Nhận xét :

\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)

Vì \(x\ge0\) nên pt a) tương đương với : \(100x+\frac{1+2+3+...+100}{101}=101x\)

\(\Leftrightarrow x=\frac{100.101}{2.101}=50\)

b) 

Tương tự câu a) , phương trình tương đương với : 

\(49x+\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{...1}{97.99}=50x\)

\(\Rightarrow x=\frac{97}{195}\)

c) 

Tương tự câu a) , phương trình tương đương với : 

\(99x+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=100x\)

\(\Rightarrow x=\frac{99}{100}\)

e. \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)

\(\Rightarrow\frac{7}{x}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=\frac{29}{45}\)

\(\Rightarrow\frac{7}{x}+\frac{1}{5}-\frac{1}{45}=\frac{29}{45}\)

\(\Rightarrow\frac{7}{x}=\frac{7}{15}\)

\(\Rightarrow x=15\)

f. \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}.\frac{22}{45}x=\frac{22}{45}\)

\(\Rightarrow\frac{11}{45}x=\frac{22}{45}\)

\(\Rightarrow x=2\)

tung từng vế một thôi

bạn nhác quá éo chịu suy nghĩ

bài này dễ vl

Bài 1:

a, \(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right)\left(5x+6\right)}=\frac{2010}{2011}\)

\(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2010}{2011}\)

\(1-\frac{1}{5x+6}=\frac{2010}{2011}\)

\(\frac{1}{5x+6}=1-\frac{2010}{2011}\)

\(\frac{1}{5x+6}=\frac{1}{2011}\)

=> 5x + 6 = 2011

    5x = 2011 - 6

    5x = 2005

    x = 2005 : 5

    x = 401

b, \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)

\(\frac{7}{x}+\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}\right)=\frac{29}{45}\)

\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)

\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)

\(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)

\(\frac{7}{x}=\frac{29}{45}-\frac{8}{45}\)

\(\frac{7}{x}=\frac{7}{15}\)

=> x = 15

c, ghi lại đề

d, ghi lại đề

Bài 2:

\(\frac{1}{n}-\frac{1}{n+a}=\frac{n+a}{n\left(n+a\right)}-\frac{n}{n\left(n+a\right)}=\frac{a}{n\left(n+a\right)}\)

\(a,\frac{x-1}{21}=\frac{3}{x+1}\)

\(\Leftrightarrow\left[x-1\right]\left[x+1\right]=63\)

\(\Leftrightarrow x^2-1=63\)

\(\Leftrightarrow x^2=64\)

\(\Leftrightarrow x^2=8^2\)

\(\Leftrightarrow x=\pm8\)

\(b,\frac{7}{x}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\left[\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}\right]=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\left[\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right]=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\left[\frac{1}{5}-\frac{1}{45}\right]=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}=\frac{21}{45}\)

\(\Leftrightarrow\frac{7}{x}=\frac{7}{15}\)

\(\Leftrightarrow x=15\)

Vậy x = 15

Bài cuối tương tự

Ta có : \(x+\frac{1}{1.5}+x+\frac{1}{5.9}+x+\frac{1}{9.13}+......+x+\frac{1}{397.401}=101x\)

\(\Leftrightarrow\left(x+x+x+......+x\right)+\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+......+\frac{1}{397.401}\right)=101x\)

\(\Leftrightarrow100x+\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+......+\frac{1}{397.401}\right)=101x\)

\(\Rightarrow x=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+......+\frac{1}{397.401}\)

\(\Rightarrow4x=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+......+\frac{4}{397.401}\)

\(\Rightarrow4x=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+.....+\frac{1}{397}-\frac{1}{401}\)

\(\Rightarrow4x=1-\frac{1}{401}\)

\(\Rightarrow4x=\frac{400}{401}\)

\(\Rightarrow x=\frac{400}{401}.\frac{1}{4}=\frac{100}{401}\)

tui biết giải, mà k biết có bao nhiêu x, bạn tính sao ra 100x vậy bạn?

Đơn giải thôi bạn chỉ cần lấy công thức tính số số hạng là ra thôi

(397 - 1) : 4 + 1 = 100 (số)

\(A=8400\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)

\(=\frac{8400}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}+\frac{4}{21.25}\right)\)

\(=2100\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\right)\)

\(=2100\left(1-\frac{1}{25}\right)\)

\(=2100\cdot\frac{24}{25}\)

\(=2016\)

\(A=8400.\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)

\(A=8400.\left(\frac{1.4}{1.5.4}+\frac{1.4}{5.9.4}+\frac{1.4}{9.13.4}+\frac{1.4}{13.17.4}+\frac{1.4}{17.21.4}+\frac{1.4}{21.25.4}\right)\)

\(A=8400.\frac{1}{4}.\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)

\(A=8400.\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\right)\)

\(A=8400.\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{25}\right)\)

\(A=8400.\frac{1}{4}.\frac{24}{25}\)

\(A=2016\)

\(A=8400.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\right)\) 

\(A=8400.\left(1-\frac{1}{25}\right)\)

\(A=8400.\frac{24}{25}=8064\)

\(->A=8064\)

a, \(\frac{1}{1.4}\)+\(\frac{1}{4.7}\)+......+\(\frac{1}{97.100}\)= |\(\frac{x}{3}\)|

\(\Rightarrow\)\(\frac{1}{3}\) ( \(\frac{3}{1.4}\)+\(\frac{3}{4.7}\)+.......+\(\frac{3}{97.100}\))= |\(\frac{x}{3}\)|

\(\Rightarrow\)\(\frac{1}{3}\) ( 1  - \(\frac{1}{4}\)+ \(\frac{1}{4}\)-\(\frac{1}{7}\)+......+\(\frac{1}{97}\)-\(\frac{1}{100}\)) = |\(\frac{x}{3}\)|

\(\Rightarrow\)\(\frac{1}{3}\) ( 1-\(\frac{1}{100}\)) = |\(\frac{x}{3}\)|

\(\Rightarrow\)\(\frac{1}{3}\) . \(\frac{99}{100}\) = |\(\frac{x}{3}\)|

\(\Rightarrow\)\(\frac{33}{100}\) = |\(\frac{x}{3}\)|

\(\Rightarrow\)\(\frac{x}{3}\)= \(\orbr{\begin{cases}\frac{33}{100}\\\frac{-33}{100}\end{cases}}\)

Với \(\frac{x}{3}\) = \(\frac{33}{100}\)

\(\Rightarrow\)100x= 33.3

 \(\Rightarrow\)100x=99

\(\Rightarrow\)x=\(\frac{99}{100}\)

Với \(\frac{x}{3}\)=\(\frac{-33}{100}\)

\(\Rightarrow\)100x=-33.3

\(\Rightarrow\)100x=-99

\(\Rightarrow\)x=\(\frac{-99}{100}\)

Vậy x=\(\orbr{\begin{cases}\frac{99}{100}\\\frac{-99}{100}\end{cases}}\)

b, \(\frac{4}{1.5}\)+ \(\frac{4}{5.9}\)+......+ \(\frac{4}{97.101}\)= |\(\frac{5x-4}{101}\)|

\(\Rightarrow\)1-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{9}\)+......+\(\frac{1}{97}\)-\(\frac{1}{101}\)= |\(\frac{5x-4}{101}\)|

\(\Rightarrow\)1-\(\frac{1}{101}\)= |\(\frac{5x-4}{101}\)

\(\Rightarrow\) \(\frac{100}{101}\)= |\(\frac{5x-4}{101}\)|

\(\Rightarrow\)\(\frac{5x-4}{101}\) =\(\orbr{\begin{cases}\frac{100}{101}\\\frac{-100}{101}\end{cases}}\)

Với \(\frac{5x-4}{101}\) =\(\frac{100}{101}\)

\(\Rightarrow\)(5x-4).101=100.101

\(\Rightarrow\)505x-404=10100

\(\Rightarrow\)505x=10504

\(\Rightarrow\)x=\(\frac{104}{5}\)

Với \(\frac{5x-4}{101}\)=\(\frac{-100}{101}\)

\(\Rightarrow\)(5x-4). 101=-100.101

\(\Rightarrow\)505x-404=-10100

\(\Rightarrow\)505x=-9696

\(\Rightarrow\)x=\(\frac{-96}{5}\)

Vậy x=\(\orbr{\begin{cases}\frac{104}{5}\\\frac{-96}{5}\end{cases}}\)