ở câu 1 ở mỗi phẫn số chúng ta cộng thêm 1, tổng là ta cộng thêm 5. Lấy 5 + -5=0. Rồi ta được tất cả tử là x+200,đặt chung ra ngoài,từ đó tính x=-200

Câu 2: => 2/42 + 2/56 + 2/72 + ... + 2/x(x+1) = 2/9

=> 2/6*7+2/7*8+...+2/x(X+1) = 2/9

\(\frac{x+1}{199}+\frac{x+2}{198}+\frac{x+3}{197}+\frac{x+4}{196}+\frac{x+5}{195}=-5\)

\(\left(\frac{x+1}{199}+1\right)+\left(\frac{x+2}{198}+1\right)+\left(\frac{x+3}{197}+1\right)+\left(\frac{x+4}{196}+1\right)+\left(\frac{x+5}{195}+1\right)=0\)

\(\frac{x+200}{199}+\frac{x+200}{198}+\frac{x+200}{197}+\frac{x+200}{196}+\frac{x+200}{195}=0\)

\(\left(x+200\right)\left(\frac{1}{199}+\frac{1}{198}+\frac{1}{197}+\frac{1}{196}+\frac{1}{195}=0\right)\)

\(x+200=0\left(Vì\frac{1}{199}+\frac{1}{198}+\frac{1}{197}+\frac{1}{196}+\frac{1}{195}\right)\)

\(\Rightarrow x=-200\)

chuyển vế rồi tách 5 thành 5 số 1` rồi nhóm vào 

\(\dfrac{x+1}{199}+\dfrac{x+2}{198}+\dfrac{x+3}{197}+\dfrac{x+4}{196}+\dfrac{x+220}{5}=0\)

\(\Leftrightarrow\left(\dfrac{x+1}{199}+1\right)+\left(\dfrac{x+2}{198}+1\right)+\left(\dfrac{x+3}{197}+1\right)+\left(\dfrac{x+4}{196}+1\right)+\dfrac{x+200}{5}+\dfrac{20}{5}-4=0\)

\(\Leftrightarrow\dfrac{x+200}{199}+\dfrac{x+200}{198}+\dfrac{x+200}{197}+\dfrac{x+200}{196}+\dfrac{x+200}{5}=0\)

\(\Leftrightarrow\left(x+200\right)\left(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+\dfrac{1}{196}+\dfrac{1}{5}\right)=0\)

\(\Leftrightarrow x=-200\)( do \(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+\dfrac{1}{196}+\dfrac{1}{5}>0\))

\(\dfrac{x+1}{199}+\dfrac{x+2}{198}+\dfrac{x+3}{197}+\dfrac{x+4}{196}+\dfrac{x+220}{5}=0\\ \Leftrightarrow\left(\dfrac{x+1}{199}+1\right)+\left(\dfrac{x+2}{198}+1\right)+\left(\dfrac{x+3}{197}+1\right)+\left(\dfrac{x+4}{196}+1\right)+\left(\dfrac{x+220}{5}-4\right)=0\\ \Leftrightarrow\dfrac{x+200}{199}+\dfrac{x+200}{198}+\dfrac{x+200}{197}+\dfrac{x+200}{196}+\dfrac{x+200}{5}=0\\ \Leftrightarrow\left(x+200\right)\left(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+\dfrac{1}{196}+\dfrac{1}{5}\right)=0\\ \Leftrightarrow x=-200\)

viết có chắc chữ giải mà cũng đúng thật vô lý

Ta có : \(\frac{x+2}{198}+\frac{x+3}{197}=\frac{x+4}{196}+\frac{x+5}{195}\)

=> \(\left(\frac{x+2}{198}+1\right)+\left(\frac{x+3}{197}+1\right)=\left(\frac{x+4}{196}+1\right)+\left(\frac{x+5}{195}+1\right)\)

=> \(\frac{x+2+198}{198}+\frac{x+3+197}{197}=\frac{x+4+196}{196}+\frac{x+5+195}{195}\)

=> \(\frac{x+200}{198}+\frac{x+200}{197}=\frac{x+200}{196}+\frac{x+200}{195}\)

=> \(\frac{x+200}{198}+\frac{x+200}{197}-\frac{x+200}{196}-\frac{x+200}{195}=0\)

=> \(\left(x+200\right)\left(\frac{1}{198}+\frac{1}{197}-\frac{1}{196}-\frac{1}{195}\right)=0\)

Ta có : \(\frac{1}{198}+\frac{1}{197}\ne\frac{1}{196}+\frac{1}{195}\)  =>    \(\frac{1}{198}+\frac{1}{197}-\frac{1}{196}-\frac{1}{195}\ne0\)

=> x + 200 = 0

=> x = -200

<=> (\(\frac{x+2}{198}\)+1) +(\(\frac{x+3}{197}\)+1) =(\(\frac{x+4}{196}\)+1) +(\(\frac{x+5}{195}\)+1)

<=> \(\frac{x+200}{198}+\frac{x+200}{197}=\frac{x+200}{196}+\frac{x+200}{195}\)

<=> \(\frac{x+200}{198}+\frac{x+200}{197}-\frac{x+200}{196}-\frac{x+200}{195}=0\)

<=> \(\left(x+200\right)\cdot\left(\frac{1}{198}+\frac{1}{197}-\frac{1}{196}-\frac{1}{195}\right)\)=0

Vì \(\frac{1}{195}>\frac{1}{196}>\frac{1}{197}>\frac{1}{198}\)

<=> \(\frac{1}{198}+\frac{1}{197}-\frac{1}{196}-\frac{1}{195}\) khác 0

<=> \(x+200=0\)

<=> x       = 

( x + 1 ) + ( x + 4 ) + ( x + 7 ) + ......... + ( x + 28 ) = 155

<=>(x+x+x...+x)+(1+4+...+28)=155

=>10x+145=155

=>10x=155-145

=>10x=10

=>x=1

làm cách khác cho nhanh hơn đi

200-199+198-197+196-195+.....+4-3+2-1

= (200-199)+(198-197)+(196-195)+.........+(4-3)+(2-1)

= 1+1+1+..........+1+1

      100 só hạng

= 100x1

=100

bằng -100 đó bạn nha

(200-199)+(198-197)+.............+(4-3)+(2-1)

=1+1+...............+1+1( có 100 số 1)

=1x100

=100

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