A)    \(x-\dfrac{2}{3}=\dfrac{4}{5}\\ x=\dfrac{4}{5}+\dfrac{2}{3}\)

           \(x=\dfrac{22}{15}\)

b)\(\dfrac{7}{9}-x=\dfrac{1}{3}\\ x=\dfrac{7}{9}-\dfrac{1}{3}\\ x=\dfrac{4}{9}\)

C)\(x:\dfrac{2}{3}=\dfrac{9}{8}\\ x=\dfrac{9}{8}x\dfrac{2}{3}\\ x=\dfrac{3}{4}\)

Câu D đâu bạn:)?

\(x^7+x^6+x^4+x^3+x^2+1\)

\(=x^4\left(x^3+x^2+1\right)+\left(x^3+x^2+1\right)\)

\(=\left(x^3+x^2+1\right)\left(x^4+1\right)\)

\(A=\dfrac{3x}{x-2}\cdot\sqrt{x^2-4x+4}\)

\(=\dfrac{3x}{x-2}\cdot\left(x-2\right)\)

=3x

\(B=\dfrac{-5y}{x+3}\cdot\sqrt{x^2+6x+9}\)

\(=\dfrac{-5y}{x+3}\cdot\left|x+3\right|\)

\(=\pm5y\)

547  + ( x-2 ) x 1/3 = 1734 

39/7     + ( x- 2 ) x 1/3  = 71/4

               ( x -2 ) x 1/3 = 71/4 - 39/7 

                ( x -2 ) x 1/3 = 341/28

                   x - 2          = 341/28 : 1/3 

                   x -2           = 341/28 x 3

                   x -2           = 1023/28

                    x              = 1023/28 + 2

                    x              = 1079/28 

            Bây giờ thì k cho mình nha

viết rõ lại được ko

a: Khi x=25 thì \(A=\dfrac{5-2}{5-3}=\dfrac{3}{2}\)

b: P=A*B

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\left(\dfrac{6x+6\sqrt{x}-12}{x+5\sqrt{x}+4}-\dfrac{5\sqrt{x}}{\sqrt{x}+4}\right)\)

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\cdot\left(\dfrac{6x+6\sqrt{x}-12}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+4\right)}-\dfrac{5\sqrt{x}}{\sqrt{x}+4}\right)\)

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\cdot\dfrac{6x+6\sqrt{x}-12-5x-5\sqrt{x}}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+\sqrt{x}-12}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}-1}=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

c: \(\sqrt{P}< =\dfrac{1}{2}\)

=>0<=P<=1/4

=>\(\left\{{}\begin{matrix}P>=0\\P-\dfrac{1}{4}< =0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{x}-2}{\sqrt{x}-1}>=0\\\dfrac{\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{1}{4}< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\\dfrac{4\left(\sqrt{x}-2\right)-\sqrt{x}+1}{4\left(\sqrt{x}-1\right)}< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\\dfrac{3\sqrt{x}-7}{\sqrt{x}-1}< =0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\1< \sqrt{x}< =\dfrac{7}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\1< x< \dfrac{49}{9}\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\x=\dfrac{49}{9}\end{matrix}\right.\)

=>\(4< =x< =\dfrac{49}{9}\)

mà x nguyên

nên \(x\in\left\{4;5\right\}\)

\(5\frac{4}{7}\) + ( x-2 ) x 1/3 = \(17\frac{3}{4}\)

39/7     + ( x- 2 ) x 1/3  = 71/4

               ( x -2 ) x 1/3 = 71/4 - 39/7 

                ( x -2 ) x 1/3 = 341/28

                   x - 2          = 341/28 : 1/3 

                   x -2           = 341/28 x 3

                   x -2           = 1023/28

                    x              = 1023/28 + 2

                    x              = 1079/28 

P/s : số lớn tek!?

Cbht!!!!!                    

\(\frac{\frac{5}{4}}{7}+\left(x-2\right)\times\frac{1}{3}=\frac{\frac{17}{3}}{4}\)

\(\frac{35}{4}+\left(x-2\right)\times\frac{1}{3}=\frac{68}{3}\)

\(\frac{35}{4}+\left(x-2\right)=68\)

\(x-2=68-8,75\)

\(x-2=59,25\)

\(x=59,25+2\)

\(x=61,25\)

`(x-1)/3+(3x-5)/2+(2x)/9+(-5x)/9`

`=(x-1)/3+(3x-5)/2+x/3`

`=(2x-2+9x-15+2x)/6`

`=(13x-17)/6`

cảm ơn bạn nhiều^^