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

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

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)\left(x-2\right)=0\)

\(\Leftrightarrow x=2,3,4\)

\(a^2-b^2-c^2+2bc\)

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

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

r sao nữa hả bạn

2)

\(\left(2x-1\right)^2-25=0\)

\(\Leftrightarrow\left(2x-1\right)^2=25\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=5\\2x-1=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

4)

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

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

\(\Leftrightarrow x=-1\)

\(f\left(x\right)=2x^3+3x^2-10x+a\)

\(f\left(x\right)\)chia hết cho \(x-2\)nên \(f\left(x\right)=\left(x-2\right).q\left(x\right)\)(\(q\left(x\right)\)là đa thức thương) 

suy ra \(f\left(2\right)=0\)

\(\Rightarrow2.2^3+3.2^2-10.2+a=0\)

\(\Leftrightarrow a=-8\)

Áp dụng định lý Bezout

\(f\left(2\right)=2.2^3+3.2^2-10.2+a=8+a\)

Mà để cho \(2x^3+3x^2-10x+a⋮x-2\)

\(\Rightarrow f\left(2\right)=0\)

\(\Rightarrow8+a=0\)

\(\Rightarrow a=-8\)

Vậy \(a=-8\) để \(2x^3+3x^2-10x+a⋮x-2\)

\(5xy.\left(3x^2-xy+4\right)\)

\(=5xy.3x^2-5xy.xy+5xy.4\)

\(=15x^3y-5x^2y+20xy\)

Vì hình thang \(ABCD\)cân có đáy \(AB\)nên \(\widehat{D}=\widehat{C}=80^o\)

\(\widehat{B}+\widehat{C}=180^o\Leftrightarrow\widehat{B}=180^o-\widehat{C}=180^o=80^o=100^o\)

em cảm ơn

\(A=27x^3+27x^2+9x+1=\left(3x\right)^3+3.\left(3x\right)^2.1+3.\left(3x\right).1^2+1^3\)

\(=\left(3x+1\right)^3=10^3=1000\)

em cảm ơn