Question

Giải pt a) x³+x-30=0

b) |x²+x+6|=x+7

c) |x+11|=x²+x+10

Answer 1

a/

\(\Leftrightarrow x^3-27+x-3=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x^2+3x+10=0\left(vn\right)\end{matrix}\right.\)

b/ Do \(x^2+x+6=\left(x+\frac{1}{2}\right)^2+\frac{23}{4}>0;\forall x\) nên pt tương đương:

\(x^2+x+6=x+7\)

\(\Leftrightarrow x^2=1\Rightarrow x=\pm1\)

c/ TH1: \(x\ge-11\)

\(\Leftrightarrow x+11=x^2+x+10\Leftrightarrow x^2=1\Rightarrow x=\pm1\)

TH2: \(x< -11\)

\(\Leftrightarrow-x-11=x^2+x+10\)

\(\Leftrightarrow x^2+2x+21=0\Leftrightarrow\left(x+1\right)^2+20=0\left(vn\right)\)