Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
c) TH1 : \(\left|2x+1\right|+\left|3x-4\right|=\left(2x+1\right)+\left(3x-4\right)=5x-3=5\)
\(\Rightarrow x=\frac{8}{5}\)
TH2 : \(\left|2x+1\right|+\left|3x-4\right|=\left(2x+1\right)+\left(-3x+4\right)=-x+5=5\)
\(\Rightarrow x=0\)
d) \(\Rightarrow\left|2x+\frac{4}{5}\right|-\left|x-\frac{3}{2}\right|=0\)
Tương tự xét 2 trường hợp như câu c. Ta sẽ tìm được x
c ) \(\left|2x+1\right|+\left|3x-4\right|=5\)
\(\Rightarrow\left[\begin{array}{nghiempt}\left|2x+1\right|=5\\\left|3x-4\right|=5\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x+1=5\\2x+1=-5\\3x-4=5\\3x-4=-5\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}2\\-3\\3\\-\frac{1}{3}\end{array}\right.\)
a) \(\frac{x-3}{x+5}=\frac{5}{7}\)
\(\Rightarrow\left(x-3\right).7=\left(x+5\right).5\)
\(\Rightarrow7x-21=5x+25\)
\(\Rightarrow7x-5x=21+25\)
\(\Rightarrow2x=46\)
\(\Rightarrow x=23\)
Vậy \(x=23\)
b) \(\frac{7}{x-1}=\frac{x+1}{9}\)
\(\Rightarrow\left(x-1\right).\left(x+1\right)=7.9\)
\(\Rightarrow\left(x-1\right)x-\left(x+1\right)=7.9\)
\(\Rightarrow x^2-x-x-1=63\)
\(\Rightarrow x^2-1=63\)
\(\Rightarrow x^2=64\)
\(\Rightarrow x=8\) hoặc \(x=-8\)
Vậy \(x=8\) hoặc \(x=-8\)
c) \(\frac{x+4}{20}=\frac{5}{x+4}\)
\(\Rightarrow\left(x+4\right)^2=100\)
\(\Rightarrow x+4=\pm10\)
+) \(x+4=10\Rightarrow x=6\)
+) \(x+4=-10\Rightarrow x=-16\)
Vậy \(x\in\left\{6;-16\right\}\)
a) \(\Rightarrow\left[\begin{array}{nghiempt}x+2\frac{1}{3}=5\frac{2}{3}\\x+2\frac{1}{3}=-5\frac{2}{3}\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{10}{3}\\x=-8\end{array}\right.\)
b) \(\Rightarrow\left[\begin{array}{nghiempt}\left(3x-2\right)+4x=9\\-\left(3x-2\right)+4x=9\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}7x-2=9\\x+2=9\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}7x=11\\x=7\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{11}{7}\\x=7\end{array}\right.\)
1) \(\left|x\right|< 4\Leftrightarrow-4< x< 4\)
2) \(\left|x+21\right|>7\Leftrightarrow\orbr{\begin{cases}x+21>7\\x+21< -7\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>-14\\x< -28\end{cases}}\)
3) \(\left|x-1\right|< 3\Leftrightarrow-3< x-1< 3\Leftrightarrow-2< x< 4\)
4) \(\left|x+1\right|>2\Leftrightarrow\orbr{\begin{cases}x+1>2\\x+1< -2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>1\\x< -3\end{cases}}\)
\(\left|x+\frac{1}{2}\right|+\left|3-y\right|=0\)
Vì \(\hept{\begin{cases}\left|x+\frac{1}{2}\right|\ge0\\\left|3-y\right|\ge0\end{cases}}\Rightarrow\)\(\left|x+\frac{1}{2}\right|+\left|3-y\right|\ge0\)
Dấu "="\(\Leftrightarrow\hept{\begin{cases}\left|x+\frac{1}{2}\right|=0\\\left|3-y\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-1}{2}\\y=3\end{cases}}\)
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
\(\frac{x+1}{2}=\frac{2}{3}\)
\(\Rightarrow3.\left(x+1\right)=2.2\)
\(\Rightarrow3x+3=4\)
\(\Rightarrow3x=4-3\)
\(\Rightarrow3x=1\)
\(\Rightarrow x=\frac{1}{3}\)
\(b,\frac{x-2}{3}=\frac{x+1}{4}\)
\(\Rightarrow4.\left(x-2\right)=3.\left(x+1\right)\)
\(\Rightarrow4x-8=3x+3\)
\(\Rightarrow4x-3x=3+8\)
\(\Rightarrow x=11\)
\(c,\frac{x-3}{x+5}=\frac{5}{7}\)
\(\Rightarrow7.\left(x-3\right)=5.\left(x+5\right)\)
\(\Rightarrow7x-21=5x+25\)
\(\Rightarrow7x-5x=25+21\)
\(\Rightarrow2x=46\)
\(\Rightarrow x=23\)
\(d,\frac{x-1}{x+2}=\frac{x-2}{x+3}\)
\(\Rightarrow\left(x-1\right)\left(x+3\right)=\left(x+2\right)\left(x-2\right)\)
\(\Rightarrow x^2+2x-3=x^2-4\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=-\frac{1}{2}\)
vì x+2\(\ne\)x+4
\(\Rightarrow\)\(\left[\begin{array}{nghiempt}x-1=0\\x-1=1\\x-1=-1\end{array}\right.\) \(\Leftrightarrow\) \(\left[\begin{array}{nghiempt}x=1\\x=2\\x=0\end{array}\right.\)
\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Leftrightarrow x+2=x+4\)
\(\Leftrightarrow0x=2\) (không thỏa mãn)
Vậy không có giá trị nào của x thỏa mãn phương trình