CÓ

2

nhu nay bn nhe a: \(\left(\right. \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} \left.\right) \cdot \frac{2}{\sqrt{x} + \sqrt{y}} + \frac{1}{x} + \frac{1}{y}\)

\(= \frac{\sqrt{x} + \sqrt{y}}{\sqrt{x y}} \cdot \frac{2}{\sqrt{x} + \sqrt{y}} + \frac{x + y}{x y}\)

\(= \frac{2}{\sqrt{x y}} + \frac{x + y}{x y} = \frac{x + y + 2 \sqrt{x y}}{x y} = \frac{\left(\left(\right. \sqrt{x} + \sqrt{y} \left.\right)\right)^{2}}{x y}\)

\(\frac{\sqrt{x^{3}} + x \cdot \sqrt{y} + y \cdot \sqrt{x} + \sqrt{y^{3}}}{\sqrt{x^{3} y} + \sqrt{x y^{3}}}\)

\(= \frac{\left(\right. x \cdot \sqrt{x} + x \cdot \sqrt{y} + y \cdot \sqrt{x} + y \cdot \sqrt{y} \left.\right)}{x \cdot \sqrt{x y} + y \cdot \sqrt{x y}} = \frac{\left(\right. x + y \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)}{\sqrt{x y} \left(\right. x + y \left.\right)} = \frac{\sqrt{x} + \sqrt{y}}{\sqrt{x y}}\)

\(P = \left[\right. \left(\right. \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} \left.\right) \cdot \frac{2}{\sqrt{x} + \sqrt{y}} + \frac{1}{x} + \frac{1}{y} \left]\right. : \left(\right. \frac{\sqrt{x^{3}} + x \cdot \sqrt{y} + y \cdot \sqrt{x} + \sqrt{y^{3}}}{\sqrt{x^{3} y} + \sqrt{x y^{3}}} \left.\right)\)

\(= \frac{\left(\left(\right. \sqrt{x} + \sqrt{y} \left.\right)\right)^{2}}{x y} : \frac{\sqrt{x} + \sqrt{y}}{\sqrt{x y}} = \frac{\sqrt{x} + \sqrt{y}}{\sqrt{x y}}\)

ok bn

(5x+1)^2=(9/8)^2

5x+1=9/8

5x=9/8-1

5x=1/8

x=1/40

vây x=1/40 tich cho mik nha

1. afford : có đủ khả năng
2. armchair : ghế bành

3

Ta có phương trình:

\(8^{x} + 342 = 7^{y}\)

Đây là một phương trình mũ hỗn hợp — hai cơ số khác nhau và có hằng số 342.

Bước 1: Thử giá trị nguyên nhỏ (nếu có)

Ta thử vài giá trị nhỏ của \(x\) và xem \(y\) có phải số nguyên không:

Thử \(x = 1\):

\(8^{1} + 342 = 8 + 342 = 350 \Rightarrow \text{Li}ệ\text{u}\&\text{nbsp}; 7^{y} = 350 ? \Rightarrow 7^{3} = 343 , \&\text{nbsp}; 7^{4} = 2401 \Rightarrow \text{Kh} \hat{\text{o}} \text{ng}\&\text{nbsp};\text{kh}ớ\text{p}\)

Thử \(x = 2\):

\(8^{2} + 342 = 64 + 342 = 406 \Rightarrow \text{Kh} \hat{\text{o}} \text{ng}\&\text{nbsp};\text{ph}ả\text{i}\&\text{nbsp};\text{l} \overset{\sim}{\text{u}} \text{y}\&\text{nbsp};\text{th}ừ\text{a}\&\text{nbsp};\text{c}ủ\text{a}\&\text{nbsp};\text{7}\)

Thử \(x = 3\):

\(8^{3} = 512 , \&\text{nbsp}; 512 + 342 = 854 \Rightarrow \text{Kh} \hat{\text{o}} \text{ng}\&\text{nbsp};\text{ph}ả\text{i}\&\text{nbsp};\text{l} \overset{\sim}{\text{u}} \text{y}\&\text{nbsp};\text{th}ừ\text{a}\&\text{nbsp};\text{c}ủ\text{a}\&\text{nbsp};\text{7}\)

Thử \(x = 4\):

\(8^{4} = 4096 , \&\text{nbsp}; 4096 + 342 = 4438 > 7^{5} = 16807 \Rightarrow \text{C} \overset{ˋ}{\text{a}} \text{ng}\&\text{nbsp};\text{xa}\)

Quay lại thử \(x = 0\):

\(8^{0} = 1 , \&\text{nbsp}; 1 + 342 = 343 = 7^{3} \Rightarrow \boxed{x = 0 , \&\text{nbsp}; y = 3} (\text{tho}ả\&\text{nbsp};\text{m} \overset{\sim}{\text{a}} \text{n})\)

✅ Đáp án:

\(\boxed{x = 0 , \&\text{nbsp}; y = 3}\)

hello