\(C=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}\)

\(C=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(C=\frac{\sqrt{x}\left(\sqrt{x}+1\right)-2\left(\sqrt{x}-1\right)-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(C=\frac{x+\sqrt{x}-2\sqrt{x}+2-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(C=\frac{\sqrt{x}\left(\sqrt{x}+1-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

\(C=\frac{\sqrt{x}}{\sqrt{x}+1}\)

P/s tham khảo nha

Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\)

\(a+b+c+ab+ac+bc=6abc\) \(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=6\)

Hay \(x+y+z+xy+yz+xz=6\)

Cần chứng minh \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=x^2+y^2+z^2\ge3\)

Ta có : \(\left(x^2+1\right)+\left(y^2+1\right)+\left(z^2+1\right)\ge2\left(x+y+z\right)\) (BĐT Cosi)

\(2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)\) (BĐT Cosi)

\(\Rightarrow3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+yz+xz\right)=12\)

\(\Rightarrow x^2+y^2+z^2\ge3\) (đpcm)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)

\(\hept{\begin{cases}2x^2+3xy+y^2=12\\x^2-xy+3y^2=11\end{cases}\Leftrightarrow\hept{\begin{cases}22x^2+3xy+11y^2=121\\x^2-xy+3y^2=121\end{cases}}}\)

\(\Rightarrow10x^2+45xy-25y^2=0\)

\(\Leftrightarrow\left(2x-y\right)\left(x+5y\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{y}{2}\\x=-5y\end{cases}}\)

Với \(x=\frac{y}{2}\)ta được \(\hept{\begin{cases}x=1\\y=2\end{cases};\hept{\begin{cases}x=-1\\y=-2\end{cases}}}\)

Với x=-5y ta được \(\hept{\begin{cases}x=\frac{-5\sqrt{3}}{2}\\y=\frac{\sqrt{3}}{3}\end{cases};\hept{\begin{cases}x=\frac{5\sqrt{3}}{3}\\y=\frac{\sqrt{3}}{3}\end{cases}}}\)

Câu hỏi của Khánh Đoàn Quốc - Toán lớp 9 - Học toán với OnlineMath

Ta có:\(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0\)

\(\Rightarrow a^3+3a^2b+3ab^2+b^3+c^3-3a^2b-3ab^2-3abc=0\)

\(\Rightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)

\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right).c+c^2\right]-3ab\left(a+b+c\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}a+b+c=0\left(loai\right)\\a=b=c\end{cases}}\)

\(\Rightarrow P=2007.2007.2007=2007^3\)

Nếu a = 0 suy ra b = 0, c = 0. 
Nếu \(a\ne0\) suy ra \(b\ne0,c\ne0\) ta có:
\(\frac{1}{a}=\frac{1+b^2}{2b^2}=\frac{1}{2b^2}+\frac{1}{2};\frac{1}{b}=\frac{1}{2c^2}+\frac{1}{2}\); \(\frac{1}{c}=\frac{1}{a^2}+1\).
Suy ra: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2b^2}+\frac{1}{2}+\frac{1}{2c^2}+\frac{1}{2}+\frac{1}{a^2}+1\)
\(\Leftrightarrow\frac{2}{a}+\frac{2}{b}+\frac{2}{c}=\frac{1}{b^2}+1+\frac{1}{c^2}+1+\frac{2}{a^2}+2\)
\(\Leftrightarrow\frac{1}{b^2}-\frac{2}{b}+1+\frac{1}{c^2}-\frac{2}{c}+1+\frac{2}{a^2}-\frac{2}{a}+2=0\)
\(\Leftrightarrow\left(\frac{1}{b}-1\right)^2+\left(\frac{1}{c}-1\right)^2+2\left(\frac{1}{a^2}-\frac{1}{a}+1\right)=0\)
\(\Leftrightarrow\left(\frac{1}{b}-1\right)^2+\left(\frac{1}{c}-1\right)^2+2\left(\frac{1}{a}-\frac{1}{2}\right)^2+\frac{3}{2}\ge\frac{3}{2}\).
Suy ra không có a, b,c tồn lại.
Có nên sửa đề chỗ \(c=\frac{2a^2}{1+a^2}\).

sao tự nhiên chui đâu ra a +b  = 5 vậy bạn? bạn có trả lời nhầm câu hỏi của ai khác khongo?