Ta có: \(\frac{2}{x^2-1}+Q=\frac{6}{x-3}-\frac{2x^2}{1-x^2}\)

=>\(Q+\frac{2}{x^2-1}=\frac{6}{x-3}+\frac{2x^2}{x^2-1}\)

=>\(Q=\frac{6}{x-3}+\frac{2x^2-2}{x^2-1}=\frac{6}{x-3}+2=\frac{6+2x-6}{x-3}=\frac{2x}{x-3}\)

\(\Leftrightarrow x.0,75+x.1,25+x.0,75+x.8,25=x+2010\)

\(\Leftrightarrow x\left(0,75+1,25+0,75+8,25\right)=x+2010\)

\(\Leftrightarrow x.11=x+2010\)

\(\Leftrightarrow x=201\)

\(\dfrac{x^3+y^3}{6}=\dfrac{x^3-2y^3}{4}\\ \Rightarrow4x^3+4y^3=6x^3-12y^3\\ \Rightarrow2x^3=16y^3\\ \Rightarrow x^3=8y^3\\ \Rightarrow x=2y\)

Mà \(x^6\cdot y^6=64\Rightarrow\left(2y\right)^6\cdot y^6=64\Rightarrow64\cdot y^{12}=64\)

\(\Rightarrow y^{12}=1\Rightarrow\left[{}\begin{matrix}y=1\Rightarrow x=2\\y=-1\Rightarrow x=-2\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(2;1\right);\left(-2;-1\right)\)

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

\(\Leftrightarrow xy-2x+y-2=1-2\)

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

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

Ta có bảng:

Vậy \(\left(x;y\right)=\left(0;1\right);\left(-2;3\right)\)

Ui anh ơi! 

a) \(\left|\dfrac{2}{7}\right|\) = \(\dfrac{2}{7}\)

b) \(\left|\dfrac{-5}{6}\right|\) = \(\dfrac{5}{6}\)

c) \(\left|4\dfrac{2}{3}\right|\) = \(4\dfrac{2}{3}\)

d) \(\left|-3,41\right|\) = \(3,41\)

Cảm ơn bạn chứ mk lớp 9 quay sang toán 7 quên hết 

a.

\(\Leftrightarrow x^2+3xy+\dfrac{9y^2}{4}=-\dfrac{3y^2}{4}+3y\)

\(\Leftrightarrow-\dfrac{3y^2}{4}+3y=\left(x+\dfrac{3y}{2}\right)^2\ge0\)

\(\Rightarrow-\dfrac{3y^2}{4}+3y\ge0\)

\(\Rightarrow3-\dfrac{3}{4}\left(y-2\right)^2\ge0\)

\(\Rightarrow\left(y-2\right)^2\le4\)

\(\Rightarrow-2\le y-2\le2\)

\(\Rightarrow0\le y\le4\)

\(\Rightarrow y=\left\{0;1;2;3;4\right\}\)

Lần lượt thế vào pt ban đầu:

Với \(y=0\Rightarrow x^2=0\Rightarrow x=0\)

Với \(y=1\Rightarrow x^2+3x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

Với \(y=2\Rightarrow x^2+6x+6=0\) ko có nghiệm nguyên ((loại)

Với \(y=3\Rightarrow x^2+9x+18=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=-6\end{matrix}\right.\)

Với \(y=4\Rightarrow x^2+12x+36=0\Rightarrow x=-6\)

b. Kiểm tra lại đề, đề bài đúng như thế này thì không giải được (có vô số nghiệm)

em cảm ơn

Bài 1:

a) \(x.\dfrac{3}{4}=\dfrac{9}{14}\)

\(\Rightarrow x=\dfrac{9}{14}:\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{6}{7}\)

b) \(x:\dfrac{5}{9}=\dfrac{3}{10}\)

\(\Rightarrow x=\dfrac{3}{10}.\dfrac{5}{9}\)

\(\Rightarrow x=\dfrac{1}{6}\)

help me

a) \(\frac{0,5}{0,2}=\frac{1,25}{0,1x}\Leftrightarrow0,1x.0,5=0,2.1,25\)

\(\Leftrightarrow0,1x.0,5=0,25\Leftrightarrow0,1x=0,5\Leftrightarrow x=5\)

b) \(x-\frac{3}{2}=2x-\frac{4}{3}\Leftrightarrow x-2x=\frac{-4}{3}+\frac{3}{2}\)
\(\Leftrightarrow x-2x=\frac{1}{6}\Leftrightarrow-x=\frac{1}{6}\Leftrightarrow x=\frac{-1}{6}\)

c) \(x+\frac{13}{14}=\frac{4}{7}\Rightarrow x=\frac{4}{7}-\frac{13}{14}\Rightarrow x=\frac{-5}{14}\)

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

\(\Leftrightarrow-3x+6=2x+1\Leftrightarrow-3x-2x=1-6\)

\(\Leftrightarrow-5x=-5\Leftrightarrow x=1\)

e) \(\left(x-1\right)^2-4=0\Leftrightarrow\left(x-1\right)^2=4\)

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

cậu có thể tham khảo bài trên ạ, nếu thấy đúng thì cho mk 1 t.i.c.k ạ, thank nhiều

\(d,-3\left(x-2\right)=2x+1\)

\(< =>-3x+6=2x+1\)

\(< =>-3x-2x+6-1=0\)

\(< =>5-5x=0\)

\(< =>5\left(1-x\right)=0< =>x=1\)

\(e,\left(x-1\right)^2-4=0\)

\(< =>\left(x-1+2\right)\left(x-1-2\right)=\left(x+1\right)\left(x-3\right)=0\)

\(< =>\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}< =>\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)