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)

các bạn giúp mình đi sáng mai mik phải nộp mất tiêu rồi

(1 - 1/2) x (1 - 1/3) x (1 - 1/4) x (1 - 1/5) x (1 - 1/6)

= 1/2 x 2/3 x 3/4 x 4/5 x 5/6

= 1/6 (tử và mẫu triệt tiêu cho nhau)

f(x)=9x³-1/3x+3x²-3x+1/3x²-1/9x³-3x²-9x+27+3x

    = 9x³-1/9x³+3x²+1/3x²-3x²-1/3-3x-9x+3x+27

   = 80/9x³+1/3x²-28/3x+27

x-[17/2-6/35]=-1/3

x-583/70=-1/3

x=-1/3+583/70

x=1679/210

vậy x=1769/210

[2/3-(x-7/4)]=9/2+5/4

[2/3-(x-7/4)]=23/4

(x-7/4)=23/4+2/3

(x-7/4)=77/12

x=77/12+7/4

x=49/6

vậy x=49/6

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

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

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

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

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

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

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

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

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

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

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

a) gọi Q(x) là thương khi chia f(x) cho g(x)

khi đó ta có dạng: f(x)=g(x).Q(x)=> f(x)=(x+3)(Q(x)   (1)

Vì (1) luôn đúng vs mọi x nên thay x=-3 vào (1) ta đc:

f(-3)= \(\left(-3\right)^3+3.\left(-3\right)^2+5.\left(-3\right)+a=0\) 0

    <=> \(-15+a=0\)

<=>a=15

Vậy vs a=15 thì f(x) chia hết cho g(x)