Mk chỉ làm bt 1 thôi nha vì máy tính mk có vấn đề
Câu 1:
a)|x-5|=2x+3
TH1:x-5=2x+3
x-5-2x-3=0
-8-x=0
x=-8
TH2:-(x-5)=2x+3
-x+5=2x+3
-x+5-2x-3=0
2-3x=0
3x=2
x=\(\frac{2}{3}\)
Vậy x=-8;\(\frac{2}{3}\)
b)3-|3x+1|=-6
|3x+1|=3-(-6)
|3x+1|=9
\(\Rightarrow\left[\begin{array}{nghiempt}3x+1=9\\3x+1=-9\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}3x=8\\3x=-10\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{8}{3}\\x=-\frac{10}{3}\end{array}\right.\)
Vậy \(x=\frac{8}{3};-\frac{10}{3}\)
Bài 1:
a)| x-5|=2x+3
=>x-5=-(2x+3) hoặc 2x+3
Xét x-5=2x+3
=>x=-8 (loại)
Xét x-5=-(2x+3)
=>x=\(\frac{2}{3}\)(thỏa mãn)
b) 3-|3x+1|=-6
=>|3x+1|=9
=>3x+1=9 hoặc -9
Xét 3x+1=9
=>3x=8
=>x=\(\frac{8}{3}\)
Xét 3x+1=-9
=>3x=-10
=>x=\(-\frac{10}{3}\)
Bài 1:
a) | x - 5 | = 2x + 3
\(\Rightarrow x-5=\pm\left(2x+3\right)\)
+) \(x-5=2x+3\)
\(\Rightarrow-5-3=2x-x\)
\(\Rightarrow-8=x\)
+) \(x-5=-\left(2x+3\right)\)
\(\Rightarrow x-5=-2x-3\)
\(\Rightarrow-5+3=-2x-x\)
\(\Rightarrow-2=-3x\)
\(\Rightarrow x=\frac{2}{3}\)
Vậy \(x=-8\) hoặc \(x=\frac{2}{3}\)
b) 3 - |3x + 1| = -6
\(\Rightarrow\left|3x+1\right|=-9\)
\(\Rightarrow3x+1=\pm9\)
+) 3x + 1 = 9
\(\Rightarrow3x=8\)
\(\Rightarrow x=\frac{8}{3}\)
+) 3x + 1 = -9
\(\Rightarrow3x=-10\)
\(\Rightarrow x=\frac{-10}{3}\)
Vậy \(x=\frac{8}{3}\) hoặc \(x=\frac{-10}{3}\)
Bài 2:
a) |a+8|<1
=>a+8<1 hoặc -1
Xét a+8< 1 <=> a<-7
Xét a+8< -1 <=> a<-9
=>-7<a<-9 <=> a=-8
b) |3a+4|<3
=>3a+4<3 hoặc -3
Xét 3a+4<3
=>3a<-1
=>a<\(-\frac{1}{3}\)
Xét 3a+4<-3
=>3a<-7
=>a<\(-\frac{7}{3}\)
\(\Rightarrow-\frac{7}{3}< x< -\frac{1}{3}\) (vì a nguyên)
=>a={-2;-1}
1. Tìm x€Q, biết
a) | x-5|=2x+3
\(=>\left[\begin{array}{nghiempt}x-5=2x+3\\x-5=-2x-3\end{array}\right.\)
\(=>\left[\begin{array}{nghiempt}x=-8\\x=\frac{2}{3}\end{array}\right.\)
b) 3-|3x+1|=-6
\(=>\left[\begin{array}{nghiempt}3x+1=9\\3x+1=-9\end{array}\right.\)
\(=>\left[\begin{array}{nghiempt}x=\frac{8}{3}\\x=\frac{10}{3}\end{array}\right.\)
2. Tìm a€Z, biết
a) |a+8|<1
=> Ia+8I = 0 => a+8 = 0 => a = -8
b) |3a+4|<3
\(=>\left[\begin{array}{nghiempt}\left|3a+4\right|=0\\\left|3a+4\right|=1\\\left|3a+4\right|=2\end{array}\right.\)
\(=>\left[\begin{array}{nghiempt}a=-\frac{4}{3}\\\left[\begin{array}{nghiempt}3a+4=1\\3a+4=-1\end{array}\right.\\\left[\begin{array}{nghiempt}3a+a=2\\3a+4=-2\end{array}\right.\end{array}\right.\)
\(=>a=-\frac{4}{3};a=-1;a=-\frac{5}{3};a=-\frac{2}{3};a=-2\)
theo đầu bài a ϵ Z => a= -1; a=-2 thỏa mãn
Vậy a= -1
hoặc a=-2
1. a) \(\left|x-5\right|=2x+3\)
* Nếu \(x-5\ge0\) \(\Rightarrow x\ge5\) .
(1) \(\Leftrightarrow x-5=2x+3\)
\(x-2x=3+5\)
\(-x=8\)
\(\Rightarrow x=-8\) ( chọn).
* Nếu x - 5 < 0 \(\Rightarrow x< 5\)
( 1) \(\Leftrightarrow x-5=-2x-3\)
x + 2x = - 3 +5
3x = 2
\(\Rightarrow x=\frac{2}{3}\) ( chọn).
Vậy \(\left[\begin{array}{nghiempt}x=8\\x=\frac{2}{3}\end{array}\right.\)
