\(ĐK:x\ne1\)

\(\Leftrightarrow\dfrac{4}{\left(x-1\right)\left(x+1\right)}=\dfrac{1-x}{x+1}+\dfrac{x+1}{x-1}\)

\(\Rightarrow4=\left(1-x\right)\left(x-1\right)+\left(x+1\right)\left(x+1\right)\)

\(\Leftrightarrow4=-\left(x-1\right)^2+\left(x+1\right)^2\)

\(\Leftrightarrow4=-x^2+2x-1+x^2+2x+1\)

\(\Leftrightarrow4x=4\)

\(\Leftrightarrow x=1\left(l\right)\)

Vậy S vô nghiệm

\(ĐK:x\ne-1;x\ne3\)

\(\Leftrightarrow\dfrac{2x}{\left(x+1\right)\left(x-3\right)}=\dfrac{x}{2\left(x+1\right)}+\dfrac{x}{2\left(x-3\right)}\)

\(\Leftrightarrow\dfrac{4x}{2\left(x+1\right)\left(x-3\right)}=\dfrac{x\left(x-3\right)+x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}\)

\(\Rightarrow4x=x^2-3x+x^2+x\)

\(\Leftrightarrow2x^2-6x=0\)

\(\Leftrightarrow2x\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(n\right)\\x=3\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{2\left(x-2\right)\left(x+2\right)-x\left(2x+3\right)}{2x\left(x-2\right)}=0\)

\(\Leftrightarrow\dfrac{2\left(x^2-4\right)-2x^2-3x}{2x\left(x-2\right)}=0\)

\(\Leftrightarrow2x^2-8-2x^2-3x=0\)

\(\Leftrightarrow-3x=8\)

\(\Leftrightarrow x=-\dfrac{8}{3}\)

cho e xin tài liệu HSG Toán vs ạ

\(Cu+\dfrac{1}{2}O_2\underrightarrow{t^o}CuO\)

\(n_{Cu}=\dfrac{m}{M}=\dfrac{6,4}{64}=0,1mol\)

\(V_{O_2}=n.22,4=0,1.\dfrac{1}{2}.22,4=1,12l\)

\(m_{sp}=n.M=0,1.80=8g\)

QF = 24 - 9 = 15

ta có: PQ // EF

\(\Leftrightarrow\dfrac{DP}{EP}=\dfrac{DQ}{FQ}\)

\(\Leftrightarrow\dfrac{x}{10,5}=\dfrac{9}{15}\)

\(\Leftrightarrow15x=10,5.9\)

\(\Leftrightarrow15x=94,5\Leftrightarrow x=6,3\)

ta có: DE // KI

\(\Leftrightarrow\dfrac{HD}{KD}=\dfrac{HE}{IE}\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{5}=\dfrac{x}{10}\)

\(\Leftrightarrow5x=10\sqrt{3}\)

\(\Leftrightarrow x=2\sqrt{3}\)

ta có: DE // BC

=> \(\dfrac{AD}{BD}=\dfrac{AE}{EC}\)

<=> \(\dfrac{4}{2}=\dfrac{6,5}{x}\)

<=> x= 6,5 . 2 =13