1 : 29 x ( 19 -13 ) - 19 x ( 29 - 13 )

= 29 x 6 - 19 x 16

= 174 - 304

=  - 130

2 : 1 - \(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

⁼ 1 - \(\frac{1}{100}\)

= \(\frac{99}{100}\)

Mk thấy phần a dễ lên bạn tự làm nha

B=(37373737.43-43434343.37):(1²+2²+3²+............+100²)

B=(37.1010101.43-43.101010101.37):(1²+2²+3²+............+100²)

B=0:(1²+2²+3²+............+100²)

B=0

Vậy B=0

Chúc bn học tốt

        \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+.....+\frac{1}{\left(x+99\right)\left(x+100\right)}\)

\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+.....+\frac{1}{x+99}-\frac{1}{x+100}\)

\(=\frac{1}{x}-\frac{1}{x+100}\)

\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+99}-\frac{1}{x+100}=\frac{1}{x}-\frac{1}{x+100}=\frac{x+100-x}{x\left(x+100\right)}=\frac{100}{x\left(x+100\right)}\)

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

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

\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{4.5}\)+\(\frac{1}{5.6}\)

A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

A=\(1-\frac{1}{3}+\frac{1}{4}-\frac{1}{6}\)

A=\(\frac{2}{3}+\frac{1}{12}\)

A=\(\frac{3}{4}\)

B=\(\left(1+\frac{1}{2}\right)\)\(\left(1+\frac{1}{3}\right)\)\(\left(1+\frac{1}{4}\right)\)...\(\left(1+\frac{1}{99}\right)\)

B=\(\frac{3}{2}\).\(\frac{4}{3}\).\(\frac{5}{4}\)...\(\frac{100}{99}\)

B=\(\frac{3.4.5...100}{2.3.4...99}\)

B=\(\frac{100}{2}\)

B=50

\(\frac{4}{x}\)=\(\frac{-y}{6}\)=0.5

\(\frac{4}{x}\)=\(\frac{-y}{6}\)=\(\frac{1}{5}\)

=> \(\frac{4}{x}\)=\(\frac{1}{5}\)=>\(x\)=\(\frac{4.5}{1}\)=9

\(\frac{-y}{6}\)=\(\frac{1}{5}\)=>\(-y\)=\(\frac{6.1}{5}\)=\(\frac{6}{5}\)=> \(y\)=\(\frac{-6}{5}\)

Vậy \(x\)= 9

\(y\)=\(\frac{-6}{5}\)

Đề bài chỉ bảo tính \(x\)nhưng mình tính cả \(y\)nếu có bài tìm cả \(y\)thì áp dụng nha

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

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