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3(x + 6) = 2(x - 5)
=> 3x + 18 = 2x - 10
=> 3x - 2x = -10 - 18
=> x = -28
vay_
c, |3x - 7| = 6
=> 3x - 7 = 6 hoac 3x - 7 = -6
=> 3x = 13 hoac 3x = 1
=> x = 13/3 hoac x = 1/3
vay_
3(x + 6) = 2(x - 5)
=> 3x + 18 = 2x - 10
=> 3x - 2x = -10 - 18
=> x = -28
Vậy x = -28
c, |3x - 7| = 6
=> 3x - 7 = 6 hoac 3x - 7 = - 6
=> 3x = 13 hoac 3x = 1
=> x = \(\frac{13}{3}\) hoac x = \(\frac{1}{3}\)
Vậy \(x\in\left\{\frac{3}{13};\frac{1}{3}\right\}\)
a) \(x-2=-6\)
\(x=-6+2\)
\(x=-4\)
b) \(15-\left(x-7\right)=-21\)
\(x-7=36\)
\(x=43\)
c) \(4.\left(3x-4\right)-2=18\)
\(4\left(3x-4\right)=20\)
\(3x-4=5\)
\(3x=9\)
\(x=3\)
d) \(\left(3x-6\right)+3=32\)
\(3x-6=29\)
\(3x=29+6\)
\(3x=35\)
\(x=\frac{35}{3}\)
e) \(\left(3x-6\right).3=32\)
\(3x-6=\frac{32}{3}\)
\(3x=\frac{32}{3}+6\)
\(3x=\frac{50}{3}\)
\(x=\frac{50}{9}\)
f) \(\left(3x-6\right):3=32\)
\(3x-6=96\)
\(3x=102\)
\(x=34\)
g) \(\left(3x-6\right)-3=32\)
\(3x-6=35\)
\(3x=41\)
\(x=\frac{41}{3}\)
h) \(\left(3x-2^4\right).7^3=2.7^4\)
\(\left(3x-2^4\right)=2.7=14\)
\(\left(3x-16\right)=14\)
\(3x=14+16=30\)
\(x=10\)
i) \(\left|x\right|=\left|-7\right|\)
\(\left|x\right|=7\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)
k) \(\left|x+1\right|=2\)
\(\Rightarrow\orbr{\begin{cases}x+1=2\\x+1=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)
l) \(\left|x-2\right|=3\)
\(\Rightarrow\orbr{\begin{cases}x-2=3\\x-2=-3\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}}\)
m) \(x+\left|-2\right|=0\)
\(x+2=0\)
\(x=-2\)
o) \(72-3\left|x+1\right|=9\)
\(3\left|x-1\right|=63\)
\(\left|x-1\right|=21\)
\(\Rightarrow\orbr{\begin{cases}x-1=21\\x-1=-21\end{cases}\Rightarrow\orbr{\begin{cases}x=22\\x=-20\end{cases}}}\)
p) Ta có: \(\left|x-1\right|=3\)
\(\Rightarrow\orbr{\begin{cases}x-1=3\\x-1=-3\end{cases}}\)
mà \(x+1< 0\)
\(\Rightarrow x-1=-3\)
\(\Rightarrow x=-2\)
q) \(\left(x-2\right)\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}}\)
hok tốt!!
a) Vì |x+1|=x+1
\(\Rightarrow x+1\ge0\)
\(\Rightarrow x\ge-1\)
KL x=-1
b) Vì |x+2|=-(x+2)
\(\Rightarrow x+2\le0\)
\(\Rightarrow x\le-2\)
c) Ta thấy \(\left|x\right|\ge0\forall x\)(1)
Để \(\left|x\right|>-3\)(2)
từ (1)và (2)
\(\Rightarrow x\in Z\)
KL \(x\in Z\)
d) Ta thấy \(\left|x\right|\ge0\forall x\)(1)
Để \(\left|x\right|< 7\)(2)
từ (1)và (2)
\(\Rightarrow-7< x< 7\)
KL -7<x<7
e) để \(\left(x-1\right)\times\left(x^2+1\right)=0\)
\(\Rightarrow\left(x-1\right)\)và \(\left(x^2+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x^2+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}}\)
KL \(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)thì (x-1)x(x2+1)=0
\(3\left(x+6\right)=2\left(x-5\right)\)
\(\Rightarrow3.x+18=2x-10\)
\(\Rightarrow3x-2x=-10-18\)
\(\Rightarrow x=-28\)
\(\left|3.x-7\right|=6\)
\(\Leftrightarrow\orbr{\begin{cases}3x-7=6\\3x-7=-6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3.x=13\\3x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{13}{3}\\x=\frac{1}{3}\end{cases}}\)
a)x.(x+1)<0 suy ra 2 số sẽ khác dấu.Ta xét 2 TH
TH1
x<0
x+1>0suy ra x>-1(loai)
TH2
x>0
x+1<0suy ra x<-1 mà x thuộc Z suy ra -1>x
Vậy x thuộc{-2;-3;...}