a, 3x + 6x - 5 = 17x
9x - 5 = 17x
9x - 17x = 5
- 8x = 5
x = -5/8
b, 8(4x + 2 ) = 20x + 11x
32x + 16 = 31x
32x - 31x = -16
x = -16
c, \(\sqrt{x}^2\) - 2x + 1 = 0
\(\left(\sqrt{x}\right)^2\) - 2x + 1 = 0
\(\left(\sqrt{x}+1\right)^2\) = 0
\(\sqrt{x+1}\) = 0
x + 1 = 0
x = -1
\(a;3x+6x-5=17x\)
\(\Leftrightarrow3x+6x-17x=5\)
\(\Leftrightarrow-8x=5\)
\(\Leftrightarrow x=\frac{-5}{8}\)
\(b;8\left(4x+2\right)=20x+11x\)
\(\Leftrightarrow32x+16=32x\)
\(\Leftrightarrow32x-32x=16\left(L\right)\)
\(c;\sqrt{x}^2-2x+1=0\)
\(\Leftrightarrow x-2x+1=0\)
\(\Leftrightarrow-x=-1\)
\(\Leftrightarrow x=1\)
TL:
a)\(3x+6x-5=17x\)
\(9x-5=17x\)
\(9x-17x=5\)
\(-8x=5\)
\(x=\frac{-5}{8}\)
Vậy.......
b) \(8\left(4x+2\right)=31x\)
\(32x+16=31x\)
\(32x-31x=-16\)
\(x=-16\)
Vậy...........
c) \(\left(\sqrt{x}\right)^2-2x+1=0\)
\(\left(\sqrt{x}-1\right)^2=0\)
\(\sqrt{x}-1=0\)
\(\sqrt{x}=1\)
\(\Rightarrow x=1\)
Vậy x=1
hc tốt
a, 3x + 6x - 5 = 17x
=> 3x + 6x - 17x = 5
=> -8x = 5
=> x = -5/8
vậy_
b, 8(4x + 2) = 20x + 11x
=> 32x + 16 = 31x
=> 32x - 31x = -16
=> x = -16
vậy_
c, \(\sqrt{x}^2-2x+1=0\)
=> x - 2x + 1 = 0
=> (x - 1)^2 = 0
=> x - 1 = 0
=> x = 1
vậy_
