Diện tích hình thang cân ABCD ạ

à thôi tôi giải được rồi :v 

Ta có : x³ + x² - 4x - 4 = 0

=> x(x + 1) - 4(x + 1) = 0

=> (x - 4)(x + 1) = 0

=> \(\orbr{\begin{cases}x-4=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-1\end{cases}}\)

Vậy \(x\in\left\{4;-1\right\}\)là giá trị cần tìm

Ta có Xét hiệu\(\left(x^3+y^3+z^3\right)-\left(x+y+z\right)=\left(x-1\right)x\left(x+1\right)+\left(y-1\right)y\left(y+1\right)+\left(z-1\right)z\left(z+1\right)\)

Vì x-1,x,x+1 là 3 số nguyên liên tiếp nên tồn tại 1 số chia hết cho 2 và 1 số chia hết cho 3

Mà (2,3)=1

\(\Rightarrow\left(x-1\right)x\left(x+1\right)⋮6\)

Lập luận tương tự,ta được:\(\hept{\begin{cases}\left(y-1\right)y\left(y+1\right)⋮6\\\left(z-1\right)z\left(z+1\right)⋮6\end{cases}}\)

\(\Rightarrow\left(x-1\right)x\left(x+1\right)+\left(y-1\right)y\left(y+1\right)+\left(z-1\right)z\left(z+1\right)⋮6\)

\(\Rightarrow\left(x^3+y^3+z^3\right)-\left(x+y+z\right)⋮6\)

Mà \(x+y+z⋮6\)

\(\Rightarrow x^3+y^3+z^3⋮6\left(ĐPCM\right)\)

\(x^3+y^3+z^3\)

=> \(\left(x+y+z\right)\). \(\left(x+y+z\right)\).\(\left(x+y+z\right)\)

Mà x , y , z chia hết cho 6

=> \(x^3+y^3+z^3\)chia hết cho 6

\(\Leftrightarrow\)\(\frac{x+1}{x-2}+\frac{8-7x}{x^2-2x}+\frac{2}{x}\left(ĐKXĐ.x\ne2.x\ne0\right)\)

\(\Leftrightarrow\)\(\frac{x\left(x+1\right)}{x\left(x-2\right)}+\frac{8-7x}{x\left(x-2\right)}+\frac{2\left(x-2\right)}{x\left(x-2\right)}\)

\(\Leftrightarrow\frac{x^2+x+8-7x+2x-4}{x\left(x-2\right)}\)

\(\Leftrightarrow\frac{x^2-4x+4}{x\left(x-2\right)}\)

\(\Leftrightarrow\frac{\left(x-2\right)^2}{x\left(x-2\right)}\Leftrightarrow\frac{x-2}{x}\)

\(\frac{x-1}{x+2}-\frac{x+1}{2-x}-\frac{x^2-2x+4}{x^2-4}\)

= \(\frac{x-1}{x+2}+\frac{x+1}{x-2}-\frac{x^2-2x+4}{x^2-4}\)

= \(\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{\left(x+2\right)\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x^2-2x+4}{\left(x-2\right)\left(x+2\right)}\)

= \(\frac{x^2-3x+2+x^2+3x+2-x^2+2x-4}{\left(x-2\right)\left(x+2\right)}=\frac{x^2+2x}{\left(x-2\right)\left(x+2\right)}=\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{x}{x-2}\)

im,em moi lop 3m :)

Mk sai từ dòng 3 nhá -- 

\(=\left(x^2-1\right)\left(\frac{2-\left(x^2-1\right)}{\left(x-1\right)\left(x+1\right)}\right)\)

\(=\frac{\left(x^2-1\right)\left(2-\left(x^2-1\right)\right)}{\left(x-1\right)\left(x+1\right)}=2-x^2+1=3-x^2\)

\(\left(x^2-1\right)\left(\frac{1}{x-1}-\frac{1}{x+1}-1\right)\)

\(=\left(x^2-1\right)\left(\frac{x+1}{\left(x-1\right)\left(x+1\right)}-\frac{x-1}{\left(x+1\right)\left(x-1\right)}-\frac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\right)\)

\(=\left(x^2-1\right)\left(\frac{-\left(x^2-1\right)}{\left(x-1\right)\left(x+1\right)}\right)\)

\(=\frac{-\left(x-1\right)^2\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}=-\left(x-1\right)\left(x+1\right)=-x^2+1\)

x² - 2x - 4 = 0

⇔ ( x² - 2x + 1 ) - 5 = 0

⇔ ( x - 1 )² - ( √5 )² = 0

⇔ ( x - 1 - √5 )( x - 1 + √5 ) = 0

⇔ x = √5 + 1 hoặc x = -√5 + 1

\(x^2-2x-4=0\)

\(\Leftrightarrow\left(x^2-2x+1\right)-5=0\)

\(\Leftrightarrow\left(x-1\right)^2-5=0\Leftrightarrow\left(x-1-\sqrt{5}\right)\left(x-1+\sqrt{5}\right)=0\)

\(\Leftrightarrow x=1\pm\sqrt{5}\)

Ta có (x³ + ax² + bx + 3) : (x² - 2x - 1) = x + a - 2 dư x(b - 2a + 5) + a + 1

Để  (x³ + ax² + bx + 3) \(⋮\) (x² - 2x - 1)

=> x(b - 2a + 5) + a + 1 = 0 \(\forall x\)

=> \(\hept{\begin{cases}b-2a+5=0\\a+1=0\end{cases}}\Rightarrow\hept{\begin{cases}b-2a=-5\\a=-1\end{cases}}\Rightarrow\hept{\begin{cases}b=-7\\a=-1\end{cases}}\)