(x+1/2).|2x-3/4||=2x-3/4 tìm x giúp mik với T_T
(ai ko trả lời là "hạnh phúc")
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a) \(\left|x\right|-\frac{3}{4}=\frac{5}{3}\)
\(\left|x\right|=\frac{5}{3}+\frac{3}{4}\)
\(\left|x\right|=\frac{29}{12}\)
\(\orbr{\begin{cases}x=\frac{29}{12}\\x=-\frac{29}{12}\end{cases}}\)
Ngô Hải Nam ơi bn trả lời giúp mik ik
bài đó là bài 4^* tìm các số nguyên x để mỗi phân số sau đây là số nguyên
Vì \(\left|x+2\right|+\left|2x+3\right|+\left|3x+4\right|\ge0\)
=> \(7x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow\left|x+2\right|+\left|2x+3\right|+\left|3x+4\right|=x+2+2x+3+3x+4\)
\(\Rightarrow6x+7=7x\)
=> x=7
=>\(\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\)
+, x+\(\frac{1}{2}\)=0 +,\(\frac{2}{3}-2x=0\)
x=\(-\frac{1}{2}\) =>\(\frac{2}{3}=2x\)
=>\(x=\frac{1}{3}\)
Vậy........
a) \(\frac{3x+2}{-4x+5}=-\frac{4}{3}\left(ĐKXĐ:x\ne\frac{5}{4}\right)\)
\(\Rightarrow3\left(3x+2\right)=-4\left(-4x+5\right)\)
\(\Leftrightarrow9x+6=16x-20\)
\(\Leftrightarrow7x=26\)
\(\Leftrightarrow x=\frac{26}{7}\)
b) \(\frac{2\left|x\right|+5}{-4x+3}=-\frac{5}{4}\)(Thôi bài sau tự tìm đkxđ nhá)
\(\Rightarrow8\left|x\right|+20=20x-15\)
\(\Leftrightarrow8\left|x\right|-20x+35\)\(\left(1\right)\)
TH1: Nếu \(x\ge0\)thì \(\left(1\right)\Leftrightarrow8x-20x+35=0\Leftrightarrow x=\frac{35}{12}\left(tm\right)\)
TH2: Nếu \(x< 0\)thì \(\left(1\right)\Leftrightarrow-8x-20x+35=0\Leftrightarrow x=\frac{35}{28}\left(ktm\right)\)
Vậy x=35/12
c)\(\frac{2x+1}{5}=\frac{3}{2x-1}\)
\(\Rightarrow4x^2-1=15\)
\(\Leftrightarrow4x^2=16\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
d)\(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)=\left(2x+1\right)\left(0,5x+2\right)\)
\(\Leftrightarrow x^2+4x+3=x^2+4,5x+2\)
\(\Leftrightarrow0,5x=1\)
\(\Leftrightarrow x=2\)
e) \(\frac{\left|6x+1\right|}{4}=\frac{2}{4}\)
\(\Leftrightarrow\left|6x+1\right|=2\)
\(\Leftrightarrow\orbr{\begin{cases}6x+1=2\\6x+1=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{6}\\x=-\frac{1}{2}\end{cases}}}\)
g)\(\frac{\left|3x-5\right|}{3}=\frac{\left|x\right|}{2}\)
\(\Leftrightarrow\frac{\left|3x-5\right|}{\left|x\right|}=\frac{3}{4}\)
\(\Leftrightarrow\left|\frac{3x-5}{x}\right|=\frac{3}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3x-5}{x}=\frac{3}{4}\\\frac{3x-5}{x}=-\frac{3}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{20}{9}\\x=\frac{4}{3}\end{cases}}}\)
Mỏi tay quá, xin tý cho sảng khoái nào!!
\(\)
a) \(\left(8\frac{4}{5}x-50\right):0,4=5\)
\(\Rightarrow\left(\frac{44}{5}x-50\right)=5.0,4\)
\(\Rightarrow\frac{44}{5}x-50=2\)
\(\Rightarrow\frac{44}{5}x=2+50\)
\(\Rightarrow\frac{44}{5}x=52\)
\(\Rightarrow x=\frac{65}{11}\)
b) \(\left(\frac{5x}{3}-3\right):15=\frac{3}{10}\)
\(\Rightarrow\frac{5x}{3}-3=\frac{3}{10}.15\)
\(\Rightarrow\frac{5x}{3}-3=\frac{45}{10}\)
\(\Rightarrow\frac{5x}{3}=\frac{45}{10}+3\)
\(\Rightarrow\frac{5x}{3}=\frac{15}{2}\)
\(\Rightarrow5x.2=3.15\)
=> 5x.2 = 45
=> 5x = 45 : 2
=> 5x = 45/2
=> x = 9/2
\(a,\Leftrightarrow\left(5x+1\right)\left(x-4\right)-\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(5x+1-x\right)=0\\ \Leftrightarrow5x\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x^2-10x-2x^2-3x=26\\ \Leftrightarrow-13x=26\\ \Leftrightarrow x=-2\\ c,\Leftrightarrow x^3+1-x^3+3x=15\\ \Leftrightarrow3x=14\\ \Leftrightarrow x=\dfrac{14}{3}\)
\(d,\Leftrightarrow x^3-5x+2x^2-10+5x-2x^2-17=0\\ \Leftrightarrow x^3-27=0\\ \Leftrightarrow x^3=27\\ \Leftrightarrow x=3\)
\(\left(x+\dfrac{1}{2}\right)\cdot\left|2x-\dfrac{3}{4}\right|=2x-\dfrac{3}{4}\)(1)
TH1: \(x>=\dfrac{3}{8}\)
(1) sẽ trở thành \(\left(x+1\right)\left(2x-\dfrac{3}{4}\right)=\left(2x-\dfrac{3}{4}\right)\)
=>\(\left(x+1\right)\left(2x-\dfrac{3}{4}\right)-\left(2x-\dfrac{3}{4}\right)=0\)
=>\(\left(2x-\dfrac{3}{4}\right)\left(x+1-1\right)=0\)
=>\(x\left(2x-\dfrac{3}{4}\right)=0\)
=>\(\left[{}\begin{matrix}x=0\left(loại\right)\\x=\dfrac{3}{8}\left(nhận\right)\end{matrix}\right.\)
TH2: \(x< \dfrac{3}{8}\)
(1) sẽ trở thành:
\(\left(x+1\right)\left(\dfrac{3}{4}-2x\right)=2x-\dfrac{3}{4}\)
=>\(\left(2x-\dfrac{3}{4}\right)\left(-x-1\right)-\left(2x-\dfrac{3}{4}\right)=0\)
=>\(\left(2x-\dfrac{3}{4}\right)\left(-x-2\right)=0\)
=>\(\left[{}\begin{matrix}x=-2\left(nhận\right)\\x=\dfrac{3}{8}\left(loại\right)\end{matrix}\right.\)
HHilhkiu