Bài 6:
a) Ta có: \(4x-10=0\)
\(\Leftrightarrow4x=10\)
\(\Leftrightarrow x=\dfrac{5}{2}\)
Vậy: \(S=\left\{\dfrac{5}{2}\right\}\)
b) Ta có: \(7-3x=9-x\)
\(\Leftrightarrow-3x+7-9+x=0\)
\(\Leftrightarrow-2x-2=0\)
\(\Leftrightarrow-2x=2\)
\(\Leftrightarrow x=-1\)
Vậy: S={-1}
c) Ta có: \(2x-\left(3-5x\right)=4\left(x+3\right)\)
\(\Leftrightarrow2x-3+5x=4x+12\)
\(\Leftrightarrow7x-3-4x-12=0\)
\(\Leftrightarrow3x-15=0\)
\(\Leftrightarrow3x=15\)
\(\Leftrightarrow x=5\)
Vậy: S={5}
d) Ta có: \(5-\left(6-x\right)=4\left(3-2x\right)\)
\(\Leftrightarrow5-6+x=12-8x\)
\(\Leftrightarrow x+11-12+8x=0\)
\(\Leftrightarrow9x-1=0\)
\(\Leftrightarrow9x=1\)
\(\Leftrightarrow x=\dfrac{1}{9}\)
Vậy: \(S=\left\{\dfrac{1}{9}\right\}\)
e) Ta có: \(4\left(x+3\right)=-7x+17\)
\(\Leftrightarrow4x+12+7x-17=0\)
\(\Leftrightarrow11x-5=0\)
\(\Leftrightarrow11x=5\)
\(\Leftrightarrow x=\dfrac{5}{11}\)
Vậy: \(S=\left\{\dfrac{5}{11}\right\}\)
B1:
a) \(3\left(2-x\right)+1=4-2x\\ \Leftrightarrow7-3x=4-2x\\ \Leftrightarrow-3x+2x=4-7\\ \Leftrightarrow-x=-3\\ \Leftrightarrow x=3\)
vậy...
b)\(3x-5=5x-1\\ \Leftrightarrow3x-5x=-1+5\\ \Leftrightarrow-2x=4\\ \Leftrightarrow x=-2\)
Vậy...
c)\(7-3x=x-5\\ \Leftrightarrow-3x-x=-5-7\\ \Leftrightarrow-4x=-12\\ \Leftrightarrow x=3\)
vậy...
d)\(5x-\left(x-1\right)=7\\ \Leftrightarrow5x-x+1=7\\ \Leftrightarrow4x=6\\ \Leftrightarrow x=\dfrac{3}{2}\)
vậy...
bạn tự làm đi nhé mình làm cả đề xong không gửi đc =))