Question

[shoppe pi pi pi pi]

giải phương trình sau

a)(x^2+x+1)(x^2+x+2)=12

b)x(x+1)(x^2+x+1)=42

c)x(x+1)(x-1)(x+2)=24

d)(x^2+1)^2+3x(x^2+1)+2x^2=0

Answer 1

[KAl(SO4)2·12H2O]

a) \(\left(x^2+x+1\right)\left(x^2+x+2\right)=12\)

\(\Leftrightarrow x^4+2x^3+4x^2+3x+2=12\)

\(\Leftrightarrow x^4+2x^3+4x^2+3x+2=12-12\)

\(\Leftrightarrow x^4+2x^3+4x^2-10=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2+x+5\right)\)

\(\Rightarrow\hept{\begin{cases}x=1\\x=2\end{cases}}\)

b) \(x\left(x+1\right)\left(x^2+x+1\right)=42\)

\(\Leftrightarrow x^4+2x^3+2x^2+x=42\)

\(\Leftrightarrow x^4+2x^3+2x^2+x=42-42\)

\(\Leftrightarrow x^4+2x^3+2x^2+x=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(x^2+x+7=0\right)\)

\(\Rightarrow\hept{\begin{cases}x=2\\x=-3\\x=-\frac{1}{2}\end{cases}}\)

c) làm tương tự b).

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

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

\(\Rightarrow\hept{\begin{cases}x=1\\x=-\frac{1}{2}\end{cases}}\)

Trình độ hơi thấp, có gì sai sót mong bạn bỏ qua cho ạ

Answer 2

[kudo shinichi]

Chỉ gợi ý thôi.

a) đặt x^2+x+1=t

=> pt <=> t(t+1)=12

tự làm nốt.

b) x(x+1)(x^2+x+1)=42

<=> (x^2+x)(x^2+x+1)=42

đặt x^2+x=t

=> pt <=>t(t+1)=42

...............................

c) x(x+1)(x-1)(x+2)=24

(x^2+x)(x^2+x-2)=24

Đặt x^2+x=t

=> pt <=> t(t-2)=24

............................

d) (x^2+1)^2+3x(x^2+1)+2x^2=0

(x^2+1)^2+x(x^2+1)+2x(x^2+1)+2x^2=0

(x^2+1)(x^2+x+1)+2x(x^2+x+1)=0

(x^2+x+1)(x^2+2x+1)=0

(x^2+x+1)(x+1)^2=0  (1)

Ta có: x^2+x+1=(x+1/2)^2+3/4>0 với mọi x

=> (1) <=> (x+1)^2=0

<=> x=-1

Vậy x=-1