tìm x biết rằng \(|x-1|+|x-3|=2x-1\)
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a: =>41-(2x-5)=720:40=18
=>2x-5=23
=>2x=28
=>x=14
b: =>100x+5050=5750
=>100x=700
=>x=7
a: =>41-(2x-5)=720:40=18
=>2x-5=23
=>2x=28
=>x=14
b: =>100x+5050=5750
=>100x=700
=>x=7
a, ( 2x - 3 )2- (2x + 1)2 = -3
4x2-12x+9-4x2+4x-1=-3
-8x-1=-3
-8x=-2
x=\(\frac{1}{4}\)
b, (5x - 1) 2 - (5x + 4)(5x - 4) = 7
25x2-10x+1-25x2+16=7
-10x+17=7
-10x=-10
x=1
c, ( x- 5)2 + (x-3)(x+3) - 2(x + 1)2=0
x2-10x+25+x2-9-2x2-4x-2=0
-14x+14=0
-14(x-1)=0
=>x-1=0
x=1
a) \(\left(2x-3\right)^2-\left(2x+1\right)^2=-3\)
\(\Leftrightarrow4x^2-12x+9-4x^2-4x-1=-3\)
\(\Leftrightarrow-16x+8=-3\)
\(\Leftrightarrow-16x=-11\)
\(\Leftrightarrow x=\frac{11}{16}\)
b)\(\left(5x-1\right)^2-\left(5x+4\right)\left(5x-4\right)=7\)
\(\Leftrightarrow25x^2-10x+1-25x^2+16=7\)
\(\Leftrightarrow-10x+17=7\)
\(\Leftrightarrow-10x=-10\)
\(\Leftrightarrow x=1\)
c)\(\left(x-5\right)^2+\left(x-3\right)\left(x+3\right)-2\left(x+1\right)^2=0\)
\(\Leftrightarrow x^2-10x+25+x^2-9-2\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow2x^2-10x-16-2x^2-4x-2=0\)
\(\Leftrightarrow-14x-18=0\)
\(\Leftrightarrow-14x=18\)
\(\Leftrightarrow x=-\frac{9}{7}\)
#H
(5x4-3x3):2x3=1/2
<=>5x4:2x3-3x3:2x3=1/2
<=>2,5x-1,5x=1/2
<=>x=1/2
Vậy x=1/2
\(\Leftrightarrow\dfrac{5}{2}x-\dfrac{3}{2}=\dfrac{1}{2}\Leftrightarrow\dfrac{5}{2}x=2\Leftrightarrow x=\dfrac{4}{5}\)
b
\(\left|6+x\right|\ge0;\left(3+y\right)^2\ge0\Rightarrow\left|6+x\right|+\left(3+y\right)^2\ge0\)
Suy ra \(\left|6+x\right|+\left(3+y\right)^2=0\)\(\Leftrightarrow\hept{\begin{cases}6+x=0\\3+y=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-6\\y=-3\end{cases}}\)
a
Ta có:\(\left|3x-12\right|=3x-12\Leftrightarrow3x-12\ge0\Leftrightarrow3x\ge12\Leftrightarrow x\ge4\)
\(\left|3x-12\right|=12-3x\Leftrightarrow3x-12< 0\Leftrightarrow3x< 12\Leftrightarrow x< 4\)
Với \(x\ge4\) ta có:
\(3x-12+4x=2x-2\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\left(KTMĐK\right)\)
Với \(x< 4\) ta có:
\(12-3x+4x=2x-2\)
\(\Rightarrow10=x\left(KTMĐK\right)\)
vi 2x-1 lẻ nen |x-1|hoac |x-3| lẻ
+)Neu |x-1|lẻ thì |x-3| lẻ(loai vi ca hai cung le)
+)neu |x-3| lẻ thì|x-1|lẻ(loai vi ca hai cung le)
vay ko co so nao tm x−1|+|x−3|=2x−1