Rút gọn:
a)A=2.l3x-1l-l5-xl
b)B=2.I2x-1I-3.I2x+3I
Tìm x
a, 2 Ix-3I-5=3
b, 2 I2x+3I + I2x+3I=6
c, 3 Ix+1I^2 + Ix+1I^2=16
a ) 2|x - 3| - 5 = 3 <=> 2|x - 3| = 8 <=> |x - 3| = 4 => x - 3 = ± 4
TH1 : x - 3 = 4 => x = 7
TH2 : x - 3 = - 4 => x = - 1
Vậy x = { - 1; 7 }
b ) 2|2x + 3| + |2x + 3| = 6 <=> 3|2x + 3| = 6 => |2x + 3| = 2 => 2x + 3 = ± 2
=> x = { - 5/2 ; - 1/2 }
c ) 3|x + 1|2 + |x + 1|2 = 16
4|x + 1|2 = 16
=> |x + 1|2 = 4 = 22 ( ko xét TH |x + 1| = - 2 vì |x + 1| ≥ 0 )
=> |x + 1| = 2 => x + 1 = ± 2 => x = { - 3; 1 }
tìm x
a,I2x-1I=x-2
b,I2x+1I=3x
c,I2x+1I=4
d,Ix-3I=1/2
e,Ix+1I+Ix-2I=0
f,I2x-1I+Ix+2I=0
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tìm giá trị x thỏa mãn I2x+3I + I2x-1I = 8/3(x+1)^2+2
tìm xa,I2x 1I x 2b,I2x 1I 3xc,I2x 1I 4d,Ix 3I 1 2e,Ix 1I Ix 2I 0f,I2x 1I Ix 2I 0
Tìm x:
a,I Ix-1I-1I=2
b,I I3x-1I-5I=2
c,I I2x-3I-x+1I=42-8
d,I(x+1)Ix-3I=x-3
a) \(\left|\left|x-1\right|-1\right|=2\Rightarrow\orbr{\begin{cases}\left|x-1\right|-1=2\\\left|x-1\right|-1=-2\end{cases}}\Rightarrow\orbr{\begin{cases}\left|x-1\right|=3\\\left|x-1\right|=-1\left(l\right)\end{cases}}\)
TH1: x - 1 = 3
x = 4
TH2: x - 1 = - 3
x = - 2
b) Tương tự câu a.
c) \(\left|\left|2x-3\right|-x+1\right|=42-8\)
\(\left|\left|2x-3\right|-x+1\right|=34\)
TH1: \(\left|2x-3\right|-x+1=34\)
\(\left|2x-3\right|-x=33\)
Với \(x\ge\frac{3}{2}\), ta có \(2x-3-x=33\Rightarrow x=36\) (tm)
Với \(x< \frac{3}{2}\), ta có \(3-2x-x+1=34\Rightarrow-3x=30\Rightarrow x=-10\left(tm\right)\)
TH2: \(\left|2x-3\right|-x+1=-34\)
\(\left|2x-3\right|-x=-35\)
Với \(x\ge\frac{3}{2}\), ta có \(2x-3-x=-35\Rightarrow x=-32\) (l)
Với \(x< \frac{3}{2}\), ta có \(3-2x-x+1=-34\Rightarrow-3x=38\Rightarrow x=\frac{38}{3}\left(l\right)\)
d) Tương tự câu c.
1)Tìm x
I2x-3I=I2x+1I
Tìm x,y biết
a,I2x-1I = Ix+3I
b,IxI+I2y-6I=0
c,Ix-2I>3
d,Ix-7I=x-7
e,I2x+6I-2x-6
Tìm x biết: Ix-3I = I2x+1I
Tìm x biết: Ix-3I = I2x+1I
| x - 3 | = | 2x + 1 |
\(\Rightarrow\orbr{\begin{cases}x-3=2x+1\\x-3=-\left(2x+1\right)\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-2x=1+3\\x+2x=-1+3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}-x=4\\3x=2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-4\\x=\frac{2}{3}\end{cases}}\)
I x - 3 I = I 2x + 1 I (1)
Xét \(x< \frac{-1}{2}\)ta co \(\left(1\right)\Leftrightarrow3-x=-2x-1\Leftrightarrow x=-4\)(thỏa mãn)
Xét \(\frac{-1}{2}\le x< 3\)ta có \(\left(1\right)\Leftrightarrow3-x=2x+1\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\)(thỏa mãn)
Xét \(x\ge3\)ta có \(\left(1\right)\Leftrightarrow x-3=2x+1\Leftrightarrow x=-4\)(loại vì \(x\ge3\))