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|1/2x| = 3 - 2x
ĐKXĐ : 3 - 2x \(\ge\)0 => 2x \(\ge\) 3 => x \(\ge\)3/2
Ta có: |1/2x| = 3 - 2x
=> \(\orbr{\begin{cases}\frac{1}{2}x=3-2x\\\frac{1}{2}x=-3+2x\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{1}{2}x+2x=3\\\frac{1}{2}x-2x=-3\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{2}x=3\\-\frac{3}{2}x=-3\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{6}{5}\left(ktm\right)\\x=2\left(tm\right)\end{cases}}\)
=> x = 2
|5x| = x - 12
ĐKXĐ : x - 12 \(\ge\)0 => x \(\ge\)12
Ta có: |5x| = x - 12
=> \(\orbr{\begin{cases}5x=x-12\\5x=-x+12\end{cases}}\)
=> \(\orbr{\begin{cases}5x-x=-12\\5x+x=12\end{cases}}\)
=> \(\orbr{\begin{cases}4x=-12\\6x=12\end{cases}}\)
=> \(\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)(ktm)
=> pt vô nghiệm
|2x - 5| = x + 1
ĐKXĐ: x + 1 \(\ge\)0 => x \(\ge\)-1
Ta có: |2x - 5| = x + 1
=> \(\orbr{\begin{cases}2x-5=x+1\\2x-5=-x-1\end{cases}}\)
=> \(\orbr{\begin{cases}2x-x=1+5\\2x+x=-1+5\end{cases}}\)
=> \(\orbr{\begin{cases}x=6\\3x=4\end{cases}}\)
=> \(\orbr{\begin{cases}x=6\\x=\frac{4}{3}\end{cases}}\)(tm)
Vậy ...
|7 - 2x| + 7 = 2x
=> |7 - 2x| = 2x - 7
ĐKXĐ: 2x - 7 \(\ge\)0 => 2x \(\ge\) 7 => x \(\ge\) 7/2
Ta có: |7 - 2x| = 2x - 7
=> \(\orbr{\begin{cases}7-2x=2x-7\\7-2x=7-2x\end{cases}}\)
=> 7 + 7 = 2x + 2x
hoặc x tùy ý (TMĐK)
=> 4x = 14 => x = 7/2
hoặc x tùy ý (Tm ĐK)
Vậy ...
a)\(\frac{1}{5}x-\frac{1}{3}=\frac{2}{4}\left(x+2\right)\)
<=>\(\frac{1}{5}x-\frac{1}{3}=\frac{2}{4}x+1\)
<=>\(-\frac{3}{10}x=\frac{4}{3}\)
<=>\(x=-\frac{40}{9}\)
b)\(\frac{5}{4}\left(x-3\right)=4+\frac{3}{2}x\)
<=>\(\frac{5}{4}x-\frac{15}{4}=4+\frac{3}{2}x\)
<=>\(-\frac{1}{4}x=\frac{31}{4}\)
<=>\(x=-31\)
c)\(\frac{5}{4}\left(x-3\right)=\frac{3}{2}\left(x+4\right)\)
<=>\(\frac{5}{4}x-\frac{15}{4}=\frac{3}{2}x+6\)
<=>\(-\frac{1}{4}x=\frac{9}{4}\)
<=>x=-9
Bài làm:
a) \(\left|\frac{1}{2}x-\frac{5}{2}\right|-1=-\frac{1}{2}\)
\(\Leftrightarrow\left|\frac{1}{2}x-\frac{5}{2}\right|=\frac{1}{2}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x-\frac{5}{2}=\frac{1}{2}\\\frac{1}{2}x-\frac{5}{2}=-\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=3\\\frac{1}{2}x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)
+ Nếu x = 6
\(\left|12-\frac{1}{3}y\right|=\frac{5}{6}\)
\(\Leftrightarrow\orbr{\begin{cases}12-\frac{1}{3}y=\frac{5}{6}\\12-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{67}{6}\\\frac{1}{3}y=\frac{77}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{67}{2}\\y=\frac{77}{2}\end{cases}}\)
+ Nếu x = 4
\(\left|8-\frac{1}{3}y\right|=\frac{5}{6}\)
\(\Leftrightarrow\orbr{\begin{cases}8-\frac{1}{3}y=\frac{5}{6}\\8-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{43}{6}\\\frac{1}{3}y=\frac{53}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{43}{2}\\y=\frac{53}{2}\end{cases}}\)
Vậy ta có 4 cặp số (x;y) thỏa mãn: \(\left(6;\frac{67}{2}\right);\left(6;\frac{77}{2}\right);\left(4;\frac{43}{2}\right);\left(4;\frac{53}{2}\right)\)
b) \(\frac{3}{2}x-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{5}{3}\)
\(\Leftrightarrow\frac{3}{2}x-\frac{1}{2}x+\frac{1}{3}=\frac{5}{3}\)
\(\Leftrightarrow x=\frac{4}{3}\)
Thay vào ta được:
\(\frac{2.\frac{4}{3}+y}{\frac{4}{3}-2y}=\frac{5}{4}\)
\(\Leftrightarrow\frac{32}{3}+4y=\frac{20}{3}-10y\)
\(\Leftrightarrow14y=-4\)
\(\Rightarrow y=-\frac{2}{7}\)
Vậy ta có 1 cặp số (x;y) thỏa mãn: \(\left(\frac{4}{3};-\frac{2}{7}\right)\)
a) \(\frac{2x-3}{4-x}=\frac{4-x}{2x-3}\)
\(\left(2x-3\right)\left(2x-3\right)=\left(4-x\right)\left(4-x\right)\)
\(\left(2x-3\right)^2=\left(4-x\right)^2\)
\(4x^2-12x+9=16-8x+x^2\)
\(4x^2-12x+9-16+8x-x^2=0\)
\(3x^2-4x-7=0\)
\(3x^2+3x-7x-7=0\)
\(3x\left(x+1\right)-7\left(x+1\right)=0\)
\(\left(x+1\right)\left(3x-7\right)=0\)
\(\hept{\begin{cases}x+1=0\\3x-7=0\end{cases}}\)
\(\hept{\begin{cases}x=-1\\x=\frac{7}{3}\end{cases}}\)
b: =>2x=16/2=8
=>x=4
a: Sửa đề: (3/2)^2x-1=(3/2)^5x-4
=>2x-1=5x-4
=>-3x=-3
=>x=1
1.
\(\left(2x+3^{ }\right)^4=16\)
⇔\(\left(2x+3\right)^4=2^4\)
⇔\(2x+3=2\)
⇔\(2x=5\)
⇔\(x=\frac{5}{2}\)
Vậy....