a) Đặt A(x)=0

\(\Leftrightarrow-4x-5=0\)

\(\Leftrightarrow-4x=5\)

hay \(x=-\dfrac{5}{4}\)

b) Đặt B(x)=0

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

\(\Leftrightarrow6x-3-2x-2=0\)

\(\Leftrightarrow4x=5\)

hay \(x=\dfrac{5}{4}\)

1: \(=x^2+1\)

3: \(=\left(x-y-z\right)^2\)

a: P(x)=6x^3-4x^2+4x-2

Q(x)=-5x^3-10x^2+6x+11

M(x)=x^3-14x^2+10x+9

b: \(C\left(x\right)=7x^4-4x^3-6x+9+3x^4-7x^3-5x^2-9x+12\)

=10x^4-11x^3-5x^2-15x+21

\(x^4+2x^3-2x^2+2x-3=0\\ \Leftrightarrow x^4+3x^3-x^3-3x^2+x^2+3x-x-3=0\\ \Leftrightarrow x^3\left(x+3\right)-x^2\left(x+3\right)+x\left(x+3\right)-\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^3-x^2+x-1\right)=0\\ \Leftrightarrow\left(x+3\right)\left[x^2\left(x-1\right)+\left(x-1\right)\right]=0\\ \Leftrightarrow\left(x+3\right)\left(x-1\right)\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-1=0\\x^2+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\left(\text{vì }x^2+1\ge1>0\right)\)

Vậy ...

\(\left(x-1\right)\left(x^2+5x-2\right)-x^3+1=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+5x-2\right)-\left(x-1\right)\left(x^2+x+1\right)=0\\ \Leftrightarrow\left(x-1\right)\left[\left(x^2+5x-2\right)-\left(x^2+x+1\right)\right]=0\\ \Leftrightarrow\left(x-1\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{4}\end{matrix}\right.\)

Vậy ...

\(x^2+\left(x+2\right)\left(11x-7\right)=4\\ \Leftrightarrow x^2-4+\left(x+2\right)\left(11x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-2\right)+\left(11x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-2+11x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\\ \Leftrightarrow3\left(x+2\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+2=0\\4x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)

Vậy ...

nghiệm đâu bạn ưi...nó là phương trình vô nghiệm hay vô số nghiệm vậy m :))

a: \(=\dfrac{x^3-3x^2-7x+x^2-3x-7}{x^2-3x-7}=x+1\)

b:\(=\dfrac{x^3+x^2+3x^2+3x+5x+5}{x+1}=x^2+3x+5\)

c:\(=\dfrac{x^3-3x^2-7x+2x^2-6x-14}{x^2-3x-7}=x+2\)

d: \(=\dfrac{x^2\left(x+5\right)+5x+25-25}{x+5}=x^2+5-\dfrac{25}{x+5}\)

a) Kết quả x + 3.          b) Kết quả  x 2  + 1.

\(\left(1-x\right)\left(5x+3\right)=\left(3x-7\right)\left(x-1\right)\)

\(< =>\left(1-x\right)\left(5x+3+3x-7\right)=0\)

\(< =>\left(1-x\right)\left(8x-4\right)=0\)

\(< =>\orbr{\begin{cases}1-x=0\\8x-4=0\end{cases}< =>\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}}\)

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

\(< =>\left(x-2\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\)

\(< =>\left(x-2\right)\left(x+1-x-2\right)=0\)

\(< =>-1\left(x-2\right)=0\)

\(< =>2-x=0< =>x=2\)