\(\frac{x^2}{2y}+\frac{y^2}{2x}+\frac{y^2}{2z}+\frac{z^2}{2y}+\frac{z^2}{2x}+\frac{x^2}{2z}\ge\frac{\left(2x+2y+2z\right)^2}{4\left(x+y+z\right)}=x+y+z\)

các  bạn ơi mình đang cần gấp . Mình chỉ còn 20 phút thui .  HUHU

              Bài làm :

1) \(\frac{x+11}{4}=\frac{2x+4}{5}\)

\(\Leftrightarrow\left(x+11\right).5=4.\left(2x+4\right)\)

\(\Leftrightarrow5x+55=8x+16\)

\(\Leftrightarrow5x-8x=16-55\)

\(\Leftrightarrow-3x=-39\)

\(\Leftrightarrow x=\frac{-39}{-3}=\frac{39}{3}=13\)

2)\(\frac{x+4}{x+10}=\frac{3}{5}\)

\(\Leftrightarrow\left(x+4\right).5=\left(x+10\right).3\)

\(\Leftrightarrow5x+20=3x+30\)

\(\Leftrightarrow5x-3x=30-20\)

\(\Leftrightarrow2x=10\)

\(\Leftrightarrow x=\frac{10}{2}=5\)

3)\(\frac{x+8}{x+14}=\frac{2}{3}\)

\(\Leftrightarrow\left(x+8\right).3=\left(x+14\right).2\)

\(\Leftrightarrow3x+24=2x+28\)

\(\Leftrightarrow3x-2x=28-24\)

\(\Leftrightarrow x=4\)

\(\left|2x-3\right|=3-2x\)

\(ĐK:x\le\dfrac{3}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3-2x\\3-2x=3-2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\0=0\left(đúng\right)\end{matrix}\right.\)

Vậy \(S=\left\{x\in R;x=\dfrac{3}{2}\right\}\)

a,\(A=\left(\frac{2x-x^2}{2\left(x^2+4\right)}-\frac{2x^2}{\left(x^2+4\right)\left(x-2\right)}\right)\left(\frac{2x+x^2\left(1-x\right)}{x^3}\right)\left(ĐKXĐ:x\ne2;x\ne0\right)\)

\(A=\frac{\left(2x-x^2\right)\left(x-2\right)-4x^2}{2\left(x^2+4\right)\left(x-2\right)}.\frac{-x^3+x^2+2x}{x^3}\)

\(=\frac{-x^3-4x}{2\left(x^2+4\right)\left(x-2\right)}.\frac{x^2-x-2}{-x^2}\)

\(=\frac{-x\left(x^2+4\right)}{2\left(x^2+4\right)\left(x-2\right)}.\frac{\left(x-2\right)\left(x+1\right)}{-x^2}=\frac{x+1}{2x}\)

b, \(A=x\Leftrightarrow\frac{x+1}{2x}=x\Rightarrow2x^2=x+1\Leftrightarrow2x^2-x-1=0\)

\(\Leftrightarrow\left(2x+1\right)\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)(thỏa mãn điều kiện)

c, \(A\in Z\Leftrightarrow\frac{x+1}{2x}\in Z\Leftrightarrow x+1⋮\left(2x\right)\)

\(\Leftrightarrow2x+2⋮2x\Leftrightarrow2⋮2x\Leftrightarrow1⋮x\Leftrightarrow x=\pm1\) (thỏa mãn ĐKXĐ)

Đặt \(x^{2\:}-2x+2=t\)

Được phương trình: \(\frac{t}{t+1}+\frac{t-1}{t}=\frac{1}{6}\)

Quy đồng và khử mẫu được: \(12t^2-6=t^2+t\)

<=> \(11t^2-t=6\)

r á. đến đó thỳ hk lm đk n~. pn xem lại đề đy na @@

thiếu xíu: đặt x^2-2x+2=t

a.

\(-\frac{2}{3}-\frac{1}{3}\times\left(2x-5\right)=\frac{3}{2}\)

\(-\frac{2}{3}-\frac{2}{3}x+\frac{5}{3}=\frac{3}{2}\)

\(\left(-\frac{2}{3}+\frac{5}{3}\right)-\frac{2}{3}x=\frac{3}{2}\)

\(\frac{3}{3}-\frac{2}{3}x=\frac{3}{2}\)

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

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

\(\frac{2}{3}x=\frac{2}{2}-\frac{3}{2}\)

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

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

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

\(x=-\frac{3}{4}\)

b.

\(\frac{1}{3}x+\frac{2}{5}\times\left(x-1\right)=0\)

\(\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)

\(x\times\left(\frac{1}{3}+\frac{2}{5}\right)=\frac{2}{5}\)

\(x\times\left(\frac{5}{15}+\frac{6}{15}\right)=\frac{2}{5}\)

\(x\times\frac{11}{15}=\frac{2}{5}\)

\(x=\frac{2}{5}\div\frac{11}{15}\)

\(x=\frac{2}{5}\times\frac{15}{11}\)

\(x=\frac{6}{11}\)

Chúc bạn học tốt

a ) \(-\frac{2}{3}-\frac{1}{3}\left(2x-5\right)=\frac{3}{2}\)

                  \(\frac{1}{3}\left(2x-5\right)=-\frac{2}{3}-\frac{3}{2}\)

                   \(\frac{1}{3}\left(2x-5\right)=\frac{-13}{6}\)

                        \(\left(2x-5\right)=-\frac{13}{6}:\frac{1}{3}\)

                         \(\left(2x-5\right)=-\frac{13}{6}.\frac{3}{1}\)

                         \(\left(2x-5\right)=-\frac{13}{2}\)

                           \(2x=-\frac{13}{2}+5\)

                            \(2x=-\frac{3}{2}\)

                              \(\Rightarrow x=-\frac{3}{2}:2\)

                              \(\Rightarrow x=-\frac{3}{2}.\frac{1}{2}\)

                              \(\Rightarrow x=-\frac{3}{4}\)