\(\left(\frac{x}{x-1}\right)^2+\left(\frac{x}{x+1}\right)^2=\frac{10}{9}\Leftrightarrow\frac{x^2}{\left(x-1\right)^2}+\frac{x^2}{\left(x+1\right)^2}=\frac{10}{9}\)

\(\Leftrightarrow\frac{x^2\left(x+1\right)^2+x^2\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)^2}=\frac{10}{9}\Leftrightarrow\frac{x^2\left[\left(x+1\right)^2-\left(x-1\right)^2\right]}{\left[\left(x-1\right)\left(x+1\right)\right]^2}=\frac{10}{9}\)

\(\Leftrightarrow\frac{x^2\left(x+1-x+1\right)\left(x+1+x-1\right)}{\left(x^2-1\right)^2}=\frac{10}{9}\Leftrightarrow\frac{x^2.2.2x}{x^4-2x^2+1}=\frac{10}{9}\)

\(\Leftrightarrow36x^3=10x^4-20x^2+10\Leftrightarrow18x^3=5x^4-10x^2+5\Leftrightarrow5x^4-18x^3-10x^2\)+5=0

đến đây tự giải tiếp

ĐK:\(x\ne1;x\ne-1\)

\(pt\Leftrightarrow\frac{x^2}{\left(x-1\right)^2}+\frac{x^2}{\left(x+1\right)^2}=\frac{10}{9}\)

\(\Leftrightarrow\frac{9x^2\left(x+1\right)^2+9x^2\left(x-1\right)^2-10\left(x-1\right)^2\left(x+1\right)^2}{9\left(x-1\right)^2\left(x+1\right)^2}=0\)

\(\Leftrightarrow9x^2\left(x+1\right)^2+9x^2\left(x-1\right)^2-10\left(x-1\right)^2\left(x+1\right)^2=0\)

\(\Leftrightarrow9x^4+18x^3+9x^2+9x^4-18x^3+9x^2-10x^4+20x^2-10=0\)

\(\Leftrightarrow8x^4+38x^2-10=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=\frac{1}{4}\\x^2=5\left(l\right)\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)

bài sai nguyên tập, mắt lé nhìn + thành -

xin lỗi :((

Mày nhìn cái chóa j

\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{3}{10}\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}=\frac{3}{10}\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+3}=\frac{3}{10}\)

\(\Leftrightarrow\frac{\left(x+3\right)-x}{x\left(x+3\right)}=\frac{3}{10}\)

\(\Leftrightarrow\frac{3}{x\left(x+3\right)}=\frac{3}{10}\)

\(\Rightarrow x\left(x+3\right)=10=2.\left(2+3\right)\)

\(\Rightarrow x=2\)

pt <=> \(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}=\frac{3}{10}\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+3}=\frac{3}{10}\)

\(\Leftrightarrow\frac{3}{x\left(x+3\right)}=\frac{3}{10}\)

\(\Leftrightarrow x^2+3x-10=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)

ĐKXĐ: x\(x\ne\)1,-1

a) pt <=> \(\left(\frac{x}{x-1}+\frac{x}{x+1}\right)^2-\frac{2x^2}{x^2-1}=\frac{10}{9}\)

<=> \(\frac{4x^4}{\left(x^2-1\right)^2}-\frac{2x^2}{x^2-1}=\frac{10}{9}\)

Đặt: t=\(\frac{2x^2}{x^2-1}\)

Pt trở thành: \(t^2-t-\frac{10}{9}=0\)\(\Leftrightarrow9t^2-9t-10=0\)<=> \(\orbr{\begin{cases}t=-\frac{1}{3}\\t=\frac{5}{6}\end{cases}}\)

Nếu: \(\frac{2x^2}{x^2-1}=-\frac{1}{3}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{\frac{1}{7}}\\x=-\sqrt{\frac{1}{7}}\end{cases}\left(tm\right)}\)

Nếu: \(\frac{2x^2}{x^2-1}=\frac{5}{6}\)(vô nghiệm)

Vậy nghiệm là ...

http://vchat.vn/pictures/service/2017/02/iit1486637364.PNG

4x^2 + 2y^2 + 2z^2 - 4xy - 2yz + 2y - 8z + 10 = (4x^2 - 4xy + y^2) + (z^2 - 6z + 9) +(y^2 + z^2 +1 -2yz -2z +2y) = (2x-y)^2 + (z-3)^2 + (y-z+1)^2 < hoặc = 0

Điều kiện:\(x\ne0\)

Đặt \(\frac{x}{3}-\frac{4}{x}=t\).Ta có:\(t^2=\left(\frac{x}{3}-\frac{4}{x}\right)^2=\frac{x^2}{9}-2.\frac{x}{3}.\frac{4}{x}+\frac{16}{x^2}=\frac{x^2}{9}+\frac{16}{x^2}-\frac{8}{3}\)

\(\Rightarrow\frac{x^2}{9}+\frac{16}{x^2}=t^2+\frac{8}{3}\).Thay vào pt ta có:\(t^2+\frac{8}{3}=\frac{10}{3}.t\)

\(\Leftrightarrow3t^2-10t+8=0\)\(\Leftrightarrow3t^2-4t-6t+8=0\)

\(\Leftrightarrow t\left(3t-4\right)-2\left(3t-4\right)=0\)

\(\Leftrightarrow\left(t-2\right)\left(3t-4\right)=0\Rightarrow\orbr{\begin{cases}t=2\\t=\frac{4}{3}\end{cases}}\)

Với \(t=2\) thì \(\frac{x^2-12}{3x}=2\Leftrightarrow x^2-12-6x=0\)\(\Rightarrow x^2-6x+9-21=0\)

\(\Leftrightarrow\left(x-3\right)^2=21\Rightarrow\orbr{\begin{cases}x-3=\sqrt{21}\\x-3=-\sqrt{21}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\sqrt{21}+3\\x=3-\sqrt{21}\end{cases}}\)

Với \(t=\frac{4}{3}\) thì \(\frac{x^2-12}{3x}=\frac{4}{3}\Leftrightarrow x^2-4x-12=0\Leftrightarrow\left(x+2\right)\left(x-6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=6\end{cases}}\)

Tập nghiệm của pt S=\(\left\{\sqrt{21}+3;3-\sqrt{21};-2;6\right\}\)