\(\Leftrightarrow\frac{-3x\left(x+1\right)}{x+\sqrt{x^2+x+1}}+x+1=0\)

\(\Leftrightarrow\left(x+1\right)\left(1-\frac{3x}{\sqrt{x^2+x+1}}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\left(1\right)\\1=\frac{3x}{\sqrt{x^2+x+1}}\left(2\right)\end{cases}}\)

PT(2)\(\Leftrightarrow\sqrt{x^2+x+1}=3x\)

\(\Leftrightarrow x^2+x+1=9x^2\)

\(\Leftrightarrow8x^2-x-1=0\)

Ta co

\(\Delta=\left(-1\right)^2-4.8.\left(-1\right)=33>0\)

\(\Rightarrow x_1=\frac{1+\sqrt{33}}{8};x_2=\frac{1-\sqrt{33}}{8}\)

Vay PT co nghiem la \(x=-1;x_1=\frac{1+\sqrt{33}}{8};x_2=\frac{1-\sqrt{33}}{8}\) 

Bạn tham khảo câu hỏi này nha ,cũng giống của bạn đó !

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\(\sqrt{x+3}+\sqrt{1-x}+8\sqrt{\left(x+3\right)\left(1-x\right)}=2\)

\(\sqrt{x+3}\left(1+8\sqrt{1-x}\right)+\sqrt{1-x}-2=0\)

\(\sqrt{x+3}\left(1+8\sqrt{1-x}\right)-\frac{x+3}{\sqrt{1-x}+2}=0\)

\(\sqrt{x+3}\left(1+8\sqrt{1-x}-\frac{\sqrt{x+3}}{\sqrt{1-x}+2}\right)=0\)

\(\Rightarrow x=-3\)

\(8\sqrt{1-x}+\frac{\sqrt{1-x}+2-\sqrt{x+3}}{\sqrt{1-x}+2}=0\)

\(\Rightarrow8\sqrt{1-x}\left(\sqrt{1-x}+2\right)+\sqrt{1-x}+\frac{1-x}{2+\sqrt{x+3}}=0\)

\(\Rightarrow\sqrt{1-x}\left(8\sqrt{1-x}+17+\frac{\sqrt{1-x}}{2+\sqrt{x+3}}\right)=0\)

\(\left(8\sqrt{1-x}+17+\frac{\sqrt{1-x}}{2+\sqrt{x+3}}\right)>0\)

\(\Rightarrow x=1\)

Vậy pt 2 nghiệm x=-3 x=1

bạn học đến đg tròn rồi à

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)  hinh nhu theo co dieu kien a,b,c  ko dong thoi = 0

<=> \(\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)

<=>  \(\frac{a+b}{ab}=\frac{c-a-b-c}{c\left(a+b+c\right)}\)

<=> \(\left(a+b\right)\left(ac+bc+c^2\right)=-ab\left(a+b\right)\)

<=> \(\left(a+b\right)\left(ac+bc+c^2\right)+ab\left(a+b\right)=0\)

<=> \(\left(a+b\right)\left(ac+bc+c^2+ab\right)=0\)

<=> \(\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)

<=> a+b=0 hoac a+c=0 hoac b+c=0

do khi luy thua a,b,c len cach so mu le la 27,41,2019 thi a,b,c ko doi dau nen \(a^{27}+b^{27}=0.hoac.b^{41}+c^{41}=0.hoac.c^{2019}+a^{2019}=0\)

P = 0 

Vay P = 0 

Study well

Ta có : \(\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}-\frac{1}{a}\Rightarrow\frac{b+c}{bc}=\frac{a-a-b-c}{a^2+ab+ac}\)

\(\Leftrightarrow\frac{b+c}{bc}=\frac{-b-c}{a^2+ab+ac}\Leftrightarrow\left(b+c\right)\left(a^2+ab+ac\right)=-\left(b+c\right)bc\)

\(\left(b+c\right)\left(a^2+ab+ac\right)+\left(b+c\right)bc=0\)

\(\Rightarrow\left(b+c\right)\left(a^2+ab+ac+bc\right)=0\)

\(\Leftrightarrow\left(b+c\right)[\left(a+b\right)a+c\left(a+b\right)]=0\)

\(\Leftrightarrow\left(b+c\right)\left(a+b\right)\left(a+c\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}b=-c\\\orbr{\begin{cases}a=-b\\c=-a\end{cases}}\end{cases}\Leftrightarrow\orbr{\begin{cases}b^{41}+c^{41}=0\\\orbr{\begin{cases}a^{27}+b^{27}=0\\c^{2019}+a^{2019}=0\end{cases}}\end{cases}}}\)\(\Leftrightarrow\orbr{\begin{cases}b=-c\\\orbr{\begin{cases}a=-b\\c=-a\end{cases}}\end{cases}\Leftrightarrow\orbr{\begin{cases}b^{41}+c^{41}=0\\\orbr{\begin{cases}a^{27}+b^{27}=0\\a^{2019}+c^{2019}=0\end{cases}}\end{cases}}}\)

\(VT=\sqrt{-\left(x-1\right)^2+3}+\sqrt{-\left(x+3\right)^2+1}< 1+\sqrt{3}\) pt vô nghiệm