a: \(2x^4-3x^3+4x+1⋮x^2-1\)

\(\Leftrightarrow2x^4-2x^2-3x^3+3x+2x^2-2+x+3⋮x^2-1\)

\(\Leftrightarrow x+3⋮x^2-1\)

\(\Leftrightarrow x^2-9⋮x^2-1\)

\(\Leftrightarrow x^2-1\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)

\(\Leftrightarrow x\in\left\{\sqrt{2};-\sqrt{2};0;\sqrt{3};-\sqrt{3};\sqrt{5};-\sqrt{5};3;-3\right\}\)

b: \(x^5+2x^4+3x^2+x-3⋮x^2+1\)

\(\Leftrightarrow x^5+x^3+2x^4+2x^2-x^3-x+x^2+1+2x-4⋮x^2+1\)

\(\Leftrightarrow2x-4⋮x^2+1\)

\(\Leftrightarrow4x^2-16⋮x^2+1\)

\(\Leftrightarrow4x^2+4-20⋮x^2+1\)

\(\Leftrightarrow x^2+1\in\left\{1;2;4;5;10;20\right\}\)

hay \(x\in\left\{0;1;-1;\sqrt{3};-\sqrt{3};2;-2;3;-3;\sqrt{19};-\sqrt{19}\right\}\)

2 câu dễ làm trước, 2 câu còn lại tối đi học về mới làm được..(giờ bận rồi)

a) ĐẶt \(x^2+3x+1=a\)

\(A=a\left(a-4\right)-5=a^2-4a-5=\left(a-5\right)\left(a+1\right)\)

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

\(=\left(x-1\right)\left(x+4\right)\left(x+1\right)\left(x+2\right)\)

c)\(C=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15\)

\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

Đặt ẩn phụ: \(t=x^2+8x+7\) rồi làm tiếp đi..

Để anh làm nốt vậy.

\(B=\left(x^2+2x\right)^2-2x^2-4x-3\)

\(B=\left(x^2+2x\right)^2-2\left(x^2+2x\right)+1-4\)

\(B=\left(x^2+2x-1\right)^2-2^2\)

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

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

___

\(D=x^2-2xy+y^2-7x+7y+12\)

\(D=\left(x-y\right)^2-7\left(x-y\right)+12\)

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

\(D=\left(x-y\right)\left(x-y-3\right)-4\left(x-y-3\right)\)

\(D=\left(x-y-3\right)\left(x-y-4\right)\)

thank you (chuẩn bị mk ra bài 2)

tự mà giải

nếu mk biết giải thì cần gì hỏi mn

\(3x^3+2x^2+2x+3=0\)

\(\Leftrightarrow3\left(x^3+1\right)+2x\left(x+1\right)=0\)

\(\Leftrightarrow3\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x^2-x+3\right)=0\)

Mà \(3x^2-x+3=3\left[\left(x-\frac{1}{6}\right)^2+\frac{35}{36}\right]>0\forall x\)

Do đó: \(x+1=0\Leftrightarrow x=-1\)

Tập nghiệm: \(S=\left\{-1\right\}\)

\(\left(x-1\right)^3+\left(2x+3\right)^3=27x^3+8\)

\(\Leftrightarrow\left[\left(x-1\right)+\left(2x+3\right)\right]\left[\left(x-1\right)^2-\left(x-1\right)\left(2x+3\right)+\left(2x+3\right)^2\right]=27x^3+8\)

\(\Leftrightarrow\left(3x+2\right)\left(x^2-2x+1-2x^2-3x+2x+3+4x^2+12x+9\right)=27x^3+8\)

\(\Leftrightarrow\left(3x+2\right)\left(3x^2+9x+13\right)=\left(3x+2\right)\left(9x^2-6x+4\right)\)

\(\Leftrightarrow\left(3x+2\right)\left(6x^2-15x-9\right)=0\)(Chuyển vế)

\(\Leftrightarrow3\left(3x+2\right)\left(2x^2-5x-3\right)=0\)

\(\Leftrightarrow3\left(3x+2\right)\left(x-3\right)\left(2x+1\right)=0\)

Tập nghiệm: \(S=\left\{-\frac{2}{3};3;-\frac{1}{2}\right\}\)