\(\left(x-1\right)+\left(x-2\right)+\left(x-3\right)+.....+\left(x-100\right)=101...">
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17 tháng 7 2019

b  \(\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{x\cdot\left(x+1\right)}=\frac{19}{100}\)

=>\(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{19}{100}\)

=>\(\frac{1}{5}-\frac{1}{x+1}\)\(=\frac{19}{100}\)

=>\(\frac{1}{x+1}=\frac{1}{5}-\frac{19}{100}\)

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

=> x+1 =100

=>x=99

17 tháng 7 2019

b) \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{19}{100}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{19}{100}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x+1}=\frac{19}{100}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{5}-\frac{19}{100}\)

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

\(\Rightarrow x+1=100\)

\(\Rightarrow x=99\)

c) \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x\left(x+2\right)}=\frac{49}{99}\)

\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{49}{99}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{49}{99}\)

\(\Rightarrow\frac{1}{x+2}=1-\frac{49}{99}\)

\(\Rightarrow\frac{1}{x+2}=\frac{50}{99}\)

\(\Rightarrow50.\left(x+2\right)=99\)

\(\Rightarrow x+2=\frac{99}{50}\)

\(\Rightarrow x=-\frac{1}{99}\)

d) Ta có : 6 = 1.6 = 2.3 = (-2) . (-3)

Lâp bảng xét 6 trường hợp: 

\(2x+1\)\(1\)\(6\)\(2\)\(3\)\(-2\)\(-3\)
\(y-2\)\(6\)\(1\)\(3\)\(2\)\(-3\)\(-2\)
\(x\)\(0\)\(\frac{5}{2}\)\(\frac{1}{2}\)\(1\)\(-\frac{3}{2}\)\(-2\)
\(y\)\(8\)\(3\)\(5\)\(4\)\(-1\)\(0\)

Vậy các cặp (x,y) \(\inℤ\)thỏa mãn là : (0;4) ; (1; 4) ; (-2 ; 0)

e) \(x^2-3xy+3y-x=1\)

\(\Rightarrow x\left(x-3y\right)+3y-x=1\)

\(\Rightarrow x\left(x-3y\right)-\left(x-3y\right)=1\)

\(\Rightarrow\left(x-3y\right)\left(x-1\right)=1\)

Lại có : 1 = 1.1 = (-1) . (-1)

Lập bảng xét các trường hợp : 

\(x-1\)\(1\)\(-1\)
\(x-3y\)\(1\)\(-1\)
\(x\)\(2\)\(0\)
\(y\)\(\frac{1}{3}\)\(\frac{1}{3}\)

Vậy các cặp(x,y) thỏa mãn là : \(\left(2;\frac{1}{3}\right);\left(0;\frac{1}{3}\right)\)

28 tháng 7 2018

a)  \(\left|x+\frac{1}{2}\right|=\frac{1}{3}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{3}\\x+\frac{1}{2}=-\frac{1}{3}\end{cases}}\) \(\Leftrightarrow\)\(\orbr{\begin{cases}x=-\frac{1}{6}\\x=-\frac{5}{6}\end{cases}}\)

Vậy....

b)  \(\left|x-\frac{1}{2}\right|=\frac{1}{3}-\frac{1}{2}\)

\(\Leftrightarrow\)\(\left|x-\frac{1}{2}\right|=-\frac{1}{6}\)   vô lí do \(\left|a\right|\ge0\)

Vậy pt vô nghiệm

c)  \(\left|x+\frac{1}{3}\right|-4=-1\)

\(\Leftrightarrow\)\(\left|x+\frac{1}{3}\right|=3\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+\frac{1}{3}=3\\x+\frac{1}{3}=-3\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{8}{3}\\x=-\frac{10}{3}\end{cases}}\)

Vậy..

28 tháng 7 2018

d)  \(\left|x-\frac{1}{5}\right|+\frac{1}{3}=\frac{1}{4}-\left|-\frac{3}{2}\right|\)

\(\Leftrightarrow\)\(\left|x-\frac{1}{5}\right|+\frac{1}{3}=-\frac{5}{4}\)

\(\Leftrightarrow\)\(\left|x-\frac{1}{5}\right|=-\frac{19}{12}\)vô lí do  \(\left|a\right|\ge0\)với mọi a

Vậy pt vô nghiệm

e)  \(\left|x-\frac{5}{2}\right|=\frac{4}{3}-\left(\frac{2}{3}-\frac{1}{2}\right)\)

\(\Leftrightarrow\)\(\left|x-\frac{5}{2}\right|=\frac{7}{6}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-\frac{5}{2}=\frac{7}{6}\\x-\frac{5}{2}=-\frac{7}{6}\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=3\frac{2}{3}\\x=\frac{4}{3}\end{cases}}\)

Vậy...

\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)

_Tần vũ_

\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)

\(\Leftrightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)

\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)

\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)

\(\Leftrightarrow3x-\frac{1}{2}=\frac{-1}{3}\)

\(\Leftrightarrow3x=\frac{1}{6}\)

\(\Leftrightarrow x=\frac{1}{18}\)

_Tần Vũ_

4 tháng 6 2016

Tìm x, biết:

3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x2;5;10;17)

2(x1)(x3) +5(x3)(x8) +12(x8)(x20) 1x20 =34 (x1;3;8;20)

x+110 +2+111 x+112 =x+113 +x+114 

x1030 +x1443 +x595 +x1488 =0

4 tháng 6 2016

Trả lời luôn à bạn

6 tháng 6 2016

a) (1/2(2/− 2x0

\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)

b) \(\left(x.6\frac{2}{7}+\frac{3}{7}\right).2\frac{1}{5}-\frac{3}{7}=-2\)

\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-2+\frac{3}{7}\)

\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-\frac{11}{7}\)

\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{11}{7}:\frac{11}{5}=-\frac{11}{7}.\frac{5}{11}\)

\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{5}{7}\)

\(\Rightarrow x.\frac{44}{7}=-\frac{5}{7}-\frac{3}{7}\)

\(\Rightarrow x.\frac{44}{7}=-\frac{8}{7}\)

\(\Rightarrow x=-\frac{8}{7}:\frac{44}{7}=-\frac{8}{7}.\frac{7}{44}\)

\(\Rightarrow x=-\frac{2}{11}\)

6 tháng 6 2016

c) \(x.3\frac{1}{4}+\left(-\frac{7}{6}\right).x-1\frac{2}{3}=\frac{5}{12}\)

\(\Rightarrow x\left(3\frac{1}{4}-\frac{7}{6}\right)=\frac{5}{12}+\frac{5}{3}\)

\(\Rightarrow x\left(\frac{13}{4}-\frac{7}{6}\right)=\frac{25}{12}\)

\(\Rightarrow x.\frac{25}{12}=\frac{25}{12}\)

\(\Rightarrow x=\frac{25}{12}:\frac{25}{12}\)

\(\Rightarrow x=1\)

d) \(5\frac{8}{17}:x+\left(-\frac{4}{17}\right):x+3\frac{1}{7}:17\frac{1}{3}=\frac{4}{11}\)

\(\Rightarrow\left(5\frac{8}{17}-\frac{4}{17}\right):x+\frac{22}{7}:\frac{52}{3}=\frac{4}{11}\)

\(\Rightarrow5\frac{4}{17}:x+\frac{33}{182}=\frac{4}{11}\)

\(\Rightarrow\frac{89}{17}:x=\frac{4}{11}-\frac{33}{182}\)

\(\Rightarrow\frac{89}{17}:x=\frac{365}{2002}\)

\(\Rightarrow x=\frac{89}{17}:\frac{365}{2002}\)

\(\Rightarrow x\approx28,7\) (số hơi lẻ)

e) \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)

\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}\)

\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}\)

\(\Rightarrow\left[\begin{array}{nghiempt}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}2x=11\\2x=-\frac{19}{2}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{11}{2}\\x=-\frac{19}{4}\end{array}\right.\)

24 tháng 1 2020

a) \(\left(x+1\right)-\frac{x+1}{3}=\frac{5\left(x+1\right)-1}{6}\)

\(\Leftrightarrow6\left(x+1\right)-2\left(x+1\right)=5\left(x+1\right)-1\)

\(\Leftrightarrow6x+6-2x-2=5x+5-1\)

\(\Leftrightarrow6x-2x-5x=5-1-6+2\)

\(\Leftrightarrow-x=0\)

\(\Leftrightarrow x=0\)

b) \(\left(1-x\right)^2+\left(x+2\right)^2=2x\left(x-3\right)-7\)

\(\Leftrightarrow1-2x+x^2+x^2+4x+4=2x^2-6x-7\)

\(\Leftrightarrow2x^2+2x+5=2x^2-6x-7\)

\(\Leftrightarrow2x+6x=-7-5\)

\(\Leftrightarrow8x=-12\)

\(\Leftrightarrow x=-\frac{3}{2}\)

c) \(2+\frac{x-2}{2}-\frac{2x-4}{3}-\frac{5}{6}\left(2-x\right)=0\)

\(\Leftrightarrow2+\frac{x}{2}-1-\frac{2}{3}x+\frac{4}{3}-\frac{5}{3}+\frac{5}{6}x=0\)

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

\(\Leftrightarrow\frac{2}{3}x=-\frac{2}{3}\)

\(\Leftrightarrow x=-1\)