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24 tháng 10 2018
1) \(\frac{x-1}{5}=\frac{x+2}{7}\)
\(\Rightarrow7x-7=5x+10\)
=> 7x - 5x = 10 + 7
2x = 17
x = 8,5
b) \(\frac{x-1}{7}=\frac{5}{x+1}\)
=> x2 - 1 = 35
x2 = 34
\(\Rightarrow x=\sqrt{34};x=-\sqrt{34}\)
c) \(\frac{x+1}{x-2}=\frac{x+2}{x+3}\)
=> x2 + 4x + 3 = x2 - 4
=> x2 - x2 + 4x + 3 = -4
4x + 3 = - 4
4x = -7
x = -7/4
19 tháng 9 2016
a, (x+1).3 = 2.2
=>3 x+3 =4
=> 3x=1
=> x=1/3
b, (x-2) .4 =(x+1).3
=>4x-8=3x+3
=>4x-3x=8+3
=>x=11
c, lam tg tu cau b
d, (x-1)(x+3)=(x+2)(x-2)
\(x^2\)+3x-x-3=\(x^2\)-2x+2x-4
x^2 +2x-3=x^2-4
x^2-x^2+2x=3-4
2x=-1
x=-0,5
22 tháng 10 2016
\(\frac{x+1}{2}=\frac{2}{3}\)
\(\Rightarrow3.\left(x+1\right)=2.2\)
\(\Rightarrow3x+3=4\)
\(\Rightarrow3x=4-3\)
\(\Rightarrow3x=1\)
\(\Rightarrow x=\frac{1}{3}\)
\(b,\frac{x-2}{3}=\frac{x+1}{4}\)
\(\Rightarrow4.\left(x-2\right)=3.\left(x+1\right)\)
\(\Rightarrow4x-8=3x+3\)
\(\Rightarrow4x-3x=3+8\)
\(\Rightarrow x=11\)
\(c,\frac{x-3}{x+5}=\frac{5}{7}\)
\(\Rightarrow7.\left(x-3\right)=5.\left(x+5\right)\)
\(\Rightarrow7x-21=5x+25\)
\(\Rightarrow7x-5x=25+21\)
\(\Rightarrow2x=46\)
\(\Rightarrow x=23\)
\(d,\frac{x-1}{x+2}=\frac{x-2}{x+3}\)
\(\Rightarrow\left(x-1\right)\left(x+3\right)=\left(x+2\right)\left(x-2\right)\)
\(\Rightarrow x^2+2x-3=x^2-4\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=-\frac{1}{2}\)
KN
5 tháng 6 2019
1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)
TB
1 tháng 1 2017
1, \(\frac{1}{2}\)- (x + \(\frac{3}{4}\))= \(\frac{5}{6}\)
(x+\(\frac{3}{4}\)) = \(\frac{5}{6}\)+ \(\frac{1}{2}\)
( x+ \(\frac{3}{4}\))=\(\frac{4}{3}\)
x = \(\frac{4}{3}\)- \(\frac{3}{4}\)
x =\(\frac{7}{12}\)
kết quả : k mk nha
XO
27 tháng 9 2020
Ta có :\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
=> \(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
=> \(2\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)
=> \(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
=> \(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
=> \(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
=> \(\frac{1}{x+1}=\frac{1}{18}\)
=> x + 1 = 18
=> x = 17
Vậy x = 17
7 tháng 8 2018
Bài 1:
a) \(x-\frac{20}{11.13}-\frac{20}{13.15}-...-\frac{20}{53.55}=\frac{3}{11}\)
\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)
\(x-\frac{20}{2}.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10.\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10\cdot\frac{4}{55}=\frac{3}{11}\)
\(x-\frac{8}{11}=\frac{3}{11}\)
\(x=1\)
b) \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(2.\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(2.\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{18}\)
=> x + 1 =18
x = 17
bài 2 ko bk lm, xl nha
Ta có : | x + 1 | + | 2 - x | ≥ | x + 1 + 2 - x | = 3
=> \(\frac{1}{\left|x+1\right|+\left|2-x\right|}\le\frac{1}{3}\)
Dấu "=" xảy ra khi ( x + 1 )( 2 - x ) ≥ 0
=> -1 ≤ x ≤ 2
Vậy MaxQ = 1/3 <=> -1 ≤ x ≤ 2