Thực hiện nhân tung ra ta có .

a.\(x^3+3x^2+3x+1-\left(x^3-3x+2\right)-3\left(x^2-9\right)=5\)

\(\Leftrightarrow6x+1-2+27=5\Leftrightarrow6x=-21\Leftrightarrow x=-\frac{7}{2}\)

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

\(\Rightarrow x^3=7\Leftrightarrow x=\sqrt[3]{7}\)

c.\(x^3+3x^2+3x+1+x^3-3x^2+3x-1=x^3+6x^2+12x+8+x^3-6x^2+12x-8\)

\(\Leftrightarrow2x^3+6x=2x^3+24x\Leftrightarrow18x=0\Leftrightarrow x=0\)

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

\(=\left(x^3+3x^2+3x+1\right)-\left(x+1\right)\left(x^2-2x+1\right)-3\left(x^2-9\right)\)

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

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

\(=6x+26\)

2x²+x-18 chia hết cho x-3

2x²-6x+6x+x-18

2x(x-3)+6(x-3)+x chia hết cho x-3

(2x+6)(x-3)+(x-3)+3 chia hết cho x-3

=>3 chia hết cho x-3 hay x-3EƯ(3)={1;-1;3;-3}

=>xE{4;2;6;0}

mk k biết biến đổi lp 8 thế này đã được chưa

x thuoc cac gt 0;2;4;6

tic

Ta có: \(x^4-30x^2+31x-30=0\) \(\Rightarrow x^4+x-30x^2+30x-30=0\)

\(\Rightarrow x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)

\(\Rightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)

\(\Rightarrow\left(x^2-x+1\right)\left(x^2+x-30\right)=0\)

Xét \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)

\(\Rightarrow x^2+x-30=0\Rightarrow x^2-5x+6x-30=0\)

\(\Rightarrow\left(x-5\right)\left(x+6\right)=0\Rightarrow\orbr{\begin{cases}x-5=0\\x+6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-6\end{cases}}}\)

Vậy x=5 hoặc x = -6

\(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)=> \(\dfrac{x^2-10x-29}{1971}-1+\dfrac{x^2-10x-27}{1973}-1=\dfrac{x^2-10x-1971}{29}-1+\dfrac{x^2-10x-1973}{27}-1\)=>\(\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}=\dfrac{x^2-10x-2000}{29}+\dfrac{x^2-10x-2000}{27}\) => \(\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)\)

=> \(x^2-10x-2000=0\)

Tự giải ra nhé hi hi

mik lm đc đến đoạn này rùi k bt lm

toàn hđt mà bạn 

a, \(\frac{x^3}{8}+\frac{3}{4}x^2y^2+\frac{3}{2}xy^4+y^6=\left(\frac{x}{2}+y^2\right)^3\)

b, \(m^3+9m^2n+27mn^2+27n^3=\left(m+3n\right)^3\)

c, \(8u^3-48u^2v+96uv^2-64v^3=\left(2y-4v\right)^3\)

d, \(\left(z-t\right)^3+15\left(z-t\right)^2+75\left(z-t\right)+125\)

\(=\left(z-t+5\right)^3\); e, \(x^3+3x^2+3x+1=\left(x+1\right)^3\)

sửa hộ mình ý c =)) do gần nhau quá nên đánh lộn 

\(\left(2u-4v\right)^3\)

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

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

\(\Rightarrow\left[\left(5x^2+3x-2\right)-\left(4x^2-3x-2\right)\right]\left[\left(5x^2+3x-2\right)+\left(4x^2-3x-2\right)\right]=0\)

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

\(\Rightarrow\left(x^2+6x\right)\left(9x^2-4\right)=0\)

\(\Rightarrow x\left(x+6\right)\left[\left(3x\right)^2-2^2\right]=0\)

\(\Rightarrow x\left(x+6\right)\left(3x-2\right)\left(3x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+6=0\\3x-2=0\\3x+2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-6\\3x=2\\3x=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-6\\x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)