Mình giải giúp bạn nha:
a, \(x^2+3\times x-6\)
Có: \(x^2+3\times x-6=0\)
\(\Rightarrow x^2+2\times x\times\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2-\left(\dfrac{3}{2}\right)^2-6=0\)
\(\Rightarrow\left(x+\dfrac{3}{2}\right)^2-\dfrac{33}{4}=0\)
\(\Rightarrow\left(x+\dfrac{3}{2}\right)^2=\dfrac{33}{4}\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{2}=\sqrt{\dfrac{33}{4}}\\x+\dfrac{3}{2}=-\sqrt{\dfrac{33}{4}}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{33}{4}}-\dfrac{3}{2}=\dfrac{-3+\sqrt{33}}{2}\\x=-\sqrt{\dfrac{33}{4}}-\dfrac{3}{2}=-\dfrac{3+\sqrt{33}}{2}\end{matrix}\right.\)
Vậy đa thức \(x^2-3x-6\) có nghiệm là \(x=\dfrac{-3+\sqrt{33}}{2};x=-\dfrac{3+\sqrt{33}}{2}\)
b, \(4\times x^2+8\times x-4\)
Cho: \(4\times x^2+8\times x-4=0\)
\(\Rightarrow\left(4\times x^2+8\times x-4\right)\times\dfrac{1}{4}=0\times\dfrac{1}{4}\)
\(4\times x^2-\dfrac{1}{4}+8\times x\times\dfrac{1}{4}-4\times\dfrac{1}{4}=0\)
\(x^2+2\times x-1=0\)
\(x^2+x+x-1=0\)
\(x\times\left(x+1\right)+\left(x+1\right)-2=0\)
\(\Rightarrow\left(x+1\right)\left(x+1\right)=2\)
\(\Rightarrow\left(x+1\right)^2=2\)
\(\Rightarrow x+1=\pm\sqrt{2}\)
TH1: \(x+1=\sqrt{2}\Rightarrow x=\sqrt{2}-1\)
TH2: \(x+1=-\sqrt{2}\Rightarrow x=-\sqrt{2}-1\)
Vậy nghiệm của đa thức \(4\times x^2+8\times x-4\) là \(x\in\left\{\sqrt{2}-1;-\sqrt{2}-1\right\}\)
c, \(x^2-x-3\)
Cho: \(x^2-x-3=0\)
\(\Rightarrow x^2-2\times x\times\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^2-3=0\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}-\dfrac{12}{4}=0\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2-\dfrac{13}{4}=0\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{13}{4}\)
\(\Rightarrow x-\dfrac{1}{2}=\pm\sqrt{\dfrac{13}{4}}\)
TH1: \(x-\dfrac{1}{2}=\sqrt{\dfrac{13}{4}}\)
\(x=\sqrt{\dfrac{13}{4}}+\dfrac{1}{2}\)
\(x=\sqrt{\dfrac{13}{2}}+\dfrac{1}{2}\)
\(x=\dfrac{\sqrt{13}+1}{2}\)
TH2: \(x-\dfrac{1}{2}=-\sqrt{\dfrac{13}{4}}\)
\(x=-\sqrt{\dfrac{13}{4}}+\dfrac{1}{2}\)
\(x=-\sqrt{\dfrac{13}{2}}+\dfrac{1}{2}\)
\(x=\dfrac{1-\sqrt{13}}{2}\)
Vậy nghiệm của đa thức \(x^2-x-3\) là \(x=\dfrac{\sqrt{13}+1}{2};x=\dfrac{1-\sqrt{13}}{2}\)
d, \(3\times x^2-x+3\times x-1\)
Cho: \(3\times x^2-x+3\times x-1=0\)
TH1: \(\Rightarrow x\times\left(3\times x-1\right)+\left(3\times x-1\right)=0\)
\(\Rightarrow\left(3\times x-1\right)^2=0\)
\(\Rightarrow3\times x-1=0\)
\(3\times x=0+1\)
\(3\times x=1\)
\(x=1\div3\)
\(x=\dfrac{1}{3}\)
TH2: \(3\times x^2-x+3\times x-1=0\)
\(\Rightarrow2\times x+3\times x-1=0\)
Mà \(3\times x-1=0\)
\(\Rightarrow2\times x=0\)
\(\Rightarrow x=0\)
Vậy đa thức có nghiệm là \(x\in\left\{0;\dfrac{1}{3}\right\}\)
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