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

\(=\left(x^2+2x+1\right)-\left(4x^2-4x+1\right)+3\left(x^2-4\right)\)

\(=\left(x^2-4x^2+3x^2\right)+\left(2x+4x\right)+\left(1-1-12\right)=6x-12=-6\)

Trả lời:

\(9x^2+6x-3=0\)

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

\(\Leftrightarrow3x^2+2x-1=0\)

\(\Leftrightarrow3x^2+3x-x-1=0\)

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

\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)

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

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

Vậy x = - 1; x = 1/3 là nghiệm của pt.

Trả lời:

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

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

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

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

\(=-\left[x\left(x+3\right)-\left(x+3\right)\right]\)

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

TL

x=-3,

x=1

HT

\(f\left(x\right)=ax^5+5x^4-9\)

Để \(f\left(x\right)\)chia hết cho \(x-1\)thì \(f\left(x\right)=q\left(x\right)\left(x-1\right)\)

suy ra \(f\left(1\right)=0\)

\(a+5-9=0\Leftrightarrow a=4\).

\(x+\frac{3}{5}=x-\frac{1}{7}\)

=>  \(x+\frac{21}{35}=x-\frac{5}{35}\)

=> \(x-x+\frac{21}{35}+\frac{5}{35}=0\)

=>  \(0x=-\frac{26}{5}\)

Vậy ko có giá trị x nào thoả mãn bài toán

\(\frac{x+3}{5}=\frac{x-1}{7}\)

\(\Leftrightarrow7\left(x+3\right)=5\left(x-1\right)\)

\(\Leftrightarrow7x+21=5x-5\)

\(\Leftrightarrow7x-5x=-5-21\)

\(\Leftrightarrow2x=-26\)

\(\Leftrightarrow x=-13\)

sorry em,anh ko biết

a b rỗng hết nhá

\(A=x^2-x+5=2^2-2+5=2+5=7\)

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

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

\(=-2\)

        (9y+7z)

\(9y^2+7yz+9y+7z\)

\(=9y^2+9y+7yz+7z\)

\(=9y\left(y+1\right)+7z\left(y+1\right)\)

\(=\left(9y+7z\right)\left(y+1\right)\)