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\(\frac{x}{1}\)= \(\frac{3}{x+2}\)
=> x (x+2) = 3
=> x+2 = 3
=> x = 1
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Với mọi \(x\in R\)ta có:
\(\left|x\right|+\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|\ge0\Leftrightarrow6x\ge0\Leftrightarrow x\ge0\)
Với \(x\ge0\)thì: \(\left|x\right|=x;\left|x+1\right|=x+1;\left|x+2\right|=x+2;\left|x+3\right|=x+3;\left|x+4\right|=x+4\)
\(pt\Leftrightarrow5x+10=6x\Leftrightarrow x=10\)
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3x-1=0 suy ra x=0,(3)
chia khoảng để giải:tự làm
nếu x<0,33 btvt
3x+1=x=1
3x-x=1-1
2x=0
x= rỗng
Nếu x>0,33 btvt
3x-1=x+1
3x-x=1+1
2x=2
x=1(nghiêm)
| 3x-1|=(x+1)
=>| 3x-1|=± x+1
Th1:3x-1=x+1 Th2:3x-1=-x+1
=>3x-x=1+1 =>3x-1=-(x-1)
=>2x=2 =>3x-1=1-x
=>x=1 =>3x+x=1+1
=>4x=2
=>x=\(\frac{1}{2}\)
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(2x+1)+(3-x)=0
=>2x+1=-3+x
=>2x+1-x=-3
=>x+1=-3
=>x=-3-1=-4
Vậy x=-4
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Đk : x khác 2
Có : x^2-1/|x-2| = x
=> x^2-1 = x.|x-2|
+, Với x < 2 => x^2-1 = x.(2-x) = 2x-x^2
=> x^2-1-(2x-x^2)=0
=> x^2-1-2x+x^2=0
=> 2x^2-2x-1 = 0
=> x=1+\(\sqrt{3}\)/2 (tm) hoặc x=1-\(\sqrt{3}\)/2 (tm)
+, Với x > 2
=> x^2-1 = x.(x-2) = x^2-2x
=> x^2-1-(x^2-2x) = 0
=> x^2-1-x^2+2x=0
=> 2x-1=0
=> 2x=1
=> x=1/2 (loại )
Vậy .............
Tk mk nha
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Ta có : \(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)
\(\Rightarrow\frac{1+2y}{18}=\frac{1+6y}{6x}=\frac{1+2y+1+6y}{18+6x}=\frac{2+8y}{18+6x}=\frac{2\left(1+4y\right)}{2\left(9+3x\right)}=\frac{1+4y}{9+3x}\)
\(\Rightarrow\frac{1+4y}{24}=\frac{1+4y}{9+3x}\)
\(\Rightarrow9+3x=24\)
\(\Rightarrow3x=24-9\)
\(\Rightarrow3x=15\)
\(\Rightarrow x=15:3\)
\(\Rightarrow x=5\)
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a) [ 2x ] = -1
=> 2x = 1
x = 1 : 2
x = 0,5
b ) [ x + 0.4 ] = 3
=> [ x + 0 ] = 3
=> x + 0 = 3 hoặc x + 0 = -3
+) x + 0 = 3
x = 3 - 0
x = 3
+) x + 0 = -3
x = ( -3 ) -0
x = -3
=> x thuộc { -3 ; 3 }
a) x = (-1) : 2 = -0,5
b) x = 3 - 0,4 = 2,6
c) x = (3 + 5) : 2/3 = 12
x.34=1
x=1/34
K mình nhé bạn
x= 1/34
one piece