Ta có : \(x^2+x+13=y^2\)

\(\Leftrightarrow4\left(x^2+x+13\right)=4y^2\)

\(\Leftrightarrow4x^2+4x+52=4y^2\)

\(\Leftrightarrow\left(4x^2+4x+1\right)-4y^2=-51\)

\(\Leftrightarrow\left(2y\right)^2-\left(2x+1\right)^2=51\)

\(\Leftrightarrow\left(2y+2x+1\right)\left(2y-2x-1\right)=51\)

Rồi xét từng trường hợp là ra nha

\(3x^2\left(x-1\right)+5x\left(1-x\right)^2\\ =-3x^2\left(1-x\right)+5x\left(1-x\right)\left(1-x\right)\\ =\left(1-x\right)\left(-3x^2+5x-5x^2\right)\\ =x\left(1-x\right)\left(-3x+5-5x\right)\\ =x\left(1-x\right)\left(5-8x\right)\)

\(3\left(x-y\right)^2+9y\left(y-x\right)^2\\ =3\left(x-y\right)\left(x-y\right)+9y\left(y-x\right)\left(y-x\right)\\ =-3\left(x-y\right)\left(y-x\right)+9y\left(y-x\right)\left(y-x\right)\\ =\left(y-x\right)\left(-3x+3y+9y^2-9xy\right)\\ =\left(y-x\right)\left[-3\left(x-y\right)+9y\left(y-x\right)\right]\\ =\left(y-x\right)\left[-3\left(x-y\right)-9y\left(x-y\right)\right]\\ =\left(y-x\right)\left(-3-9y\right)\left(x-y\right)\\ =-3\left(y-x\right)\left(1+3y\right)\left(x-y\right)\)

\(3\left(x-y\right)^2+9y\left(y-x\right)\\ =3\left(x-y\right)\left(x-y\right)-9y\left(x-y\right)\\ =\left(x-y\right)\left(3x-3y-9y\right)\\ =\left(x-y\right)\left(3x-12y\right)\\ =3\left(x-y\right)\left(x-4y\right)\)

trình bày giúp mình nha...thanks nhìu!!!

mình cho ai trả lời đầu tiên nhá!!!

(x - 13 + y)² + (x - 6 - y)² ≥ 0 + 0 = 0

Vì dấu "=" xảy ra nên x - 13 + y = 0 và x - 6 - y = 0

x + y = 13 và x - y = 6

x = (13 - 6) : 2 = 3,5

y = 13 - 3,5 = 9,5

Vậy x = 3,5 và y = 9,5

(\(x\) - 13 + y)² + (\(x\) - 6 - y)² = 0

(\(x\) - 13 + y)² ≥ 0 ∀ \(x;y\)

(\(x-6-y\))² ≥ 0 ∀ \(x;y\)

⇒(\(x-13+y\))² + (\(x\) - 6- y)² = 0

⇔ \(\left\{{}\begin{matrix}x-13+y=0\\x-6-y=0\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}x-6-y=0\\x-13+y+x-6-y=0\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}y=x-6\\2x=19\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}x=\dfrac{19}{2}\\y=\dfrac{19}{2}-6\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}x=\dfrac{19}{2}\\y=\dfrac{7}{2}\end{matrix}\right.\)

𝓥𝓲̀ \(\left(x-13+y\right)^2\ge0;\left(x-6-y\right)^2\ge0\)

\(\Rightarrow\left(x-13+y\right)^2+\left(x-6-y\right)^2\ge0\)

𝓓𝓪̂́𝓾 𝓫𝓪̆̀𝓷𝓰 𝔁𝓪̉𝔂 𝓻𝓪 𝓴𝓱𝓲 \(\left(x-13+y\right)^2=0;\left(x-6-y\right)^2=0\) 

\(\Rightarrow\left(x-13+y\right)^2=0\)                             \(\Rightarrow\left(x-6-y\right)^2=0\)

\(x-13+y=0\)                                      \(x-6-y=0\)

\(x+y=13\)                                            \(x-y=6\)

\(\Rightarrow\)𝔁 𝓵𝓪̀ 1 𝓼𝓸̂́  𝓵𝓸̛́𝓷 𝓱𝓸̛𝓷 𝔂 𝓫𝓸̛̉𝓲 𝓿𝓲̀ 𝓴𝓱𝓲 𝔁-𝔂 𝓴𝓮̂́𝓽 𝓺𝓾𝓪̉ 𝓵𝓪̀ 1 𝓼𝓸̂́ 𝓷𝓰𝓾𝔂𝓮̂𝓷 𝓭𝓾̛𝓸̛𝓷𝓰

\(\Rightarrow x=\left(13+6\right)\div2=9,5\)                                  

\(\Rightarrow y=13-9,5=3,5\) 

𝓥𝓪̣̂𝔂 𝔁=9,5 𝓿𝓪̀ 𝔂=3,5                          

(\(x\) -13 +y)² + (\(x\) - 6 - y)² = 0

(\(x-13+y\))² ≥0; (\(x\) - 6 - y)² ≥ 0∀ \(x;y\)

⇒(\(x-13+y\))² + (\(x-6-y\))² = 0

⇔ \(\left\{{}\begin{matrix}x-13+y=0\\x-6-y=0\end{matrix}\right.\)

⇒ -13 - 6 + 2\(x\) = 0 ⇒ \(x\) = \(\dfrac{19}{2}\) ⇒ y = \(\dfrac{19}{2}\) - 6 ⇒ y = \(\dfrac{7}{2}\)

Vậy (\(x\);y) = (\(\dfrac{19}{2}\); \(\dfrac{7}{2}\))

\(\left(x-13+y\right)^2+\left(x-6-y\right)^2=0\left(1\right)\)

Ta có :

\(\left\{{}\begin{matrix}\left(x-13+y\right)^2\ge0,\forall x;y\in R\\\left(x-6-y\right)^2\ge0,\forall x;y\in R\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}\left(x-13+y\right)^2=0\\\left(x-6-y\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-13+y=0\\x-6-y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=19\\y=x-6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{2}\\y=\dfrac{19}{2}-6=\dfrac{7}{2}\end{matrix}\right.\)

Vậy \(\left\{{}\begin{matrix}x=\dfrac{19}{2}\\y=\dfrac{7}{2}\end{matrix}\right.\) thoả mãn đề bài

chịu thôi

   x³ + x²y - y + x + x³y² - x³ + x²y

= x³ - x³ + x²y + x²y - y + x + x³y²

= 2x²y - y + x + x³y²

thanks Hàn Băng Dii 😄😄😄

cứu mình đi mấy bạn ,mai nộp rồi

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