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

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

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

\(\Leftrightarrow\left(2x^2+2x+1\right)^2=4y^2-11\)

\(\Leftrightarrow\left(2x^2+2x+1\right)^2-\left(2y\right)^2=11\)

\(\Leftrightarrow\hept{\begin{cases}\left(2x^2+2x+1\right)^2=25\\\left(2y\right)^2=26\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=1;x=-2\\y=\pm3\end{cases}}\)

oke bạn

oke bạn

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 \(P=2\)

\(\frac{2x-1}{2}-1=\frac{x^2+x-3}{x-1}-\frac{5x-2}{2-2x}\)ĐKXĐ : \(x\ne1\)

\(\Leftrightarrow\frac{\left(2x-1\right)\left(x-1\right)}{2\left(x-1\right)}-\frac{2\left(x-1\right)}{2\left(x-1\right)}=\frac{2x^2+2x-6}{2\left(x-1\right)}-\frac{5x-2}{2\left(x-1\right)}\)

\(\Rightarrow2x^2-3x+1-2x+2=2x^2+2x-6-5x+2\)

\(\Leftrightarrow2x^2-5x+3=2x^2-3x-4\Leftrightarrow-2x+7=0\Leftrightarrow x=\frac{7}{2}\)( tmđk )

Vậy tập nghiệm của phương trình là S = { 7/2 }

\(\frac{2x-3}{4}-x+2=\frac{x-1}{3}\)

\(\Leftrightarrow\frac{2x-3}{4}-\frac{x-2}{1}=\frac{x-1}{3}\)

\(\Leftrightarrow\frac{6x-9-12x+24}{12}=\frac{4x-4}{12}\)

\(\Rightarrow-6x+15=4x-4\Leftrightarrow-10x=-19\Leftrightarrow x=\frac{19}{10}\)

Vậy tập nghiệm của phương trình là S = { 19/10 } 

19/10 nha tui tính rùi