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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
\(\frac{x+1}{2}=\frac{2}{3}\)
\(\Rightarrow3.\left(x+1\right)=2.2\)
\(\Rightarrow3x+3=4\)
\(\Rightarrow3x=4-3\)
\(\Rightarrow3x=1\)
\(\Rightarrow x=\frac{1}{3}\)
\(b,\frac{x-2}{3}=\frac{x+1}{4}\)
\(\Rightarrow4.\left(x-2\right)=3.\left(x+1\right)\)
\(\Rightarrow4x-8=3x+3\)
\(\Rightarrow4x-3x=3+8\)
\(\Rightarrow x=11\)
\(c,\frac{x-3}{x+5}=\frac{5}{7}\)
\(\Rightarrow7.\left(x-3\right)=5.\left(x+5\right)\)
\(\Rightarrow7x-21=5x+25\)
\(\Rightarrow7x-5x=25+21\)
\(\Rightarrow2x=46\)
\(\Rightarrow x=23\)
\(d,\frac{x-1}{x+2}=\frac{x-2}{x+3}\)
\(\Rightarrow\left(x-1\right)\left(x+3\right)=\left(x+2\right)\left(x-2\right)\)
\(\Rightarrow x^2+2x-3=x^2-4\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=-\frac{1}{2}\)
14 tháng 9 2017
\(a,\frac{-1}{2}+\left(x-3\right):\frac{-1}{2}=-1\frac{2}{3}.\)
\(\Rightarrow\left(x-3\right):\frac{-1}{2}=-1\frac{2}{3}-\frac{-1}{2}=\frac{-7}{6}\)
\(\Rightarrow x-3=\frac{-7}{6}\cdot\frac{-1}{2}=\frac{7}{12}\)
\(\Rightarrow x=\frac{7}{12}+3=3\frac{7}{12}\)
\(b.2,25+\frac{3}{2}:\left(x-5\right)=2\frac{1}{2}\)
\(\Rightarrow\frac{3}{2}:\left(x-5\right)=2\frac{1}{2}-2,25=\frac{1}{4}\)
\(\Rightarrow x-5=\frac{3}{2}:\frac{1}{4}=6\)
\(\Rightarrow x=6+5=11\)
\(c,\left(\frac{1}{3}-x\right)^2=\frac{1}{4}=\left(\frac{1}{2}\right)^2=\left(-\frac{1}{2}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}\frac{1}{3}-x=\frac{1}{2}\\\frac{1}{3}-x=-\frac{1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}-\frac{1}{2}=-\frac{1}{6}\\x=\frac{1}{3}-\frac{-1}{2}=\frac{5}{6}\end{cases}}\)
\(d,\frac{3}{2}+\frac{x-1}{3}=1\)
\(\Rightarrow\frac{x-1}{3}=1-\frac{3}{2}=-\frac{1}{2}\)
\(\Rightarrow x-1=-\frac{1}{2}\cdot3=-\frac{3}{2}\)
\(\Rightarrow x=-\frac{3}{2}+1=\frac{1}{2}\)
\(e,-\frac{6}{8}+\frac{x}{12}=\frac{5}{6}\)
\(\Rightarrow\frac{x}{12}=\frac{5}{6}-\frac{-6}{8}=\frac{19}{12}\)
\(\Rightarrow x=19\)
\(g,\frac{1}{2}-\frac{1}{3}\left(x-2\right)=-\frac{2}{3}\)
\(\Rightarrow-\frac{1}{3}\left(x-2\right)=-\frac{2}{3}-\frac{1}{2}=-\frac{7}{6}\)
\(\Rightarrow x-2=\frac{-7}{6}:\frac{-1}{3}=\frac{7}{2}\)
\(\Rightarrow x=\frac{7}{2}+2=2\frac{7}{2}\)
\(h,\frac{5}{2}\left(x+1\right)-\frac{1}{2}=3\frac{1}{2}\)
\(\Rightarrow\frac{5}{2}\left(x+1\right)=3\frac{1}{2}-\frac{1}{2}=3\)
\(\Rightarrow x+1=3:\frac{5}{2}=\frac{6}{5}\)
\(\Rightarrow x=\frac{6}{5}-1=\frac{1}{5}\)
\(k,\frac{x}{3}-\frac{1}{2}=-2\left(x+1\right)+3\)
\(\Rightarrow x\cdot\frac{1}{3}-\frac{1}{2}=-2x-2+3\)
\(\Rightarrow\frac{1}{3}x+2x=-2+3+\frac{1}{2}\)
\(\Rightarrow\frac{7}{3}x=\frac{3}{2}\Rightarrow x=\frac{3}{2}:\frac{7}{2}=\frac{3}{7}\)
18 tháng 6 2019
1) \(\frac{x+1}{15}+\frac{x+2}{14}=\frac{x+3}{13}+\frac{x+4}{12}\)
\(\Leftrightarrow\frac{x+16}{15}+\frac{x+16}{14}-\frac{x+16}{13}-\frac{x+16}{12}=0\)
\(\Leftrightarrow\left(x+16\right)\left(\frac{1}{15}+\frac{1}{14}-\frac{1}{13}-\frac{1}{12}\right)=0\)
\(\Leftrightarrow x=-16\)
2)3)4) tương tự
Gợi ý : 2) cộng 3 vào cả hai vế
3)4) cộng 2 vào cả hai vế
5) \(\frac{x+1}{20}+\frac{x+2}{19}+\frac{x+3}{18}=-3\)
\(\Leftrightarrow\frac{x+21}{20}+\frac{x+21}{19}+\frac{x+21}{18}=0\)
\(\Leftrightarrow\left(x+21\right)\left(\frac{1}{20}+\frac{1}{19}+\frac{1}{18}\right)=0\)
\(\Leftrightarrow x=-21\)
6) sửa VT = 4 rồi tương tự câu 5)
7 tháng 8 2018
\(\frac{x+2}{x+6}=\frac{3}{x+1}\)
\(\Rightarrow\left(x+2\right)\left(x+1\right)=3\left(x+6\right)\)
\(\Rightarrow x^2+x+2x+2=3x+18\)
\(\Rightarrow x^2+x+2x-3x=18-2\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x=\pm4\)
các phần còn lại tương tự :)
7 tháng 8 2018
a)\(\frac{x+2}{x+6}\) =\(\frac{3}{x+1}\)
<=>\(\frac{\left(x+2\right)\left(x+1\right)}{\left(x+6\right)\left(x+1\right)}\) =\(\frac{3\left(x+6\right)}{\left(x+1\right)\left(x+6\right)}\)
=> ( x+2) ( x+1) = 3(x+6)
<=> x2 +3x +3 = 3x +18
<=> x2 +3x -3x = 18 -3
<=> x2 = 15
=> x = \(\sqrt{15}\)
Vậy x=\(\sqrt{15}\)
b)
\(\frac{x+3}{x-2}=\frac{x+2}{x-1}\)(ĐKXĐ: x khác 2 và 1)
\(\Leftrightarrow\frac{\left(x+3\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}=\frac{\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)}\)
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=\left(x+2\right)\left(x-2\right)\)
\(\Leftrightarrow x^2-x+3x-3=x^2-2x+2x-4\)
\(\Leftrightarrow x^2-x+3x-x^2-2x+2x=3-4\)
\(\Leftrightarrow2x=-1\Leftrightarrow x=-\frac{1}{2}\)
\(\frac{x+3}{x-2}=\frac{x+2}{x-1}\)
<=> ( x + 3 ) ( x - 1 ) = ( x + 2 ) ( x - 2 )
<=> x2 + 2x - 3 = x2 - 4
<=> x2 + 2x - 3 - x2 + 4 = 0
<=> 2x + 1 = 0
<=> x = - 1/2