\(\dfrac{1}{x-1}+\dfrac{-2}{3}\left(\dfrac{3}{4}-\dfrac{6}{5}\right)=\dfrac{5}{2-2x}\)

\(\dfrac{1}{x-1}+\dfrac{-2}{3}\times\dfrac{-9}{20}=\dfrac{5}{2-2x}\)

\(\dfrac{1}{x-1}+\dfrac{3}{10}=\dfrac{5}{2-2x}\)

\(\dfrac{10+3x-3}{10x-10}=\dfrac{5}{2-2x}\)

\(\dfrac{7+3x}{10x-10}=\dfrac{5}{2-2x}\)

\(\left(7+3x\right)\left(2-2x\right)=5\left(10x-10\right)\)

\(7\left(2-2x\right)+3x\left(2-2x\right)=50x-50\)

\(14-14x+6x-6x^2=50x-50\)

\(14-8x-6x^2=50x-50\)

\(14-8x-6x^2+50=50x\)

\(36+8x-6x^2=50x\)

\(36=50x+6x^2-8x\)

\(36=42x+6x^2\)

\(\dfrac{36}{6}=\dfrac{42x+6x^2}{6}\)

\(6=7x+x^2\)

\(6=x\left(7+x\right)\)

còn lại mình chịu

D

Chu vi hình tròn là

\(P=d\cdot3,14=35\cdot3,14=109,9\left(cm\right)\)

Bán kính hình tròn là

 \(r=\dfrac{d}{2}=\dfrac{35}{2}=17,5\left(cm\right)\)

Diện tích hình tròn là

\(S=r^2\cdot3,14=\left(17,5\right)^2\cdot3,14=961,625\left(cm^2\right)\)

\(\dfrac{6\cdot3\cdot2\cdot4\cdot5\cdot15}{3\cdot2\cdot2\cdot3\cdot20\cdot3\cdot5}\)

\(=\dfrac{4\cdot5}{20}=\dfrac{20}{20}=1\)

\(\dfrac{x}{3}=\dfrac{y}{5};5y=3z\) và \(x+y+z=98\)

\(\Rightarrow\dfrac{x}{3}=\dfrac{y}{5};\dfrac{y}{3}=\dfrac{z}{5}\Rightarrow\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{25}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{25}=\dfrac{x+y+z}{9+15+25}=\dfrac{98}{49}=2\)

\(\dfrac{x}{9}=2\Rightarrow x=2\cdot9=18\)

\(\dfrac{y}{15}=2\Rightarrow x=2\cdot15=30\)

\(\dfrac{z}{25}=2\Rightarrow x=2\cdot25=50\)

Vậy: x=18; y=30 và z=50

\(x^2\cdot y+x\cdot y-x=4\)

\(x\cdot y\cdot\left(x+1\right)-x=4\)

\(x\cdot y\cdot\left(x+1\right)-x-1=4-1\)

\(xy\cdot\left(x+1\right)-\left(x+1\right)=3\)

\(\left(x+1\right)\left(xy-1\right)=3\)

⇒ \((x + 1) ; (xy - 1)\) là ước của 3

⇒ \(\text{(x + 1) ; (xy - 1)}\in\left\{\pm1;\pm3\right\}\)

Ta có:

TH1:

\(\left[{}\begin{matrix}x+1=1\\xy-1=3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1-1\\xy=3+1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\xy=4\end{matrix}\right.\Rightarrow y=\dfrac{xy}{x}=\dfrac{4}{0}\) (vô nghiệm)

TH2:

\(\left[{}\begin{matrix}x+1=-1\\xy-1=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1-1\\xy=-3+1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-2\\xy=-2\end{matrix}\right.\)\(\Rightarrow y=\dfrac{xy}{x}=\dfrac{-2}{-2}=1\) (chọn)

TH3:

\(\left[{}\begin{matrix}x+1=3\\xy-1=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3-1\\xy=1+1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\xy=2\end{matrix}\right.\)\(\Rightarrow y=\dfrac{xy}{x}=\dfrac{2}{2}=1\) (chọn)

TH4:

\(\left[{}\begin{matrix}x+1=-3\\xy-1=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-3-1\\xy=-1+1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-4\\xy=0\end{matrix}\right.\)\(\Rightarrow y=\dfrac{xy}{x}=\dfrac{0}{-4}=0\)(vô nghiệm)

Vậy (x,y)ϵ{(-2 ; 1);(2 ; 1)}

\(B=\dfrac{2\cdot8^4\cdot27^2+4\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)

\(B=\dfrac{2\cdot\left(2^3\right)^4\cdot\left(3^3\right)^2+2^2\cdot2^9\cdot3^9}{2^7\cdot2^7\cdot3^7+2^7\cdot2^2\cdot10\cdot\left(3^2\right)^4}\)

\(B=\dfrac{2\cdot2^{12}\cdot3^6+2^{11}\cdot3^9}{2^{14}\cdot3^7+2^9\cdot10\cdot3^8}\)

\(B=\dfrac{2^{11}\cdot2^2\cdot3^6+2^{11}\cdot3^6\cdot3^3}{2^{11}\cdot2^3\cdot3^6\cdot3+\dfrac{2^{11}}{2^2}\cdot10\cdot3^6\cdot3^2}\)

\(B=\dfrac{\left(2^{11}\cdot3^6\right)\left(2^2+3^3\right)}{\left(2^{11}\cdot3^6\right)\left(2^3\cdot3\right)+2^{11}\cdot\dfrac{1}{2^2}\cdot10\cdot3^6\cdot3^2}\)

\(B=\dfrac{\left(2^{11}\cdot3^6\right)\left(2^2+3^3\right)}{\left(2^{11}\cdot3^6\right)\left(2^3\cdot3+\dfrac{1}{2^2}\cdot10\cdot3^2\right)}\)

\(B=\dfrac{2^2+3^3}{2^3\cdot3+\dfrac{1}{2^2}\cdot10\cdot3^2}\)

\(B=\dfrac{4+27}{8\cdot3+\dfrac{1}{4}\cdot10\cdot9}\)

\(B=\dfrac{31}{24+\dfrac{1}{4}\cdot90}\)

\(B=\dfrac{31}{24+\dfrac{45}{2}}\)

\(B=\dfrac{31}{\dfrac{48}{2}+\dfrac{45}{2}}\)

\(B=\dfrac{31}{\dfrac{93}{2}}\)

\(B=31\div\dfrac{93}{2}\)

\(B=31\times\dfrac{2}{93}\)

\(B=\dfrac{2}{3}\)

\(\dfrac{x-3}{x+5}=\dfrac{5}{7}\left(x\ne-5\right)\)

\(\Rightarrow7\left(x-3\right)=5\left(x+5\right)\)

\(\Rightarrow7x-21=5x+25\)

\(\Rightarrow7x-21-5x-25=0\)

\(7x-5x-21-25=0\)

\(2x-46=0\)

\(2x=0+46=46\)

\(x=\dfrac{46}{2}=23\left(tmđk\right)\)

\(\dfrac{x-1}{2}+\dfrac{x}{2}=x:2+\dfrac{1}{2}\)

\(\dfrac{x-1+x}{2}=\dfrac{x}{2}+\dfrac{1}{2}\)

\(\dfrac{2x-1}{2}=\dfrac{x+1}{2}\)

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

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

\(x=2\)