Mình chưa học phương trình nên giải theo cách của lớp dưới thôi :)))

Vì \(\hept{\begin{cases}345⋮5\\5y^2⋮5\end{cases}}\Rightarrow3x^2⋮5\)

Mà \(\left(3;5\right)=1\Rightarrow x^2⋮5\Rightarrow x⋮5\)

Lại có \(3x^2\le345\Rightarrow x^2\le115\Rightarrow\left|x\right|\le10\)

Mà \(x⋮5\Rightarrow x\in\left\{0;\pm5;\pm10\right\}\)

\(\Rightarrow y^2=\frac{345-3.25}{5}=54\)không phải số chính phương

\(\Rightarrow y^2=\frac{345-3.100}{5}=9\Rightarrow y=\pm3\)

Vậy \(\left(x;y\right)\in\left\{\left(10;3\right);\left(10;-3\right);\left(-10;3\right);\left(-10;-3\right)\right\}\)

\(3x^2+5y^2=345=>x^2=\frac{345-5y^2}{3}=>x=\sqrt{\frac{345-5y^2}{3}}\)

MODE 7 (TABLE) nhập \(f\left(x\right)=\sqrt{\frac{345-5X^2}{3}}\)

start -9 end: 9 ,step=1

tìm đc \(\left(x;y\right)=\left(10;3\right);\left(3;10\right);\left(-10;-3\right);\left(-3;-10\right)\)

đây là sử dụng máy tính casio

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mấy dạng này nên sử dụng google không tính phí =))

+)Câu hỏi của Nguyễn Hương Ly - Toán lớp 8 - Học toán với OnlineMath

+)Tìm nghiệm nguyên của phương trình : 3*x^2 + 5*y^2 = 345? | Yahoo Hỏi & Đáp

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Ta có:  

x² + 2y² + 3xy + 3x + 5y = 15

<=> x² + 2y² + 3xy + 3x + 5y + 2 = 17

<=> (x² + xy + 2x) + (2xy + 2y² + 4y) + (x + y + 2) = 17

<=> (x + y + 2)(x + 2y + 1) = 17

=> (x + y + 2, x + 2y + 1) = (1,17; 17,1; - 1,-17; -17,-1)

Giải ra là tìm được x,y nhé

VeryVery good.Thanks. I will give 1  for you.Love

thanks nhìu

Ta có: 5y² chia hết cho 5; 345 chia hết cho 5.

Vậy: 3x² phải chia hết cho 5.

=> x chia hết cho 5

Trường hợp 1: x = 0

=> PT vô nghiệm.

Trường hợp 2: x = 5

=> PT vô nghiệm

Trường hợp 3: x = 10

=> PT có nghiệm x = 10; y = 3

Trường hợp 4: x >= 15

=> VT > VP

=> PT có nghiệm duy nhất: x = 10, y = 3.