a) \(x^4-2x^3+2x^2-2x+a=\left(x^2-2x+1\right)\left(x^2+bx+c\right)\) (sửa đề)

\(\Leftrightarrow x^2\left(x^2-2x+1\right)+\left(x^2-2x+1\right)+a-1=\left(x^2-2x+1\right)\left(x^2+bx+c\right)\)

\(\Leftrightarrow\left(x^2-2x+1\right)\left(x^2+1\right)+a-1=\left(x^2-2x+1\right)\left(x^2+bx+c\right)\)

\(\Rightarrow\left\{{}\begin{matrix}a-1=0\\b=0\\c=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=c=1\\b=0\end{matrix}\right.\)

b) \(x^3+3x^2-x-3=\left(x-2\right)\left(x^2+bx+c\right)+a\)

\(\Leftrightarrow x^3-2x^2+5x^2-10x+9x-18+15=\left(x-2\right)\left(x^2+bx+c\right)+a\)

\(\Leftrightarrow x^2\left(x-2\right)+5x\left(x-2\right)+9\left(x-2\right)+15=\left(x-2\right)\left(x^2+bx+c\right)+a\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+5x+9\right)+15=\left(x-2\right)\left(x^2+bx+c\right)+a\)

\(\Rightarrow\left\{{}\begin{matrix}a=15\\b=5\\c=9\end{matrix}\right.\)

#$\mathtt{Toru}$

a) Chu vi của miếng bánh là:

\(\dfrac{5x}{2}+8x+4y^2=\dfrac{5x+16x}{2}+4y^2=\dfrac{21x+8y^2}{2}\)

b) Chu vi của miếng bánh là:

\(\dfrac{21\cdot4+8\cdot3^2}{2}=78\left(cm\right)\)

c) Chu vi của miếng bánh là:

\(\dfrac{21\cdot1,5+8\cdot2,34^2}{2}=37,6524\left(cm\right)\)

d) Diện tích của miếng bánh là:

\(\dfrac{1}{2}\cdot8x\cdot\left(2,5x+1\right)=4x\left(2,5x+1\right)=10x^2+4x\)

Ta có: AC > BC > AB 

\(BC^2+AB^2=20^2+15^2=625\) (1)

\(AC^2=25^2=625\) (2)

Từ (1) và (2) \(\Rightarrow BC^2+AB^2=AC^2\) 

Vậy ΔABC là tam giác vuông tại C

Ta có: 

\(\left\{{}\begin{matrix}AB=15\left(cm\right)\\BC=20\left(cm\right)\\AC=25\left(cm\right)\end{matrix}\right.\)(giả thiết)

\(\Rightarrow\left\{{}\begin{matrix}AB^2+BC^2=15^2+20^2=225+400=625\left(cm\right)\\AC^2=25^2=625\left(cm\right)\end{matrix}\right.\)

\(\Rightarrow AB^2+BC^2=AC^2\)

\(\Rightarrow\Delta ABC\) vuông (theo định lý Pi-ta-go đảo)

Vậy \(\Delta ABC\) là tam giác vuông.

\(4\dfrac{7}{5741}\cdot\dfrac{1}{3759}-\dfrac{4}{3759}\cdot1\dfrac{2}{5741}+\dfrac{1}{3759}+\dfrac{1}{3759\cdot5741}\\ =\dfrac{22971}{5741}\cdot\dfrac{1}{3759}-\dfrac{1}{3759}\cdot\dfrac{22972}{5741}+\dfrac{1}{3759}\cdot\dfrac{5741}{5741}+\dfrac{1}{3759}\cdot\dfrac{1}{5741}\\ =\dfrac{1}{3759}\cdot\left(\dfrac{22971}{5741}-\dfrac{22972}{5741}+\dfrac{5741}{5741}+\dfrac{1}{5741}\right)\\ =\dfrac{1}{3759}\cdot\dfrac{5741}{5741}=\dfrac{1}{3759}\cdot1=\dfrac{1}{3759}\)

Với `x=-19 => 1-x=20`

Thay `1-x=20` vào B, ta được:

`B=x^{10}+(1-x).x^9+(1-x).x^8+...+(1-x).x^2+(1-x).x+20`

`=x^{10}+x^9-x^{10}+x^8-x^9+...+x^2-x^3+x-x^2+20`

`=x+20=-19+20=1`

\(x^4+6x^3+7x^2-6x+1\\ =\left(x^4+6x^3+9x^2\right)-2x^2-6x+1\\ =\left[\left(x^2\right)^2+2\cdot x^2\cdot3x+\left(3x\right)^2\right]-2\left(x^2+3x\right)+1\\ =\left(x^2+3x\right)^2-2\left(x^2+3x\right)+1^2\\=\left(x^2+3x-1\right)^2\)

1 He seldom studies Maths because he does not like it
2 To make this soup, you should mix it with beef and wine

1. He seldom studies maths because he does not like it.

2. To make this soup, you should mix it with beef and wine.

\(g)\dfrac{x}{xy+y^2}-\dfrac{y}{x^2+xy}\\ =\dfrac{x}{y\left(x+y\right)}-\dfrac{y}{x\left(x+y\right)}\\ =\dfrac{x^2}{xy\left(x+y\right)}-\dfrac{y^2}{xy\left(x+y\right)}\\ =\dfrac{x^2-y^2}{xy\left(x+y\right)}\\ =\dfrac{\left(x+y\right)\left(x-y\right)}{xy\left(x+y\right)}\\ =\dfrac{x-y}{xy}\) 

h) 

\(\dfrac{x^2+4}{x^2-4}-\dfrac{x}{x+2}-\dfrac{x}{2-x}\\ =\dfrac{x^2+4}{\left(x+2\right)\left(x-2\right)}-\dfrac{x}{x+2}+\dfrac{x}{x-2}\\ =\dfrac{x^2+4}{\left(x+2\right)\left(x-2\right)}-\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\\ =\dfrac{x^2+4-x^2+2x+x^2+2x}{\left(x+2\right)\left(x-2\right)}\\ =\dfrac{x^2+4x+4}{\left(x+2\right)\left(x-2\right)}\\ =\dfrac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}\\ =\dfrac{x+2}{x-2}\) 

i) 

\(\dfrac{5}{6x-6}+\dfrac{9}{14x-14}+\dfrac{6}{7x-7}\\ =\dfrac{5}{6\left(x-1\right)}+\dfrac{9}{14\left(x-1\right)}+\dfrac{6}{7\left(x-1\right)}\\ =\dfrac{7\cdot5}{42\left(x-1\right)}+\dfrac{3\cdot9}{42\left(x-1\right)}+\dfrac{6\cdot6}{42\left(x-1\right)}\\ =\dfrac{35+27+36}{42\left(x-1\right)}\\ =\dfrac{98}{42\left(x-1\right)}\\ =\dfrac{7}{3\left(x-1\right)}\)

\(x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128\\ =\left[\left(x+6\right)\left(x+4\right)\right]\left[x\left(x+10\right)\right]+128\\ =\left(x^2+6x+4x+24\right)\left(x^2+10x\right)+128\\ =\left(x^2+10x+24\right)\left(x^2+10x\right)+128\\ =\left(x^2+10x\right)^2+24\left(x^2+10x\right)+128\\ =\left(x^2+10x\right)^2+8\left(x^2+10x\right)+16\left(x^2+10x\right)+128\\ =\left(x^2+10x\right)\left[\left(x^2+10x\right)+8\right]+16\left[\left(x^2+10x\right)+8\right]\\ =\left(x^2+10x+8\right)\left(x^2+10x+16\right)\)

\(\left(-2x^5+x^4-3x^3\right):2x^3\)
\(=-x^2+\dfrac{1}{2}x-\dfrac{3}{2}\)

Mình nghĩ đề như này đúng kh bạn? \(\left(-2x^5+x^4-3x^3\right):\left(2x^3\right)\), còn đề như trên thì thực hiện chia 2 rồi nhân x mũ 3 bạn nhé.

\(\left(-2x^5+x^4-3x^3\right):\left(2x^3\right)\\ =\dfrac{-2x^5}{2x^3}+\dfrac{x^4}{2x^3}-\dfrac{3x^3}{2x^3}\\ =-x^2+\dfrac{x}{2}-\dfrac{3}{2}\)