a) \(\left(x-1\right)^2=0\)
=> \(x-1=0\)
=> \(x=1\)
b) \(x\left(2x+1\right)=0\)
=> \(\orbr{\begin{cases}x=0\\2x+1=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=-\frac{1}{2}\end{cases}}\)
c) \(\left(-12\right)^2.x=56+10.13x\)
=> \(144x=56+130x\)
=> \(144x-130x=56\)
=> \(14x=56\)
=> \(x=56:14\)
=> \(x=4\)
d) \(x-\left(17-x\right)=x-7\)
=> \(x-17+x=x-7\)
=> \(2x-17=x-7\)
=> \(2x-x=-7+17\)
=> \(x=10\)
a) (x-1)2 = 0
=> x - 1 = 0
=> x = 1
vậy_
b) x(2x+1)=0
=> x = 0 hoặc 2x + 1 = 0
=> x = 0 hoặc 2x = -1
=> x = 0 hoặc x = -1/2
vậy_
c) (-12)^2.x=56+10.13.x
=> 144x = 56 + 130x
=> 144x - 130x = 56
=> 14x = 56
=> x = 4
vậy_
d) x-(17-x)=x-7
=> x - 17 + x = x - 7
=> x + x - x = -7 + 17
=>x = 10
vậy_
e) (x+4) ⋮ (x+1)
=> x + 1 + 3 ⋮ x + 1
=> 3 ⋮ x + 1
=> x + 1 thuộc Ư(3)
=> x + 1 thuộc {-1; 1; -3; 3}
=> x thuộc {-2; 0; -4; 2}
vậy_
g) (4x+3)⋮(x-2)
=> 4x - 4 + 7 ⋮ x - 2
=> 2(x - 2) + 7 ⋮ x - 2
=> 7 ⋮ x - 2
=> x - 2 thuộc Ư(7)
=> x - 2 thuộc {-1; 1; -7; 7}
=> x thuộc {1; 3; -5; 9}
vậy_
#)Giải :
a)\(\left(x-1\right)^2=0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
b)\(x\left(2x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\2x+1\end{cases}=0}\Rightarrow x=0\)
c)\(\left(-12\right)^2.x=56+10.13x\)
\(\Rightarrow144x=56+130x\)
\(\Rightarrow14x=56\)
\(\Rightarrow x=4\)
d)\(x-\left(17-x\right)=x-7\)
\(\Rightarrow x-17+x=x-7\)
\(\Rightarrow x+x-17=x-7\)
\(\Rightarrow x+x-x=-7+17\)
\(\Rightarrow x=10\)
\(a,\left(x-1\right)^2=0\)
\(x-1=0\)
\(x=1\)
\(b,x\left(2x+1\right)=0\)
\(x=0\)
hoặc
\(2x+1=0\)
\(2x=-1\)
\(x\)\(=-1:2\)
\(x=-\frac{1}{2}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{2}\end{cases}}\)
\(c,\left(-12\right)^2.x=56+10.13x\)
\(\Rightarrow144x=56+130x\)
\(\Rightarrow14x=56\)
\(\Rightarrow x=4\)
\(d,x-\left(17-x\right)=x-7\)
\(2x-17=x-7\)
\(2x-x=-7+17\)
\(x=10\)
e,\(\left(x+4\right)⋮\left(x+1\right)\)
\(x+1+3⋮x+1\)
\(3⋮x+1\)
\(\Rightarrow x+1\inƯ\left(3\right)=\left\{\mp1;\mp3\right\}\)
Ta có bảng
| x+1 | -1 | 1 | -3 | 3 |
| x | -2 | 0 | -4 | 2 |
g,\(\left(4x+3\right)⋮\left(x-2\right)\)
\(4x-4+7⋮x-2\)
\(2\left(x-2\right)+7⋮x-2\)
\(7⋮x-2\)
\(\Rightarrow x-2\inƯ\left(7\right)=\left\{\mp1;\mp7\right\}\)
Ta có bảng
| x-2 | -1 | 1 | -7 | 7 |
| x | 1 | 3 | -5 | 9 |
Lần xong đăng 1 lần 2->3 câu thui nha :)) mk giải mãi ,mới xong
