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1/ \(\frac{1}{3x}:\frac{2}{3}=1\)
<=> \(\frac{3}{3×2×x}=\:1\)
<=> \(\frac{1}{2x}=1\)<=> x = \(\frac{1}{2}\)
X+1/3 = 3/ 4
X = 3/4 -1/3
X = 5/12
X - 2/5 = 5/7
X = 5/7 -2/5
X = 9/35
-X - 2/3 = 6/7
-X = 6/7 - 2/3
-X = 4/21
4/7 - X = 1/3
X = 4/7 - 1/3
X = 5/21
k mk nha bn!!!! thank bn nhìu nha
x + 1/3 = 3/4
x = 3/4 - 1/3
x = 5/12
x-2/4=5/7
x=5/7+2/5=
x=39/35
tườn tự nhé
giúp tớ nhé
tớ bị trừ 590 điểm
cảm ơn trước
a, \(\left(3x-5\right)\left(x+1\right)-\left(3x-1\right)\left(x+1\right)=x-4\)
\(\Leftrightarrow\left(x+1\right)\left(3x-5-3x+1\right)=x-4\Leftrightarrow-4\left(x+1\right)=x-4\)
\(\Leftrightarrow-4x-4=x-4\Leftrightarrow-4x-x=0\Leftrightarrow x=0\)
b, \(\left(x-2\right)\left(x+3\right)-\left(x+4\right)\left(x-7\right)=5-x\)
\(\Leftrightarrow x^2+x-6-x^2-3x+28=5-x\Leftrightarrow-2x+22=5-x\Leftrightarrow x=17\)
c, thiếu đề
d, \(3\left(x-7\right)\left(x+7\right)-\left(x-1\right)\left(3x+2\right)=13\)
\(\Leftrightarrow3x^2-147-3x^2+x+2=13\Leftrightarrow x=11+147=158\)
a.\(3x^2-2x-5-\left(3x^2+2x-1\right)=x-4\)
\(\Leftrightarrow-5x=0\Leftrightarrow x=0\)
b.\(x^2+x-6-\left(x^2-3x-28\right)=5-x\)
\(\Leftrightarrow5x=-17\Leftrightarrow x=-\frac{17}{5}\)
c.\(5\left(x^2-10x+21\right)-\left(5x^2-9x-2\right)=0\)
\(\Leftrightarrow-41x+107=0\Leftrightarrow x=\frac{107}{41}\)
d.\(3\left(x^2-49\right)-\left(3x^2-x-2\right)=13\Leftrightarrow x=158\)
a)\(\left(\frac{3}{5}\right)^5\times x=\left(\frac{3}{7}\right)^7\)
\(\Leftrightarrow\frac{3^5}{5^5}\times x=\frac{3^7}{7^7}\)
\(\Leftrightarrow x=\frac{3^7}{7^7}:\frac{3^5}{5^5}\)
\(\Leftrightarrow x=\frac{3^7\times5^5}{7^7\times3^5}\)
\(\Leftrightarrow x=\frac{3^2\times5^5}{7^7}\)
b)\(\left(\frac{-1}{3}\right)^3\times x=\frac{1}{81}\)
\(\Leftrightarrow\frac{\left(-1\right)^3}{3^3}\times x=\frac{1}{3^4}\)
\(\Leftrightarrow x=\frac{1}{3^4}:\frac{-1}{3^3}\)
\(\Leftrightarrow x=\frac{1\times3^3}{3^4\times\left(-1\right)}\)
\(\Leftrightarrow x=\frac{1}{-3}\)
c)\(\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{3}+\frac{1}{2}\)
\(\Leftrightarrow x=\frac{5}{6}\)
d)\(\Leftrightarrow\left(x+\frac{1}{2}\right)^4=\left(\frac{2}{3}\right)^4\)
\(\Leftrightarrow x+\frac{1}{2}=\frac{2}{3}\)
\(\Leftrightarrow x=\frac{2}{3}-\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{6}\)
a) \(\frac{x}{x+1}=\frac{x+5}{x+7}\)
\(=>x\left(x+7\right)=\left(x+1\right).\left(x+5\right)\)
\(=>x^2+7x=x^2+6x+5\)
\(=>x^2+7x-x^2-6x-5=0\)
\(=>x-5=0\)
\(=>x=5\)
vay \(x=5\)
b) \(\frac{x+7}{x+4}=\frac{x-1}{x-2}\)
\(=>\left(x+7\right)\left(x-2\right)=\left(x+4\right)\left(x-1\right)\)
\(=>x^2+5x-14=x^2+3x-4\)
\(=>x^2+5x-14-x^2-3x+4=0\)
\(=>2x-10=0\)
\(=>2\left(x-5\right)=0\)
\(=>x-5=0\)
\(=>x=5\)
vay \(x=5\)
c) \(\frac{x+2}{x-2}=\frac{x-3}{x+3}\)
\(=>\left(x+2\right)\left(x+3\right)=\left(x-2\right)\left(x-3\right)\)
\(=>x^2+5x+6=x^2-7x+6\)
\(=>x^2+5x+6-x^2+7x-6=0\)
\(=>12x=0\)
\(=>x=0\)
vay \(x=0\)
a,xet cac th sau
x<1'=>1-x+4+x=4=>3-2x=4
=>2x=-1=>x=-1/2
th2 1<x,<5
=>x-1+4+x=4<=>3=4(vo li)
vay x=-1/2
| x | = \(\frac{1}{3}-\frac{2}{7}\)
| x | = \(\frac{1}{21}\)
\(\Rightarrow x=\)\(\frac{1}{21}\)hoặc\(-\frac{1}{21}\)