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

Đặt x-1 = a (a thuộc N*) (1)

2x + 3 =b ( b thuộc N*) (2)

=> (x-1) + (2x+3) = 3x+2

Ta có a\(^3\)+ b\(^3\)=( a+b)\(^3\)

=> a\(^3\) + b\(^3\)= a\(^3\)+ 3a\(^2\)b + 3ab\(^2\)+ b\(^3\)

=> 3a\(^2\)b + 3ab\(^2\)=0

=> 3ab(a+b) = 0

=> a=0 hoặc b = 0

+) Thay a=0 vào (1), ta có: x-1=0 <=> x=1

+) Thay b=0 vào (2) ta có 2x+3 =0 <=> x=\(\dfrac{-3}{2}\)

Vậy nghiệm của pt là 1; \(\dfrac{-3}{2}\)

a,\(x\left(x+1\right)\left(x^2+x+2\right)\)

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

ĐẶT X^2+X=A\(\Rightarrow\left(x^2+x\right)\left(x^2+x+2\right)=a\left(a+2\right)=42\)

\(\Rightarrow a=\pm1,\pm2,\pm3,\pm6,\pm7,\pm42\)

SUY RA TÌM ĐC X

b,

a) \(x\left(x+1\right)\left(x^2+x-2\right)=48\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=48\)

Đặt \(x^2+x=t\Rightarrow t\left(t-2\right)=48\Leftrightarrow t^2-2t-48=0\Leftrightarrow\orbr{\begin{cases}x=-8\\x=6\end{cases}}\)

Với x = -8, ta có: \(x^2+x=-8\Leftrightarrow x^2+x+8=0\) (Vô nghiệm)

Với x = 6, ta có: \(x^2+x=6\Leftrightarrow x^2+x-6=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

Vậy phương trình có tập nghiệm \(S=\left\{-3;2\right\}\)

b) \(\left(x-1\right)^3+\left(2x+3\right)^3=27x^3+8\)

\(\Leftrightarrow\left(x-1+2x+3\right)\left[\left(x-1\right)^2-\left(x-1\right)\left(2x+3\right)+\left(2x+3\right)^2\right]=\left(3x+2\right)\left(9x^2-6x+4\right)\)

\(\Leftrightarrow\left(3x+2\right)\left(3x^2+9x+13\right)=\left(3x+2\right)\left(9x^2-6x+4\right)\)

\(\Leftrightarrow\left(3x+2\right)\left(3x^2+9x+13-9x^2+6x-4\right)=0\)

\(\Leftrightarrow\left(3x+2\right)\left(-6x^2+15x+9\right)=0\)

TH1: \(3x+2=0\Leftrightarrow x=-\frac{2}{3}\)

TH2: \(-6x^2+15x+9=0\Leftrightarrow\left(x-3\right)\left(-6x-3\right)=0\)

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

xin loi la( 2x-1)3 nha mn

cách 1: khai triển ra rồi làm 

cách 2: đặt \(x+1=a\)và \(2x-1=b\)\(\Rightarrow a+b=3x\)

phương đã cho \(\Leftrightarrow a^3+b^3=\left(a+b\right)^3\)

phần còn lại tự giải nhé

Bài làm

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

\(\Leftrightarrow\left(x^3+3x^2+3x+1\right)+8x^3=27x^3\)

\(\Leftrightarrow x^3+3x^2+3x+1+8x^3=27x^3\)

\(\Leftrightarrow\left(x^3+8x^3-27x^3\right)+3x^2+3x+1=0\)

\(\Leftrightarrow-18x^3+3x^2+3x+1=0\)

tự giải nốt. 

Ta có: \(\left(x-1\right)^3+\left(2x+3\right)^3=27x^3+8\)

\(\Leftrightarrow x^3-3x^2+3x-1+8x^3+36x^2+54x+27-27x^3-8=0\)

\(\Leftrightarrow-18x^3+33x^2+57x+18=0\)

\(\Leftrightarrow-18x^3+54x^2-21x^2+63x-6x+18=0\)

\(\Leftrightarrow-18x^2\left(x-3\right)-21x\left(x-3\right)-6\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(-18x^2-21x-6\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(-18x^2+9x+12x-6\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left[-9x\left(2x-1\right)+6\left(2x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x-1\right)\left(-9x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x-1=0\\-9x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=1\\-9x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{3;\dfrac{1}{2};\dfrac{2}{3}\right\}\)

pt⇔x3−3x2+3x−1+8x3+36x2+54x+27=27x3+8

⇔18x3−33x2−57x−18=0

⇔(3x+2)(6x2−15x−9)=0

⇔3(3x+2)(2x+1)(x−3)=0

⇔x∈{−12,−23,3}

Giải pt :

a) \(2x\left(x+5\right)-\left(x-3\right)^2=x^2+6\)

\(\Leftrightarrow2x^2+10x-x^2+6x-9-x^2-6=0\)

\(\Leftrightarrow16x-15=0\)

\(\Leftrightarrow x=\frac{15}{16}\)

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

\(\Leftrightarrow2x-18=2x\)

\(\Leftrightarrow-18=0\)( vô lí )

=> x thuộc rỗng

c)d) tương tự

e) \(\frac{5x-2}{6}+\frac{3-4x}{2}=2-\frac{x+7}{3}\)

\(\Leftrightarrow\frac{5x-2}{6}+\frac{9-12x}{6}=\frac{12}{6}-\frac{2x+14}{6}\)

\(\Leftrightarrow5x-2+9-12x=12-2x-14\)

\(\Leftrightarrow-5x+9=0\)

\(\Leftrightarrow x=\frac{9}{5}\)

f) \(\frac{2x-1}{2}=\frac{2x+1}{4}-\frac{1-2x}{8}\)

\(\Leftrightarrow\frac{4\left(2x-1\right)}{8}=\frac{2\left(2x+1\right)}{8}-\frac{1-2x}{8}\)

\(\Leftrightarrow8x-4=4x+2-1+2x\)

\(\Leftrightarrow2x-5=0\)

\(\Leftrightarrow x=\frac{5}{2}\)

Tìm x :

a) \(3x^3-27x=0\)

\(\Leftrightarrow3x\left(x^2-9\right)=0\)

\(\Leftrightarrow3x\left(x-3\right)\left(x+3\right)=0\)

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

b) \(2x^3-12x^2+18x=0\)

\(\Leftrightarrow2x\left(x^2-6x+9\right)=0\)

\(\Leftrightarrow2x\left(x-3\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)