\(\left(5x+7\right)-\left(x-3\right)^2\)

\(=5x+7-\left(x^2-6x+9\right)\)

\(=5x+7-x^2+6x-9\)

\(=-x^2+11x-2\)

\(=-1.\left(x^2-11x+2\right)\)

\(y^3-\frac{1}{8}=0\Leftrightarrow y^3-\left(\frac{1}{2}\right)^3=0\)

\(\Leftrightarrow\left(y-\frac{1}{2}\right)\left(y^2+\frac{1}{2}y+\frac{1}{4}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y-\frac{1}{2}=0\\y^2+\frac{1}{2}y+\frac{1}{4}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=\frac{1}{2}\\y^2+2.\frac{1}{4}y+\frac{1}{16}+\frac{3}{16}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=\frac{1}{2}\\\left(y+\frac{1}{4}\right)^2+\frac{3}{16}=0\left(vôlí\right)\end{cases}}\)

Vậy \(y=\frac{1}{2}\)

Bài 1: Phân tích các đa thức sau thành nhân tử

\(36a^4-y^2\)

\(=\left(6a\right)^2-y^2\)

\(=\left(6a^2-y\right).\left(6a^2+y\right)\)

\(6x^2+x-2\)

\(=6x^2+4x-3x-2\)

\(=2x.\left(3x+2\right)-\left(3x+2\right)\)

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

Bài 2: Tìm x, biết

\(x.\left(x-4\right)+1=3x-5\)

\(\Rightarrow x^2-4x+1=3x-5\)

\(\Rightarrow x^2-4x+1-3x+5=0\)

\(\Rightarrow x^2-7x+6=0\)

\(\Rightarrow x^2-6x-x+6=0\)

\(\Rightarrow x.\left(x-6\right)-\left(x-6\right)=0\)

\(\Rightarrow\left(x-6\right).\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-6=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=1\end{cases}}\)

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

\(\Rightarrow\left(2x^3-3x^2\right)-\left(2x-3\right)=0\)

\(\Rightarrow x^2.\left(2x-3\right)-\left(2x-3\right)=0\)

\(\Rightarrow\left(x^2-1\right).\left(2x-3\right)=0\)

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

Trường hợp 1: \(x-1=0\Rightarrow x=1\)

Trường hợp 2: \(x+1=0\Rightarrow x=-1\)

Trường hợp 3: \(2x-3=0\Rightarrow x=\frac{3}{2}\)

Bài 3:

a)

\(A=x^3-9x^2+27x-27\)

\(=x^3-3x^2.3x.3^x-3^3\)

\(=\left(x-3\right)^3\)

Thay vào ta được 

\(A=\left(1-3\right)^3\)

\(=\left(-2\right)^3\)

\(=-8\)

Vậy \(A=-8\) khi \(x=1\)

b)

 

Vậy đa thức thương là \(2x^2+5x+9\)

Vậy đa thức dư là \(a-18\)

hhi bài mình hoi bẩn sorry nha

\(16x-5x^2-3\)

\(=-5x^2+16x-3\)

\(=-5x^2+15x+x-3\)

\(=\left(-5x^2+15x\right)+\left(x-3\right)\)

\(=-5x.\left(x-3\right)+\left(x-3\right)\)

\(=\left(-5x+1\right).\left(x-3\right)\)

\(2x^2+7x+5\)

\(=2x^2+2x+5x+5\)

\(=\left(2x^2+2x\right)+\left(5x+5\right)\)

\(=2x.\left(x+1\right)+5.\left(x+1\right)\)

\(=\left(2x+5\right).\left(x+1\right)\)

\(2x^2+3x+5\) (Bạn xem lại đề nhé.)

\(x^3-3x^2+1-3x\)

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

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

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

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

\(x^2-4x-5\)

\(=x^2-5x+x-5\)

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

\(=x.\left(x-5\right)+\left(x-5\right)\)

\(=\left(x+1\right).\left(x-5\right)\)

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

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

\(=\left(a^2-2a+1\right).\left(a^2+2a+1\right)\)

\(=\left(a-1\right)^2.\left(a+1\right)^2\)

\(x^3-0,25x=0\)

\(\Rightarrow x.\left(x^2-0,25\right)=0\)

\(\Rightarrow x.\left(x-0,5\right).\left(x+0,5\right)=0\)

Trường hợp 1: \(x=0\)

Trường hợp 2: \(x-0,5=0\Rightarrow x=0,5\)

Trường hợp 3: \(x+0,5=0\Rightarrow x=-0,5\)

\(2x.\left(3x-5\right)-\left(5-3x\right)=0\)

\(\Rightarrow2x.\left(3x-5\right)+\left(3x-5\right)=0\)

\(\Rightarrow\left(3x-5\right).\left(2x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-5=0\\2x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{-1}{2}\end{cases}}\)

\(49x^2+14x+1=0\)

\(\Rightarrow\left(7x+1\right)=0\)

\(\Rightarrow7x=-1\)

\(\Rightarrow x=\frac{-1}{7}\)

\(x^2+4x+3\)

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

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

\(=x.\left(x+1\right)+3.\left(x+1\right)\)

\(=\left(x+3\right).\left(x+1\right)\)