=1/1-1/2+1/2 -1/3+1/3-1/4....... 1/56 -1/57+1/57-1/58

=  1/1-1/58

= 57/58

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{56.57}+\frac{1}{57.58}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{56}-\frac{1}{57}+\frac{1}{57}-\frac{1}{58}\)

\(=\frac{1}{1}-\frac{1}{58}\)

\(=\frac{57}{58}\)

=1/1‐1/2+1/2 ‐1/3+1/3‐1/4....... 1/56 ‐1/57+1/57‐1/58
= 1/1‐1/58
= 57/58

2) Ta có: \(\dfrac{59-x}{19}+\dfrac{58-x}{18}=\dfrac{57-x}{17}+\dfrac{56-x}{16}\)

\(\Leftrightarrow\dfrac{59-x}{19}-1+\dfrac{58-x}{18}-1=\dfrac{57-x}{17}-1=\dfrac{56-x}{16}-1\)

\(\Leftrightarrow\dfrac{40-x}{19}+\dfrac{40-x}{18}-\dfrac{40-x}{17}-\dfrac{40-x}{16}=0\)

\(\Leftrightarrow\left(40-x\right)\left(\dfrac{1}{19}+\dfrac{1}{18}-\dfrac{1}{17}-\dfrac{1}{16}\right)=0\)

mà \(\dfrac{1}{19}+\dfrac{1}{18}-\dfrac{1}{17}-\dfrac{1}{16}\ne0\)

nên 40-x=0

hay x=40

Vậy: x=40

cho mk hỏi 4x-  5/8 hay (4x-5)/8   =6-x/2 phải ko

a) Ta có: \(7-\left(2x+4\right)=-\left(x+4\right)\)

\(\Leftrightarrow7-2x-4=-x-4\)

\(\Leftrightarrow-2x+3+x+4=0\)

\(\Leftrightarrow-x+7=0\)

\(\Leftrightarrow-x=-7\)

hay x=7

Vậy: S={7}

b) Ta có: \(\dfrac{2+x}{5}-0.5x=\dfrac{1-2x}{4}+0.25\)

\(\Leftrightarrow\dfrac{4\left(2+x\right)}{20}-\dfrac{0.5x\cdot20}{20}=\dfrac{5\left(1-2x\right)}{20}+\dfrac{20\cdot0.25}{20}\)

\(\Leftrightarrow4\left(2+x\right)-10x=5\left(1-2x\right)+5\)

\(\Leftrightarrow8+4x-10x=5-10x+5\)

\(\Leftrightarrow-6x+8=-10x+10\)

\(\Leftrightarrow-6x+8+10x-10=0\)

\(\Leftrightarrow4x-2=0\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

d) Ta có: \(\dfrac{x-1}{59}+\dfrac{x-2}{58}+\dfrac{x-3}{57}=\dfrac{x-59}{1}+\dfrac{x-58}{2}+\dfrac{x-57}{3}\)

\(\Leftrightarrow\dfrac{x-1}{59}-1+\dfrac{x-2}{58}-1+\dfrac{x-3}{57}-1=\dfrac{x-59}{1}-1+\dfrac{x-58}{2}-1+\dfrac{x-57}{3}-1\)

\(\Leftrightarrow\dfrac{x-60}{59}+\dfrac{x-60}{58}+\dfrac{x-60}{57}=\dfrac{x-60}{1}+\dfrac{x-60}{2}+\dfrac{x-60}{3}\)

\(\Leftrightarrow\left(x-60\right)\left(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}\right)-\left(x-60\right)\left(1+\dfrac{1}{2}+\dfrac{1}{3}\right)=0\)

\(\Leftrightarrow\left(x-60\right)\left(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}-1-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)

mà \(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}-1-\dfrac{1}{2}-\dfrac{1}{3}\ne0\)

nên x-60=0

hay x=60

Vậy: S={60}

Phương pháp giải:

- Đặt tính: Viết phép tính sao cho các chữ số cùng hàng thẳng cột với nhau.

- Tính: Trừ các số lần lượt từ phải sang trái.

Lời giải chi tiết:

a)

c)

18B

19B

a ) x = 5 6 ; b ) x = 8 7 ; c ) x = − 11 14 − ; d ) x = − 13 12

\(\left(\dfrac{x+1}{55}+\dfrac{x+2}{56}+\dfrac{x+3}{57}+\dfrac{x+4}{58}\right)-4=0\)

<=>\(\dfrac{x+1}{55}+\dfrac{x+2}{56}+\dfrac{x+3}{57}+\dfrac{x+4}{58}=4\)

<=>\(\dfrac{x+1}{55}-1+\dfrac{x+2}{56}-1+\dfrac{x+3}{57}+\dfrac{x+4}{58}-1=4-4\)

<=>\(\dfrac{x+1}{55}-\dfrac{55}{55}+\dfrac{x+2}{56}-\dfrac{56}{56}+\dfrac{x+3}{57}-\dfrac{57}{57}+\dfrac{x+4}{58}-\dfrac{58}{58}=0\)

<=>\(\dfrac{x-54}{55}+\dfrac{x-54}{56}+\dfrac{x-54}{57}+\dfrac{x-54}{58}=0\)

<=>\(\left(x-54\right)\left(\dfrac{1}{55}+\dfrac{1}{56}+\dfrac{1}{57}+\dfrac{1}{58}\right)=0\)

<=>x-54=0

<=>x=54

vậy phương trình có tập nghiệm là S={54}

=>\(\left(\dfrac{x+1}{55}-1\right)+\left(\dfrac{x+2}{56}-1\right)+\left(\dfrac{x+3}{57}-1\right)+\left(\dfrac{x+4}{58}-1\right)=0\)

=>x-54=0

=>x=54

Ta có: \(\dfrac{x-1}{59}+\dfrac{x-2}{58}+\dfrac{x-3}{57}=\dfrac{x-4}{56}+\dfrac{x-5}{55}+\dfrac{x-6}{54}\)

\(\Leftrightarrow\dfrac{x-60}{59}+\dfrac{x-60}{58}+\dfrac{x-60}{57}-\dfrac{x-60}{56}-\dfrac{x-60}{55}-\dfrac{x-60}{54}=0\)

\(\Leftrightarrow x-60=0\)

hay x=60

Áp dụng tính chất “Trong tam giác vuông, hai góc nhọn phụ nhau” ta có:

+ Hình 55:

+ Hình 56:

+ Hình 57 :

+ Hình 58: