Question

[♥❖︵VŇ♔༺ℒεɠεηɗ༻❖︵VŇ♔[English Club]...]

các bạn giả giúp mik bài này nha

Cho : P(x)=\(^{x^3-3x^2+6x-8}\)

Q(x)=\(-6x^2+x^3-8+12\)

Tìm x để Q(x) - P(x) = 0

giúp mik nha 

mik sẽ tik cho 5 tik 

Answer 1

[Kiệt Nguyễn]

\(Q\left(x\right)-P\left(x\right)=0\)

\(\Leftrightarrow\left(-6x^2+x^3-8+12\right)-\left(x^3-3x^2+6x-8\right)=0\)

\(\Leftrightarrow\left(-6x^2+x^3+4\right)-\left(x^3-3x^2+6x-8\right)=0\)

\(\Leftrightarrow-6x^2+x^3+4-x^3+3x^2-6x+8=0\)

\(\Leftrightarrow-3x^2-6x+12=0\)

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

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

\(\Leftrightarrow x^2+2x+1=5\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+1=\sqrt{5}\\x+1=-\sqrt{5}\end{cases}}\Leftrightarrow x=\pm\sqrt{5}-1\)

Answer 2

[zZz Cool Kid_new zZz]

\(P\left(x\right)-Q\left(x\right)=\left(x^3-3x^2+6x-8\right)-\left(-6x^2+x^3-8+12\right)\)

\(P\left(x\right)-Q\left(x\right)=\left(x^3-3x^2+6x-8\right)-\left(-6x^2+x^3+4\right)\)

\(P\left(x\right)-Q\left(x\right)=x^3-3x^2+6x-8+6x^2-x^3-4\)

\(P\left(x\right)-Q\left(x\right)=3x^2+6x-4\)

Ta cần phân tích \(3x^2+6x-4\) thành nhân tử

Ta có:\(P\left(x\right)-Q\left(x\right)=-\frac{1}{3}\left(-9x^2-18x+12\right)\)

\(=-\frac{1}{3}\left[21-\left(9x^2+18x+9\right)\right]\)

\(=-\frac{1}{3}\left[21-\left(3x+3\right)^2\right]\)

\(=-\frac{1}{3}\left(\sqrt{21}-3x-3\right)\left(\sqrt{21}+3x+3\right)\)

\(\Rightarrow x=\frac{\sqrt{21}-3}{3};x=\frac{-\sqrt{21}-3}{3}\)