\(A=\left[\frac{6x^2}{x^3-1}-\frac{2x-2}{x^2+x+1}-\frac{1}{x-1}\right]:\frac{x^2+9}{\left(x-1\right)\left(9-4x\right)}\)

\(=\left[\frac{6x^2}{x^3-1}-\frac{\left(2x-2\right)\left(x-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}-\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right]\cdot\frac{\left(x-1\right)\left(9-4x\right)}{x^2+9}\)

\(=\frac{6x^2-\left(2x^2-4x+2\right)-x^2-x-1}{\left(x^2+x+1\right)\left(x-1\right)}\cdot\frac{\left(x-1\right)\left(9-4x\right)}{x^2+9}\)

\(=\frac{5x^2-2x^2+4x-2-x-1}{\left(x^2+x+1\right)}\cdot\frac{\left(9-4x\right)}{x^2+9}\)

\(=\frac{3x^2+3x-3}{\left(x^2+x+1\right)}\cdot\frac{\left(9-4x\right)}{x^2+9}\)

Biểu thức A bạn viết đúng chưa?

Ta có: 3x-(-17)=14+2x

\(\Leftrightarrow3x+17-14-2x=0\)

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=-3\)

Vậy: x=-3

3x-(-17) = 14+ 2x

3x+17 = 14+2x 

3x-2x=14-17

x=-3 

ta có: ( x+2).(y-3) = 5 = 5.1 = ( -5). (-1)

TH1:

* x+2 = 5 => x = 3 ( TM)

y - 3 = 1 => y= 4 (TM)

* x+2 = 1 => x = - 1 ( TM)

y- 3 = 5 => y = 8 (TM)

TH2:

* x + 2 = -1 => x = -3 ( TM)

y - 3 = - 5 => y = - 2 ( TM)

* x + 2 = -5 => x = - 7 ( TM)

y - 3 = - 1 => y = 2 (TM)

KL: \(\left(x;y\right)\in\left\{\left(3;4\right);\left(-1;8\right);\left(-3;-2\right);\left(-7;2\right)\right\}\)

Câu hỏi của le ngoc han - Toán lớp 7 - Học toán với OnlineMath

\(-\frac{17}{21}:\left(\frac{5}{4}-\frac{2}{5}\right)< x+\frac{4}{7}< 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\)

\(\Leftrightarrow-\frac{17}{21}:\frac{17}{20}< x+\frac{4}{7}< \frac{12}{12}-\frac{6}{12}+\frac{4}{12}-\frac{3}{12}\)

\(\Leftrightarrow-\frac{17}{21}.\frac{20}{17}< x+\frac{4}{7}< \frac{7}{12}\)

\(\Leftrightarrow-\frac{20}{21}< x+\frac{4}{7}< \frac{7}{12}\)

\(\Leftrightarrow-\frac{20}{21}< x< \frac{1}{84}\)

\(\Leftrightarrow-\frac{80}{84}< x< \frac{1}{84}\)

\(\Leftrightarrow-80< x< 1\Leftrightarrow x\in\left\{-79;-78;...;0\right\}\)

mà để Giá trị nguyên lớn nhất của x

\(\Rightarrow x=-1\)

1,x=6

3,x=-9