a.\(-x^2+2\text{x}-2\le-1\Leftrightarrow-(x-1)^2-1\le-1\)

Do \((x-1)^2\ge0\)

\(\Rightarrow-(x-1)^2\le0\)

\(\Rightarrow-(x-1)^2-1\le-1\)

\(x^3+xy-3x-y=5\)

\(\Leftrightarrow x^3-3x-5=y\left(1-x\right)\)

Với \(x=1\)không thỏa mãn. 

Với \(x\ne1\): 

\(y=\frac{x^3-3x-5}{1-x}=\frac{\left(x-1\right)\left(x^2+x-2\right)-7}{1-x}=-\left(x^2+x-2\right)+\frac{7}{x-1}\)

Để \(y\inℤ\)thì \(\frac{7}{x-1}\inℤ\Leftrightarrow x-1\inƯ\left(7\right)=\left\{-7,-1,1,7\right\}\)

\(\Leftrightarrow x\in\left\{-6,0,2,8\right\}\)

Ta có các bộ \(\left(x,y\right)\)thỏa mãn là: \(\left(-6,-29\right),\left(0,-5\right),\left(2,3\right),\left(8,-69\right)\).

ta có: x⁶ +27=(x²)³ +3³

                    =(x²+3)(x⁴  - 3x² +9)

em ms hok lớp 1

Phân tích mẫu \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)

\(=a^2\left(b-c\right)+b^2c-ab^2+c^2a-c^2b\)

\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b^2-c^2\right)\)

\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b+c\right)\left(b-c\right)\)

\(=\left(b-c\right)\left(a^2+bc-ab-ac\right)=\left(b-c\right)\left[a\left(a-c\right)-b\left(a-c\right)\right]\)

\(=\left(b-c\right)\left(a-b\right)\left(a-c\right)=-\left(b-c\right)\left(a-b\right)\left(c-a\right)\)

Đặt b - c = x, c - a = y, a - b = z

=> x + y + z = b - c + c - a + a - b = 0

Từ x+y+z=0 => x³+y³+z³=3xyz (tự c/m)

=>\(A=\frac{x^3+y^3+z^3}{-xyz}=\frac{3xyz}{-xyz}=-3\)

em ms hok lớp 1

em ms hok lớp 1

\(A=\left(n-1\right)\left(3-2n\right)-n\left(n+5\right)\)

\(=3n-2n^2-3+2n-\left(n^2+5n\right)\)

\(=3n-2n^2-3+2n-n^2-5n\)

\(=\left(3n-5n+2n\right)-\left(2n^2-n^2\right)-3\)

\(=-3\)

\(\Rightarrowđpcm\)

em ms hok lớp 1

\(A=x^3+2x^2-2x-12=\left(x^3-2x^2\right)+\left(4x^2-8x\right)+\left(6x-12\right)\)

\(=x^2\left(x-2\right)+4x\left(x-2\right)+6\left(x-2\right)=\left(x-2\right)\left(x^2+4x+6\right)\)

Vậy B = x²+4x+6

tôi chưa học về cái này nên chua trả lời được

xin lỗi nhé

em ms hok lớp 1

giúp mình =)) @@