trả lời nhanh hộ mình với cảm ơn :(

theo cách tính tổng (bn có thể xem lại ở toán 7 hay 6 j đấy) thì bt trên bằng 1/x - 1/(x+5)

từ đó tính tiếp nha bn

Bài 4:

1: \(\left(x-1\right)\left(x^2+x+1\right)-x^3-6x=11\)

=>\(x^3-1-x^3-6x=11\)

=>-6x-1=11

=>-6x=11+1=12

=>\(x=\dfrac{12}{-6}=-2\)

2: \(16x^2-\left(3x-4\right)^2=0\)

=>\(\left(4x\right)^2-\left(3x-4\right)^2=0\)

=>\(\left(4x-3x+4\right)\left(4x+3x-4\right)=0\)

=>(x+4)(7x-4)=0

=>\(\left[{}\begin{matrix}x+4=0\\7x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{4}{7}\end{matrix}\right.\)

3: \(x^3-x^2-3x+3=0\)

=>\(\left(x^3-x^2\right)-\left(3x-3\right)=0\)

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

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

=>\(\left[{}\begin{matrix}x-1=0\\x^2-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)

4: \(\dfrac{x-1}{x+2}=\dfrac{x+2}{x+1}\)(ĐKXĐ: \(x\notin\left\{-2;-1\right\}\))

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

=>\(x^2+4x+4=x^2-1\)

=>4x+4=-1

=>4x=-5

=>\(x=-\dfrac{5}{4}\left(nhận\right)\)

5: ĐKXĐ: \(x\notin\left\{0;-1\right\}\)

\(\dfrac{1}{x}+\dfrac{2}{x+1}=0\)

=>\(\dfrac{x+1+2x}{x\left(x+1\right)}=0\)

=>3x+1=0

=>3x=-1

=>\(x=-\dfrac{1}{3}\left(nhận\right)\)

6: ĐKXĐ: \(x\notin\left\{0;3\right\}\)

\(\dfrac{9-x^2}{x}:\left(x-3\right)=1\)

=>\(\dfrac{-\left(x^2-9\right)}{x\left(x-3\right)}=1\)

=>\(\dfrac{-\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)}=1\)

=>\(\dfrac{-x-3}{x}=1\)

=>-x-3=x

=>-2x=3

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

a,

=x³+1-(x³-1)

=x³+1-x³+1

=2

b,

tự làm nha bạn

a: \(x-2-\frac{x^2-10}{x+2}\)

\(=\frac{\left(x-2\right)\left(x+2\right)-x^2+10}{x+2}\)

\(=\frac{x^2-4-x^2+10}{x+2}=\frac{6}{x+2}\)

b: \(\frac{x}{y^2-xy}-\frac{y}{xy-x^2}\)

\(=\frac{-x}{y\left(x-y\right)}+\frac{y}{x\left(x-y\right)}=\frac{-x^2+y^2}{xy\left(x-y\right)}=\frac{-\left(x-y\right)\left(x+y\right)}{xy\left(x-y\right)}\)

\(=\frac{-x-y}{xy}\)

c: \(\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}\)

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

\(=\frac{1}{x-5}-\frac{1}{x}=\frac{x-\left(x-5\right)}{x\left(x-5\right)}=\frac{5}{x\left(x-5\right)}\)

Câu hỏi của Best Friend Forever - Toán lớp 7 - Học toán với OnlineMath

1: \(x\left(1-x\right)+\left(x-1\right)^2\)

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

=-x+1

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

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

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

\(=x^2+4x+4-x^2+2x+3=6x+7\)

5: \(\left(x-2\right)^2+\left(x-1\right)\left(x+5\right)\)

\(=x^2-4x+4+x^2+5x-x-5\)

\(=2x^2-1\)

7: \(\left(1-2x\right)\left(5-3x\right)+\left(4-x\right)^2\)

\(=\left(2x-1\right)\left(3x-5\right)+\left(x-4\right)^2\)

\(=6x^2-10x-3x+5+x^2-8x+16\)

\(=7x^2-21x+21\)

9: \(\left(x+1\right)^2+\left(x-2\right)\left(x+2\right)-4x\)

\(=x^2+2x+1+x^2-4-4x\)

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

11: \(\left(x+4\right)^2+\left(x+5\right)\left(x-5\right)-2x\left(x+1\right)\)

\(=x_{}^2+8x+16+x^2-25-2x^2-2x\)

=6x-9

13: \(\left(x-1\right)^2-2\left(x+3\right)\left(x-3\right)+4x\left(x-4\right)\)

\(=x^2-2x+1-2\left(x^2-9\right)+4x^2-16x\)

\(=5x^2-18x+1-2x^2+18=3x^2-18x+19\)

2: \(\left(x-3\right)^2-x^2+10x-7\)

\(=x^2-6x+9-x^2+10x-7\)

=4x+2

4: \(\left(x+4\right)\left(x-2\right)-\left(x-3\right)^2\)

\(=x^2-2x+4x-8-\left(x^2-6x+9\right)\)

\(=x^2+2x-8-x^2+6x-9=8x-17\)

6: (x-3)(x+3)-x(x+23)

\(=x^2-9-x^2-23x\)

=-23x-9

8: (x-2)(x+2)-(x-3)(x+1)

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

\(=x^2-4-\left(x^2-2x-3\right)\)

\(=x^2-4-x^2+2x+3=2x-1\)

10: \(\left(x+2\right)^2-\left(x+3\right)\left(x-3\right)+10\)

\(=x^2+4x+4-\left(x^2-9\right)+10\)

\(=x^2+4x+14-x^2+9=4x+23\)

12: \(\left(x-1\right)^2-\left(x-4\right)\left(x+4\right)+\left(x+3\right)^2\)

\(=x^2-2x+1-\left(x^2-16\right)+x^2+6x+9\)

\(=2x^2+4x+10-x^2+16=x^2+4x+26\)

a: \(=\dfrac{5\left(x+2\right)}{10xy^2}\cdot\dfrac{12x}{x+2}=\dfrac{60x}{10xy^2}=\dfrac{6}{y^2}\)

b: \(=\dfrac{x-4}{3x-1}\cdot\dfrac{3\left(3x-1\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{3}{x+4}\)

c: \(=\dfrac{2\left(2x+1\right)}{\left(x+4\right)^2}\cdot\dfrac{\left(x+4\right)}{3\left(x+3\right)}=\dfrac{2\left(2x+1\right)}{3\left(x+3\right)\left(x+4\right)}\)

d: \(=\dfrac{5\left(x-1\right)}{3\left(x+1\right)}\cdot\dfrac{x+1}{x-1}=\dfrac{5}{3}\)

a/ \(\left(2x+3\right)\left(x-5\right)-\left(x-1\right)^2+x\left(7-x\right)\)

\(=2x^2-2x-15-x^2+2x-1+7x-x^2\)

\(=7x-16\)

b, = x² - 16 - ( x³ - 3³ ) : ( x - 3 )

= x² - 16 - \([\) ( x - 3 ) ( x² + 3x + 9 ) \(]\) : ( x - 3 )

= x² - 16 - ( x² + 3x + 9 )

= x² - 16 - x² - 3x - 9

= -25 - 3x