ko biết

a) \(3-2x>4\)

\(\Leftrightarrow-2x>1\)

\(\Leftrightarrow x< \frac{-1}{2}\)

b) \(\frac{2}{3-x}-\frac{9}{3+x}=\frac{1}{2}\)ĐKXĐ : \(x\pm3\)

\(\Leftrightarrow\frac{-4\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)}-\frac{18\left(x-3\right)}{2\left(x-3\right)\left(x+3\right)}=\frac{\left(x-3\right)\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow-4x-13-18x+54=x^2-9\)

\(\Leftrightarrow x^2+22x-50=0\)

\(\Leftrightarrow x^2+2\cdot x\cdot11+11^2-171=0\)

\(\Leftrightarrow\left(x+11\right)^2=\left(\pm\sqrt{171}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{171}-11\\x=-\sqrt{171}-11\end{cases}}\)( thỏa )

Vậy....

\(a,\)\(3-2x>4\)

\(\Rightarrow-2x>1\)

\(\Rightarrow x< \frac{-1}{2}\)

\(D=\frac{2x+1}{x-3}=\frac{2x-6}{x-3}+\frac{7}{x-3}=2+\frac{7}{x-3}\in Z\Leftrightarrow x-3\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)

\(\Leftrightarrow x\in\left\{-4;2;4;10\right\}\)

D= \(\frac{2x+1}{x-3}=2+\frac{7}{x-3}\)

để D dương thì x-3 là uocs của 7=(-1,1,-7,7)

xét từng TH: 

x-3=-1=> x=2

x-3=1=>x=4

x-3=-7=>x=-4

x-3=7=>x=10

các giá trị x là 2,4,-4,10

\(\dfrac{2}{3}\times\left(x+\dfrac{4}{5}\right)=\dfrac{-1}{3}\\ x+\dfrac{4}{5}=\dfrac{-1}{3}:\dfrac{2}{3}\\ x+\dfrac{4}{5}=\dfrac{-1}{3}\times\dfrac{3}{2}\\ x+\dfrac{4}{5}=\dfrac{-1}{2}\\ x=\dfrac{-1}{2}-\dfrac{4}{5}\\ x=\dfrac{-5}{10}-\dfrac{8}{10}\\ x=\dfrac{-13}{10}\)

\(\dfrac{2}{3}.\left(x+\dfrac{4}{5}\right)=-\dfrac{1}{3}\)

    \(\left(x+\dfrac{4}{5}\right)=\left(-\dfrac{1}{3}:\dfrac{2}{3}\right)\)

    \(\left(x+\dfrac{4}{5}\right)=\left(-\dfrac{1}{3}.\dfrac{3}{2}\right)=-\dfrac{1}{2}\)

      \(x\)            \(=\left(-\dfrac{1}{2}\right)-\dfrac{4}{5}\)

      \(x\)            \(=\left(-\dfrac{1}{2}\right)+\left(-\dfrac{4}{5}\right)\)

      \(x\)            \(=-\dfrac{13}{10}\)

\(\left(x-\frac{1}{2}\right)^2=0\)

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

\(\Leftrightarrow x-\frac{1}{2}=0\)

\(\Leftrightarrow x=\frac{1}{2}\)

Vậy x = 1/2

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

\(\Leftrightarrow\left(x-2\right)^2=1^2\)

\(\Leftrightarrow x-2=1\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)

Vậy x = 3 hoặc x = 1

\(\left(2x-1\right)^3=-8\)

\(\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)

\(\Leftrightarrow2x-1=-2\)

<=> 2x = -1

<=> x = -0,5

Vậy x = -0,5

\(\left(x-\frac{1}{2}\right)^2=0\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1+2\\x=-1+2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

Vậy\(x\in\left\{3;1\right\}\)
\(\left(2x-1\right)^3=-8\)

\(\left(2x-1\right)^3=\left(-2\right)^3\)

\(2x-1=-2\)

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

\(2x=-1\)

\(x=-1\times2\)

\(x=-2\)

\(x\left(\frac{1}{2}\right)^2=\frac{1}{16}\)

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

\(\Leftrightarrow\orbr{\begin{cases}x\frac{1}{2}=\frac{1}{4}\\x\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}:\frac{1}{2}\\x=-\frac{1}{4}:\frac{1}{2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}}\)

\(49\times7^x=2401\)

   \(7^x=2401:49\)

\(7^x=49\)

\(7^x=7^2\)

\(\Rightarrow x=2\)

\(2x^x-15=17\)

\(2x^x=17+15\)

\(2x^x=32\)

\(2x^x=2^5\)

\(\Rightarrow x\times x=5\)

\(\Rightarrow x^2=5\)

\(2^x+3\times2^x=32\)

\(2^x\times\left(3+1\right)=32\)

\(2^x\times4=32\)

\(2^x=32:4\)

\(2^x=8\)

\(2^x=2^3\)

\(\Rightarrow x=3\)

\(2x\left(x-\frac{1}{7}\right)=0\)

\(x\left(x-\frac{1}{7}\right)=0:2\)

\(x\left(x-\frac{1}{7}\right)=0\)

\(x\times x-x\times\frac{1}{7}=0\)

\(x^2-x\frac{1}{7}=0\)

\(x-x=0:\frac{1}{7}\)

\(x-x=0\)

\(\Rightarrow x=0\)

\(x-\dfrac{1}{4}=\dfrac{5}{3}\times\dfrac{4}{9}\\ \Rightarrow x-\dfrac{1}{4}=\dfrac{20}{27}\\ \Rightarrow x=\dfrac{20}{27}+\dfrac{1}{4}\\ \Rightarrow x=\dfrac{107}{108}\)

=>x-1/4=20/27

hay x=107/108

cảm  ơn  bạn 

7/48 - (1/2 x 2 + 1/6 x 4 + 1/8 x 5 + 1/12 x 7 + 1/14 x 8) : x = 0

7/48 - (1 + 2/3 + 5/8 + 7/12 + 4/7) : x = 0 (đã rút gọn)

7/48 - (336/336 + 224/336 + 210/336 + 196/336 + 192/336) : x = 0 (quy đồng)

7/48 - 193/56 : x  = 0

193/56 : x = 0 + 7/48

193/56 : x = 7/48

              x = 193/56 : 7/48

              x = 1158/49

\(\dfrac{2x+1}{3}=\dfrac{4}{5}\Leftrightarrow5\left(2x+1\right)=4.3\Leftrightarrow10x+5=12\Leftrightarrow x=\dfrac{7}{10}\)

\(\Leftrightarrow10x+5=12\)

hay x=7/10

\(A=\frac{1}{3.5}+\frac{1}{5.7}+..+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{100}{609}\\ \)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+..+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{1}{3}-\frac{1}{2x+3}\)\(=\frac{2x}{3\left(2x+3\right)}\)

\(A=\frac{x}{3\left(2x+3\right)}=\frac{100}{609}=\frac{100}{3.203}=\frac{100}{3\left(2.100+3\right)}\)\(\Rightarrow x=100\)