thay anh dai dien kieu j day mik vs

vao thong tin tai khoan o cho hinh tam giac ben canh ten cua ban roi an vao doi anh hien thi .xong

T a   c ó : x 2 + y 2 = a 2 + b 2 ⇔ x 2 - a 2 = b 2 - y 2 ⇔ x - a x + a = b - y b + y M à   x + y   = a + b ⇔ x - a = b - y   n ê n   t a   c ó x - a x + a = x - a b + y ⇔ x - a x + a - x - a b + y = 0 ⇔ x - a x + a - b - y = 0 ⇔ x - a = 0 x + a - b - y = 0 ⇔ x = a x - y = b - a

+) Với x = a thay vào x + y = a + b ta có: a + y = a + b

Suy ra y = b

Do đó:   x n + y n = a n + b n

+) Với x - y = b - a suy ra x = b - a + y thay vào x + y = a + b ta có:

 b - a + y  + y = a + b

2y = 2a

y = a

Suy ra x - a = b - a hay x = b

Do đó:  x n + y n = b n + a n = a n + b n

Vậy  x n + y n = a n + b n

Đáp án cần chọn là C

Áp dụng tính chất các dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=\dfrac{x+y+z}{a+b+c}=\dfrac{x+y+z}{1}\)

\(x=a\left(x+y+z\right)=x^2=a^2.\left(x+y+z\right)^2\)

\(y=b\left(x+y+z\right)=y^2=b^2\left(x+y+z\right)^2\)

\(z=c\left(x+y+z\right)=z^2=c^2.\left(x+y+z\right)^2\)

\(\Rightarrow x^2+y^2+z^2=a^2\left(x+y+z\right)^2+b^2\left(x+y+z\right)^2+c^2\left(x+y+z\right)^2\)

                         \(=\left(x+y+z\right)^2\left(a^2+b^2+c^2\right)=\left(x+y+z\right)^2\) (do \(a^2+b^2+c^2=1\))

lol

a. Trừ vế theo vế \(\left(1\right)\) cho \(\left(2\right)\) ta được \(x^2-y^2=4x-4y\)

\(\Leftrightarrow\left(x-y\right)\left(x+y-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=4-y\end{matrix}\right.\)

TH1: \(x=y\)

Phương trình \(\left(1\right)\) tương đương:

\(x^2=2x\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y=0\\x=y=2\end{matrix}\right.\)

TH2: \(x=4-y\)

Phương trình \(\left(2\right)\) tương đương:

\(y^2=4y-4\)

\(\Leftrightarrow y^2-4y+4=0\)

\(\Leftrightarrow\left(y-2\right)^2=0\)

\(\Leftrightarrow y=2\)

\(\Rightarrow x=2\)

Vậy hệ đã cho có nghiệm \(\left(x;y\right)\in\left\{\left(0;0\right);\left(2;2\right)\right\}\)

b. \(\left\{{}\begin{matrix}x+y+xy=5\\x^2+y^2=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy=5-\left(x+y\right)\\\left(x+y\right)^2-2xy=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy=5-\left(x+y\right)\\\left(x+y\right)^2-10+2\left(x+y\right)=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy=5-\left(x+y\right)\\\left(x+y\right)^2+2\left(x+y\right)-15=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy=5-\left(x+y\right)\\\left(x+y+5\right)\left(x+y-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy=5-\left(x+y\right)\\\left[{}\begin{matrix}x+y=-5\\x+y=3\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y=-5\\xy=10\end{matrix}\right.\\\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x+y=-5\\xy=10\end{matrix}\right.\Leftrightarrow\) vô nghiệm

TH2: \(\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\end{matrix}\right.\)

Vậy ...

Ta có x + y = a + b 

=> (x + y)² = (a + b)² 

=> x² + y² + 2xy = a² + b² + 2ab 

=> xy = ab

Lại có x + y = a + b

=> (x  + y)³ = (a + b)³ 

=> x³ + 3x²y + 3xy² + y³ = a³ + 3a²b + 3ab² + b³ 

=> x³ + y³ + 3xy(x + y) = a³ + b³ + 3ab(a + b)

=> x³ + y³ = a³ + b³ (vì x + y = a + b ; xy = ab)