5 => n = 5 + 1 = 6

d)  (2⁵)^-n.(2⁴)ⁿ = 2¹¹ => 2^-5n. 2⁴ⁿ = 2¹¹ => 2^-5n+4n = 2¹¹ => 2^-n = 2¹¹ => -n = 11 => n = -11

\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{\left(y+z+1\right)+\left(x+z+2\right)+\left(x+y-3\right)}{x+y+z}=2\)

Suy ra: \(\frac{1}{x+y+z}=2\Rightarrow x+y+z=\frac{1}{2}\)(*)

Ta có: \(\frac{y+z+1}{x}=2\Leftrightarrow y+z+1=2x\Leftrightarrow x+y+z+1=3x\Leftrightarrow\frac{1}{2}+1=3x\Leftrightarrow x=\frac{1}{2}\)

\(\frac{x+z+2}{y}=2\Leftrightarrow x+z+2=2y\Leftrightarrow x+y+z+2=3y\Leftrightarrow\frac{1}{2}+2=3y\Leftrightarrow y=\frac{5}{6}\)

Từ (*) suy ra: \(z=\frac{1}{2}-x-y=\frac{1}{2}-\frac{1}{2}-\frac{5}{6}\Leftrightarrow z=-\frac{5}{6}\)

Vậy \(x=\frac{1}{2};y=\frac{5}{6};z=-\frac{5}{6}\)

\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{\left(y+z+1\right)+\left(x+z+2\right)+\left(x+y-3\right)}{x+y+z}=2\)

\(\Rightarrow\frac{1}{x+y+z}=2\Rightarrow x+y+z=\frac{1}{2}\left(\cdot\right)\)

Ta có : \(\frac{y+z+1}{x}=2\Leftrightarrow y+z+1=2x\Rightarrow x+y+z+1=3x\Rightarrow\frac{1}{2}+1=3x\Leftrightarrow x=\frac{1}{2}\)

\(\frac{x+z+2}{y}=2\Leftrightarrow x+z+2=2y\Leftrightarrow x+y+z+2=3y\Leftrightarrow\frac{1}{2}+2=3y\Leftrightarrow y=\frac{5}{6}\)

Từ \(\left(\cdot\right)\Rightarrow z=\frac{1}{2}-x-y=\frac{1}{2}-\frac{1}{2}-\frac{5}{6}\Leftrightarrow z=-\frac{5}{6}\)

Vậy \(x=\frac{1}{2};y=\frac{5}{6};z=-\frac{5}{6}\)

\(a^2=bc\Rightarrow\frac{a}{c}=\frac{b}{a}=\frac{a+b}{c+a}=\frac{a-b}{c-a}\)=>(a+b)(c-a)=(c+a)(a-b)

\(\Rightarrow\frac{a+b}{a-b}=\frac{c+a}{c-a}\)

=>đpcm

\(A=1+3+3^2+3^3+...+3^{101}\)

\(3A=3+3^2+3^3+3^4+...+3^{101}\)

\(3A-A=\left(3+3^2+3^3+3^4+...+3^{101}\right)-\left(1+3+3^2+3^3+...+3^{100}\right)\)

\(2A=3^{101}-1\)

\(A=\left(3^{101}-1\right):2\)

Thu gọn tổng sau:

A=1+3+3²+3³+...+3¹⁰⁰ 

B= 2¹⁰⁰-2⁹⁹-2⁹⁸-2⁹⁷-...-2²-2

C= 3¹⁰⁰-3⁹⁹+3⁹⁸-3⁹⁷-...+3²-3+1 

\(\frac{a\frac{1}{b}}{b\frac{1}{a}}=\frac{a}{b}\text{ (}=\frac{\frac{ab+1}{b}}{\frac{ba+1}{a}}=\frac{a}{b}\text{)}\)

Rất nhiều tỉ số như vậy, không có gì là đặc biệt cả.

Ta có: 1,5. 4,8 = 2. 3,6  

Do đó có 4 tỉ lệ thức:

\(\frac{1,5}{2}=\frac{3,6}{4,8};\frac{1,5}{3,6}=\frac{2}{4,8};\frac{4,8}{2}=\frac{3,6}{1,5};\frac{4,8}{3,6}=\frac{2}{1,5}\)

mjk k rảnh để làm cái này

sorry bn nha

gọi số tiền mỗi người nhận được lần lượt là x; y ; z => x + y + z = 3280 000

Vì số tiền chia tỉ lệ với số nông cụ nên ta có:  x: y : z = 96 : 120 : 112 = 12 : 15 : 14

=> \(\frac{x}{12}=\frac{y}{15}=\frac{z}{14}=\frac{x+y+z}{12+15+14}=\frac{3280000}{41}=80000\)

=> x = 80 000 x 12 = 960 000

y = 80 000 x 15 = 1200 000

z = 80 000 x 14  = 1 120 000 

Vậy...

vì sao lại ra được 12,15,14

a ) 20 : x = (-12) : 15

=> 20 . 15 = -12x

=> 300 = -12x

=> x = 300 : (-12) = -25

b) \(\frac{1}{2}:1\frac{1}{4}=x:3\frac{1}{3}\)

=> \(\frac{1}{2}:\frac{5}{4}=x:\frac{10}{3}\)

=> \(\frac{1}{2}\cdot\frac{10}{3}=\frac{5}{4}x\)

=> \(\frac{5}{3}=\frac{5}{4}x\)

=> \(x=\frac{5}{3}:\frac{5}{4}=\frac{4}{3}\)

c) \(\frac{-4,4}{9,9}=\frac{x}{1,89}\)

\(\Rightarrow-4,4\cdot1,89=9,9x\)

=> \(-8,316=9,9x\)

=> \(x=-8,316:9,9=0,84\)

a) \(\frac{790^4}{79^4}=\frac{79^4.10^4}{79^4}=10^4=10000\)

b) \(\frac{3^2}{0,375^2}=\frac{0,375^2.8^2}{0,375^2}=8^2=64\)

c) \(3^2.\frac{1}{243}.81^2.\frac{1}{3^3}=3^2.3^{-5}.3^8.3^{-3}=3^2=9\)

d) \(\left(4.2^5\right):\left(2^3.\frac{1}{16}\right)=2^7:\left(2^3.2^{-4}\right)=2^7:2^{-1}=2^7:\frac{1}{2}=2^8\)