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
\(\frac{x+1}{2}=\frac{2}{3}\)
\(\Rightarrow3.\left(x+1\right)=2.2\)
\(\Rightarrow3x+3=4\)
\(\Rightarrow3x=4-3\)
\(\Rightarrow3x=1\)
\(\Rightarrow x=\frac{1}{3}\)
\(b,\frac{x-2}{3}=\frac{x+1}{4}\)
\(\Rightarrow4.\left(x-2\right)=3.\left(x+1\right)\)
\(\Rightarrow4x-8=3x+3\)
\(\Rightarrow4x-3x=3+8\)
\(\Rightarrow x=11\)
\(c,\frac{x-3}{x+5}=\frac{5}{7}\)
\(\Rightarrow7.\left(x-3\right)=5.\left(x+5\right)\)
\(\Rightarrow7x-21=5x+25\)
\(\Rightarrow7x-5x=25+21\)
\(\Rightarrow2x=46\)
\(\Rightarrow x=23\)
\(d,\frac{x-1}{x+2}=\frac{x-2}{x+3}\)
\(\Rightarrow\left(x-1\right)\left(x+3\right)=\left(x+2\right)\left(x-2\right)\)
\(\Rightarrow x^2+2x-3=x^2-4\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=-\frac{1}{2}\)
a, \(\frac{x+1}{2}=\frac{2}{3}\Rightarrow3\left(x+1\right)=4\)
=> \(3x+3=4\Rightarrow3x=1\Rightarrow x=\frac{1}{3}\)
b, \(\frac{x-2}{3}=\frac{x+1}{4}\Rightarrow4\left(x-2\right)=3\left(x+1\right)\)
\(\Rightarrow4x-8=3x+3\Rightarrow4x-3x=-8+3\Rightarrow x=-5\)
c, \(\frac{x-3}{x+5}=\frac{5}{7}\Rightarrow7\left(x-3\right)=5\left(x+5\right)\)
=> \(7x-21=5x+25\Rightarrow7x-5x=-21+25\)
=> \(2x=4\Rightarrow x=2\)
Câu d mik ko bít lm.....Chúc bn hok tốt
\(a,\frac{x+1}{2}=\frac{2}{3}\)
\(\Leftrightarrow\left(x+1\right).3=2.2\)
\(\Leftrightarrow3x+3=4\)
\(\Leftrightarrow3x=4-3\)
\(\Leftrightarrow3x=1\)
\(\Leftrightarrow x=1:3\)
\(\Leftrightarrow x=\frac{1}{3}\)
Vậy : \(x=\frac{1}{3}\)
\(b,\frac{x-2}{3}=\frac{x+1}{4}\)
\(\Leftrightarrow\left(x-2\right).4=\left(x+1\right).3\)
\(\Leftrightarrow4x-8=3x+3\)
\(\Leftrightarrow4x-3x=3+8\)
\(\Leftrightarrow x=11\)
Vậy : \(x=11\)
\(c,\frac{x-3}{x+5}=\frac{5}{7}\)
\(\Leftrightarrow\left(x-3\right).7=\left(x+5\right).5\)
\(\Leftrightarrow7x-21=5x+25\)
\(\Leftrightarrow7x-5x=25+21\)
\(\Leftrightarrow2x=46\)
\(\Leftrightarrow x=46:2\)
\(\Leftrightarrow x=23\)
Vậy : \(x=23\)
\(d,\frac{x-1}{x+2}=\frac{x-2}{x+3}\)
\(\Leftrightarrow\left(x-1\right).\left(x+3\right)=\left(x-2\right).\left(x+2\right)\)
\(\Leftrightarrow x^2+2x-3=x^2-4\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=-1:2\)
\(\Leftrightarrow x=-\frac{1}{2}\)
Vậy : \(x=-\frac{1}{2}\)
Vậy :\(x=-\frac{1}{2}\)
Rất vui vì giúp đc bạn !!!
