tìm x : -4x2 - 3x + 5 = 0
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\(x^4-y^2+2x^3+2x^2+x+3=0\)
\(\Leftrightarrow\left(x^4+2x^3+x^2\right)+\left(x^2+x\right)=y^2-3\)
\(\Leftrightarrow\left(x^2+x\right)^2+\left(x^2+x\right)=y^2-3\)
\(\Leftrightarrow\left(2x^2+2x+1\right)^2=4y^2-11\)
\(\Leftrightarrow\left(2x^2+2x+1\right)^2-\left(2y\right)^2=11\)
\(\Leftrightarrow\hept{\begin{cases}\left(2x^2+2x+1\right)^2=25\\\left(2y\right)^2=26\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=1;x=-2\\y=\pm3\end{cases}}\)
Ta có : \(\frac{x}{a}+\frac{y}{b}=1\)
\(\Rightarrow\left(\frac{x}{a}+\frac{y}{b}\right)^2=1\)
\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+2.\frac{xy}{ab}=1\)
\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}=1-2.\frac{xy}{ab}=1-2\left(-2\right)=5\)
\(\Rightarrow\frac{x^3}{a^3}+\frac{y^3}{b^3}=\left(\frac{x}{a}+\frac{y}{b}\right)\left(\frac{x^2}{a^2}-2.\frac{xy}{ab}+\frac{y^2}{b^2}\right)\)
\(=1\left(5+4\right)=9\)
Vậy \(\frac{x^3}{a^3}+\frac{y^3}{b^3}=9\)
Từ \(a^{100}+b^{100}=a^{101}+b^{101}=a^{102}+b^{102}\)
\(\Rightarrow a^{100}+b^{100}+a^{102}+b^{102}=2\left(a^{101}+b^{101}\right)\)
\(\Rightarrow a^{100}+b^{100}+a^{102}+b^{102}-2\left(a^{101}+b^{101}\right)=0\)
\(\Rightarrow\left(a^{102}-2a^{101}+a^{100}\right)+\left(b^{102}-2b^{101}+b^{100}\right)=0\)
\(\Rightarrow\left(a^{51}-a^{50}\right)^2+\left(b^{51}-b^{50}\right)^2=0\left(1\right)\)
Vif \(\hept{\begin{cases}\left(a^{51}-a^{50}\right)^2\ge0\forall a\\\left(b^{51}-b^{50}\right)^2\ge0\forall b\end{cases}}\)
\(\Rightarrow\left(a^{51}-a^{50}\right)^2+\left(b^{51}-b^{50}\right)^2\ge0\forall a,b\left(2\right)\)
Tứ (1) và (2) :
\(\Rightarrow\hept{\begin{cases}\left(a^{51}-a^{50}\right)^2=0\\\left(b^{51}-b^{50}\right)^2=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}a^{51}-a^{50}=0\\b^{51}-b^{50}=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}a^{51}=a^{50}\\b^{51}=b^{50}\end{cases}}\)
Vì a,b là các số thực dương nên \(a=b=1\)
\(\Rightarrow P=a^{2007}+b^{2007}=1^{2007}+1^{2007}=1+1=2\)
Vậy \(P=2\)
\(-4x^2-3x+5=0\)
\(\Leftrightarrow4x^2+3x-5=0\)
\(\Leftrightarrow4\left(x^2+\frac{3}{4}x-\frac{5}{4}\right)=0\)
\(\Leftrightarrow x^2+\frac{3}{4}x-\frac{5}{4}=0\)
\(\Leftrightarrow\left(x^2+2.x.\frac{3}{8}+\frac{9}{64}\right)-\frac{89}{64}=0\)
\(\Leftrightarrow\left(x+\frac{3}{8}\right)^2=\frac{89}{64}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{3}{8}=\frac{\sqrt{89}}{8}\\x+\frac{3}{8}=\frac{-\sqrt{89}}{8}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{89}-3}{8}\\x=\frac{-\sqrt{89}-3}{8}\end{cases}}\)
Vậy phương trình có tập nghiệm : \(S=\left\{\frac{\pm\sqrt{89}-3}{8}\right\}\)