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\(a.\)\(\left|2x-3\right|=x-1\) \(\left(Đk:x-1\ge0\Leftrightarrow x\ge1\right)\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=x-1\\2x-3=1-x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-x=3-1\\2x+x=1+3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\3x=4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{4}{3}\end{cases}}\)( T/m điều kiện )
\(b.\)\(\left|2x-1\right|=x+4\) \(\left(Đk:x+4\ge0\Leftrightarrow x\ge-4\right)\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=x+4\\2x-1=-x-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-x=1+4\\2x+x=1-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\3x=3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}\) (T/m điều kiện )
\(c.\)\(\left|x-3\right|=x-4\) \(\left(Đk:x-4\ge0\Leftrightarrow x\le4\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=x-4\\x-3=4-x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-x=3+4\\x+x=4+3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}0=7\\2x=7\end{cases}}\)
\(\Leftrightarrow x=\frac{7}{2}\)( T/m điều kiện )
\(d.\)\(\left|2x-8\right|+4x=10\)
\(\Leftrightarrow\left|2x-8\right|=10-4x\) \(\left(Đk:10-4x\ge0\Leftrightarrow4x\le10\Leftrightarrow x\le\frac{5}{2}\right)\)
\(\Leftrightarrow\orbr{\begin{cases}2x-8=10-4x\\2x-8=4x-10\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x+4x=10+8\\2x-4x=8-10\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}6x=18\\-2x=-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=18:6\\x=\left(-2\right):\left(-2\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=1\end{cases}}\)
Câu a, b đúng rồi :))
Câu c. Em sai điều kiện.
Câu d: Em sai đáp án : x = 3 với x =1 nha!
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\(|x^2+4|=4x\Rightarrow\orbr{\begin{cases}x^2+4=4x\Rightarrow x^2-4x+4=0\Rightarrow\left(x-2\right)^2=0\Rightarrow x-2=0\Rightarrow x=2\\x^2+4=-4x\Rightarrow x^2+4x+4=0\Rightarrow\left(x+2\right)^2=0\Rightarrow x+2=0\Rightarrow x=-2\end{cases}}\)
\(|2-4x|=2x-1\Rightarrow\orbr{\begin{cases}2-4x=2x-1\Rightarrow-4x-2x=-1-2\Rightarrow-6x=-3\Rightarrow x=\frac{1}{2}\\2-4x=-2x+1\Rightarrow-4x+2x=1-2\Rightarrow-2x=-1\Rightarrow x=\frac{1}{2}\end{cases}}\)
| x2 + 4 | = 4x
\(\Rightarrow\) x2 + 4 = \(\pm\)4x
TH1: x2 + 4 = 4x
\(\Rightarrow\)x2 +4 - 4x = 0
\(\Rightarrow\)( x -2 )2 = 0
\(\Rightarrow\)x - 2 = 0
\(\Rightarrow\) x= 2
| 2 - 4x | = 2x + 1
\(\Rightarrow\)2 - 4x = \(\pm\) 2x + 1
TH1 : Tự làm tiếp nha :))
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a) \(\left|2x-3\right|-\dfrac{5}{2}=\dfrac{1}{3}\)
\(\left|2x-3\right|=\dfrac{1}{3}+\dfrac{5}{2}=\dfrac{2}{6}+\dfrac{15}{6}\)
\(\left|2x-3\right|=\dfrac{17}{6}\)
\(+)2x-3=\dfrac{17}{6}\Rightarrow2x=\dfrac{35}{6}\Rightarrow x=\dfrac{35}{12}\)
\(+)2x-3=\dfrac{-17}{6}\Rightarrow2x=\dfrac{1}{6}\Rightarrow x=\dfrac{1}{12}\)
vậy...
\(\left|x-1\right|+3x=1\\ \Rightarrow\left|x-1\right|=1-3x\\ \Rightarrow\left\{{}\begin{matrix}x-1=1-3x\\x-1=-1+3x\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}4x=2\\-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=0\end{matrix}\right.\)
Dấu ngoặc vuông nhé
thánh bấm nhầm
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a, \({\mid x^2 + 4\mid}=4x\) (ĐK: x\(\geq\)0)
\(\implies \)\(x^2 +4= 4x\)
hoặc \(x^2+4=-4x\)
\(\implies\)\(x^2-4x+4=0\)
hoặc \(x^2+4x+4=0\)
\(\implies\)x=2 (t/m)
hoặc x=-2 (ko t/m)
Vậy x=2
b, \(\mid2-4x\mid=2x+1\)
(ĐK: \(x\geq-1/2\))
\(\implies\) 2 -4x =2x+1
hoặc 2 -4x = -2x-1
\(\implies\)x= 1/6 (t/m)
hoặc x= 3/2 (t/m)
Vậy x=1/6 hoặc x=3/2
c,\(\mid\mid x\mid-7\mid=x+5\) (đk: \(x\geq-5\) )
TH1: \(\mid x \mid -7= x+5\) \(\implies\)\(\mid x \mid =x+12 \) (đk:\(x\geq -12\) )
\(\implies\)x = x+12
hoặc -x =x+12
\(\implies\)vô nghiệm
hoặc x = -6 (ko t/m)
TH2: \(\mid x \mid -7= -x-5\) \(\implies\) \(\mid x \mid =-x+2\) (đk: \(x\leq2\) )
\(\implies\)x = -x+2
hoặc -x = -x+2
\(\implies\)x=1 (t/m)
hoặc vô nghiệm
Vậy x=1
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b) 2003 - | x - 2003 | = x
=> 2003 - x = | x - 2003 |
=> \(2003-x=\orbr{\begin{cases}x-2003\\2003-x\end{cases}}\)
\(\Rightarrow x=\orbr{\begin{cases}2003-x+2003\\2003-2003+x\end{cases}}\)
\(\Rightarrow x=\orbr{\begin{cases}4006-x\\0+x=x\end{cases}}\)
\(\Rightarrow x=4006-x\)
\(\Rightarrow4006=2x\Rightarrow x=4006:2=2003\)
c) Ta có : \(\left|2x-3\right|\ge0;\left|2x+4\right|\ge0\)
\(\Rightarrow\left|2x-3\right|+\left|2x+4\right|=3-2x+4+2x\)
\(=3+4=7\)
Thay \(\left|2x-3\right|=7\)
\(\Rightarrow2x-3=\orbr{\begin{cases}7\\-7\end{cases}}\Rightarrow2x=\orbr{\begin{cases}10\\-4\end{cases}}\Rightarrow x=\orbr{\begin{cases}5\\-2\end{cases}}\)
Thay \(\left|2x+4\right|=7\)
\(\Rightarrow2x+4=\orbr{\begin{cases}7\\-7\end{cases}}\Rightarrow2x=\orbr{\begin{cases}3\\-11\end{cases}}\Rightarrow x=\orbr{\begin{cases}\frac{3}{2}\\\frac{-11}{2}\end{cases}}\)
Vậy \(x\in\left(5;-2;\frac{3}{2};\frac{-11}{2}\right)\)
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\(\frac{7^{x+2}+7^{x+1}+7x}{57}=\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)
\(\Rightarrow\frac{7x\left(7^2+7^1+1\right)}{57}=\frac{5^{2x}\left(1+5^1+5^3\right)}{131}\)
\(\Rightarrow\frac{7x\left(49+7+1\right)}{57}=\frac{5^{2x}\left(1+5+125\right)}{131}\)
\(\Rightarrow\frac{7x.57}{57}=\frac{5^{2x}.131}{131}\)
\(\Rightarrow7x=25x\)
\(\Rightarrow x=0\)
\(\left(4x-3\right)^4=\left(4x-3\right)^2\)
\(\Rightarrow\left(4x-3\right)^4-\left(4x-3\right)^2=0\)
\(\Rightarrow\left(4x-3\right)^2\left[\left(4x-3\right)^2-1\right]=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(4x-3\right)^2=0\\\left(4x-3\right)^2=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}4x-3=0\\4x-3=-1\\4x-3=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\x=\frac{1}{2}\\x=1\end{cases}}\)
Ta có : |4 + 2x| = - 4x
=> - 4x \(\ge\) 0
=> -x \(\ge\)0
=> x \(\le\)0
\(\orbr{\begin{cases}4+2x=4x\\4+2x=-4x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}4x-2x=4\\4x+2x=-4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=4\\6x=-4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\left(\text{loại}\right)\\x=-\frac{2}{3}\left(\text{TM}\right)\end{cases}}\)
Vậy \(x=-\frac{2}{3}\)
|4 + 2x| = -4x
4 + 2x = -4x hoặc -(4 + 2x) = -4x
4 = -4x - 2x -4 - 2x = -4x
4 = -6x -4 = -4x + 2x
-2/3 = x -4 = -2x
x = -2/3 2 = x
x = 2