Ta có : 

\(\frac{21^2.14.125}{35^3.6}=\frac{3^2.7^2.2.7.5^3}{5^3.7^3.2.3}=\frac{2.3^2.5^3.7^3}{2.3.5^3.7^3}=\frac{3}{1}=3\)

Vậy \(\frac{21^2.14.125}{35^3.6}=3\)

\(\frac{21^2.14.125}{35^3.6}\)= \(\frac{21^2.2.7.125}{42875.2.3}\)= \(\frac{21^2.7.125}{125.343.3}\)= \(\frac{21^2.7.125}{125.7.49.3}\)= \(\frac{21^2}{49.3}\)= \(\frac{441}{147}\)

Mình làm rồi nhưng bạn thử tính lại cho chắc nha

Chúc bạn học tốt!

= 3 nha bạn

Tk mk nha

`@` `\text {Ans}`

`\downarrow`

\(\dfrac{21^2\cdot14\cdot125}{35^2\cdot125}\)

`=`\(\dfrac{3^2\cdot7^2\cdot2\cdot7\cdot5^2}{5^2\cdot7^2\cdot5^2}\)

`=`\(\dfrac{3^2\cdot2\cdot7\cdot5^2\cdot7^2}{5^2\cdot5^2\cdot7^2}\)

`=`\(\dfrac{3^2\cdot2\cdot7}{5^2}=\dfrac{126}{25}\)

\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne2\end{cases}}\)

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

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

\(\Leftrightarrow x^2+2x-x+2=2\)\(\Leftrightarrow x^2+x=0\)\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

So sánh với ĐKXĐ ta thấy: \(x=0\)không thoả mãn

Vậy tập nghiệm của phương trình là \(S=\left\{-1\right\}\)

Ta có: \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\)

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

   \(\Leftrightarrow\frac{x^2+2x-x+2}{x^2-2x}=\frac{2}{x^2-2x}\)

    \(\Rightarrow x^2+x+2=2\)

   \(\Leftrightarrow x^2+x=0\)

   \(\Leftrightarrow x.\left(x+1\right)=0\)

   \(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

Vậy \(S=\left\{-1;0\right\}\)

\(ĐKXĐ:x\ne2;x\ne0\)

\(pt\Leftrightarrow\frac{x^2+2x-x+1}{x^2-2x}=\frac{2}{x^2-2x}\)

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

\(\Leftrightarrow x^2+x-1=0\)

\(\Delta=1^2+4.1=5\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-1+\sqrt{5}}{2}\\x=\frac{-1-\sqrt{5}}{2}\end{cases}}\)

Đkxđ: \(\hept{\begin{cases}x\ne2\\x\ne0\end{cases}}\)

\(\frac{x+3}{x-2}+\frac{x+2}{x}=2\) 

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

\(\Rightarrow x\left(x+3\right)+\left(x-2\right)\left(x+2\right)=2x\left(x-2\right)\)

\(\Leftrightarrow x^2+3x+x^2-4=2x^2-4x\)

\(\Leftrightarrow x^2+3x+x^2-2x^2+4x=4\)

\(\Leftrightarrow7x=4\)

\(\Leftrightarrow x=\frac{4}{7}\)

Mình thiếu điều kiện xác định ^_^

Cho mình bổ xung thêm

\(ĐKXĐ:x\ne\pm1\)

và mình sửa lại nữa là: \(\orbr{\begin{cases}x=-1\left(L\right)\\x=-3\left(TM\right)\end{cases}}\)

Vậy \(S=\left\{-3\right\}\)

\(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{x^2+3}{1-x^2}\) đkxđ \(x\ne\pm1\)

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

\(\Leftrightarrow x^2+2x+1-x^2-2x-1+x^2+3=0\)

\(\Leftrightarrow x^2+3=0\)

\(\Leftrightarrow x^2=-3\)

\(\Leftrightarrow x\in\varnothing\)

\(ĐKXĐ:x\ne\pm1\)

\(pt\Leftrightarrow\frac{\left(x+1\right)^2-\left(x-1\right)^2}{x^2-1}=\frac{-x^2-3}{x^2-1}\)

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

\(\Leftrightarrow x^2+2x+1-x^2+2x-1=-x^2-3\)

\(\Leftrightarrow x^2+4x+3=0\)

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

các bn lm đến đâu cx dc miễn là lm hộ mk cái ạ, ai đang lm vào nhắn tin vs mk để mk bít nha

a; \(-\dfrac{8}{3}+\dfrac{7}{5}-\dfrac{71}{15}< x< -\dfrac{13}{7}+\dfrac{19}{14}-\dfrac{7}{2}\)

              -\(\dfrac{19}{15}\) - \(\dfrac{71}{15}\) < \(x\) < -\(\dfrac{1}{2}\) - \(\dfrac{7}{2}\)

              -6 < \(x\) < -4

             vì \(x\) \(\in\) Z nên \(x\) = -5

- \(\dfrac{19}{6}+\dfrac{-15}{2}+\dfrac{11}{3}\) < \(x\) < - \(\dfrac{5}{4}\) + \(\dfrac{19}{12}\) - \(\dfrac{10}{3}\)

- \(\dfrac{32}{3}+\dfrac{11}{3}< x< \) \(\dfrac{1}{3}\) - \(\dfrac{10}{3}\)

-7 < \(x\) < - 3

Vì \(x\in\) Z nên \(x\in\) {-6;-5;-4}

\(A=\frac{4^6.3^4.9^5}{6^{12}}=\frac{\left(2^2\right)^6.3^3.\left(3^2\right)^5}{6^{12}}\)

\(=\frac{2^{12}.3^3.3^{10}}{6^{12}}=3^{13}.3^{12}=3^{25}\)

\(A=\frac{4^6.3^4.9^5}{6^{12}}\)

\(A=\frac{2^6.2^6.3^4.3^5.3^5}{2^{12}.3^{12}}\)

\(A=\frac{3^3.3^5}{1}\)

\(A=3^8\)
 

\(B=\frac{21^2.14.125}{35^3.6}\)

\(B=\frac{3^2.7^2.2.7.5^3}{5^3.7^3.2.3}\)

\(B=\frac{3.1.1.1.1}{1.1.1.1}\)

\(B=3\)