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\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x=-\frac{13}{4}\)
\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)
\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x=-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{1}{2}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)
\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)
\(c,\frac{1}{3}x+\frac{2}{5}(x+1)=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)
\(\Leftrightarrow x=-\frac{6}{11}\)
d,e,f Tương tự
(2x-3/7)(2x^2+18)=0 => 2x-3/7=0 hoặc 2x^2+18=0 => 2x=3/7 hoặc 2x^2=-18(loại vì 2x^2 >= 0)
=>x=3/7 / 2=> x=3/7*1/2=>x=3/14
Vậy : x=3/14
f(1/2)=2*1/2-3=-2
y=f(x)=2x -3
Suy ra 2x-3=-5
2x= -5+3
2x=-2
Vậy x= -1
a) |2x-3|+x=21
|2x-3|=21-x
\(\Rightarrow\)\(\orbr{\begin{cases}2x-3=21-x\\2x-3=-\left(21-x\right)\end{cases}}\)
TH1: 2x-3=21-x
2x-x=21+3
x=24
TH2: 2x-3=-(21-x)
2x-3 = -21+x
2x-x=-21+3
x=-18
Vậy x \(\varepsilon\){-18;24}
Ta có :\(\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right).\left(2x-2\right)=\left(-\frac{3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)
=> \(\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right).\left(2x-2\right)=-\frac{1}{2}\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\)
=> \(2x-2=-\frac{1}{2}\)
=> \(2x=\frac{3}{2}\)
=> \(x=\frac{3}{4}\)
+) Nếu \(x\le-\frac{1}{2}\Leftrightarrow\left|2x-3\right|=3-2x\)
\(\left|2x+1\right|=-2x-1\)
\(pt\Leftrightarrow3-2x-2x-1=4\)
( nhập vào máy )
\(\Leftrightarrow x=-0,5\left(tm\right)\)
+) Nếu \(-\frac{1}{2}< x< \frac{3}{2}\)thì | 2x - 3 | = 3 - 2x
| 2x + 1 | = 2x + 1
\(pt\Leftrightarrow3-2x+2x+1=4\)
\(\Leftrightarrow x=-0,5\)( loại )
+) Nếu \(x\ge\frac{3}{2}\)thì | 2x - 3 | = 2x - 3
| 2x + 1 | = 2x + 1
\(pt\Leftrightarrow2x-3+2x+1=4\)
\(\Leftrightarrow x=1,5\left(tm\right)\)
Vậy \(x\in\left\{-0,5;1,5\right\}\)