4.x+1=21
(-3).x-1=8
(-1005).(x+2)=0
(8+x).(6-x)=0
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Giải:
a) \(\left|48-3x\right|=0\)
\(\Leftrightarrow48-3x=0\)
\(\Leftrightarrow3x=48\)
\(\Leftrightarrow x=\dfrac{48}{3}=16\)
Vậy x = 16.
b) \(\left|-x-7\right|=24\)
\(\Leftrightarrow\left[{}\begin{matrix}-x-7=24\\-x-7=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=31\\-x=-17\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-31\\x=17\end{matrix}\right.\)
Vậy \(x=-31\) hoặc \(x=17\).
c) \(\left|4-x\right|=21\)
\(\Leftrightarrow\left[{}\begin{matrix}4-x=21\\4-x=-21\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-17\\x=25\end{matrix}\right.\)
Vậy \(x=-17\) hoặc \(x=25\).
d) \(\left|x+8\right|+12=0\)
\(\Leftrightarrow\left|x+8\right|=-12\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=-12\\x+8=-\left(-12\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-20\\x=4\end{matrix}\right.\)
Vậy \(x=-20\) hoặc \(x=4\).
e) \(6-\left|x\right|=2\)
\(\Leftrightarrow\left|x\right|=6-2=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
Vậy \(x=4\) hoặc \(x=-4\).
Chúc bạn học tốt!
1. |48 - 3x| = 0.
\(\Leftrightarrow\) 48 - 3x = 0.
\(\Leftrightarrow\) 3x = 48.
\(\Leftrightarrow\) x = \(\dfrac{48}{3}=16.\)
Vậy x = 16.
2. |-x - 7| = 24.
\(\Leftrightarrow\left[{}\begin{matrix}-x-7=24.\\-x-7=-24.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=31.\\-x=-17.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-31.\\x=17.\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-31.\\x=17.\end{matrix}\right.\)
3. |4 - x| = 21.
\(\Leftrightarrow\left[{}\begin{matrix}4-x=21.\\4-x=-21.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-17.\\x=25.\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-17.\\x=25.\end{matrix}\right.\)
4. |x + 8| + 12 = 0.
|x + 8| = 0 - 12.
|x + 8| = -12.
\(\Leftrightarrow\left[{}\begin{matrix}x+8=-12.\\x+8=-\left(-12\right).\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=-12.\\x+8=12.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-20.\\x=4.\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-20.\\x=4.\end{matrix}\right.\)
5. 6 - |x| = 2.
|x| = 6 - 2.
|x| = 4.
\(\Leftrightarrow\left[{}\begin{matrix}x=4.\\x=-4.\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=4.\\x=-4.\end{matrix}\right.\)
\(\left(4-3x\right)\left(10x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)
\(\left(7-2x\right)\left(4+8x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)
rồi thực hiện đến hết ...
Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>
\(\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)
\(2x^2-7x+3=4x^2+4x-3\)
\(2x^2-7x+3-4x^2-4x+3=0\)
\(-2x^2-11x+6=0\)
\(2x^2+11x-6=0\)
\(2x^2+12x-x-6=0\)
\(2x\left(x+6\right)-\left(x+6\right)=0\)
\(\left(x+6\right)\left(2x-1\right)=0\)
\(x+6=0\Leftrightarrow x=-6\)
\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
\(3x-2x^2=0\)
\(x\left(2x-3\right)=0\)
\(x=0\)
\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Tự lm tiếp nha
a) \(\left|4-x\right|+2x=3\)
<=> \(\left|4-x\right|=3-2x\)
<=> \(\orbr{\begin{cases}4-x=3-2x\left(x\le4\right)\\x-4=3-2x\left(x>4\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-1\left(tm\right)\\3x=7\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-1\\x=\frac{7}{3}\left(ktm\right)\end{cases}}\)
Vậy x = -1
b) \(\left|x-7\right|+2x+5=6\)
<=> \(\left|x-7\right|=1-2x\)
<=> \(\orbr{\begin{cases}x-7=1-2x\left(đk:x\ge7\right)\\x-7=2x-1\left(đk:x< 7\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}3x=8\\x=-6\left(tm\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{8}{3}\left(ktm\right)\\x=-6\left(tm\right)\end{cases}}\)
Vậy x = -6
c) \(3x-\left|2x+1\right|=2\)
<=> \(\left|2x+1\right|=3x-2\)
<=> \(\orbr{\begin{cases}2x+1=3x-2\left(đk:x\ge-\frac{1}{2}\right)\\2x+1=2-3x\left(đk:x< -\frac{1}{2}\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}x=3\left(tm\right)\\5x=1\end{cases}}\)
<=> \(\orbr{\begin{cases}x=3\\x=\frac{1}{5}\left(ktm\right)\end{cases}}\)
Vậy x = 3
d) \(\left|x+2\right|-x=2\)
<=> \(\left|x+2\right|=x+2\)
<=> \(\orbr{\begin{cases}x+2=x+2\left(đk:x\ge-2\right)\\x+2=-x-2\left(x< -2\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}0x=0\\2x=-4\end{cases}}\)
<=> 0x = 0 (luôn đúng) và x = -2 (ktm)
Vậy x \(\ge\)-2
e) \(\left|x-3\right|=21\)
<=> \(\orbr{\begin{cases}x-3=21\\3-x=21\end{cases}}\)
<=> \(\orbr{\begin{cases}x=24\\x=-18\end{cases}}\)
Vậy x = 24 hoặc x = -18
f) \(\left|2x+3\right|-\left|x-3\right|=0\)
<=> \(\left|2x+3\right|=\left|x-3\right|\)
<=> \(\orbr{\begin{cases}2x+3=x-3\\2x+3=3-x\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\3x=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=0\end{cases}}\)
Vậy x thuộc {-6; 0}
g) Ta có: \(\left|x+\frac{1}{8}\right|\ge0\forall x\)
\(\left|x+\frac{2}{8}\right|\ge0\forall x\)
\(\left|x+\frac{5}{8}\right|\ge0\forall x\)
=> VT = \(\left|x+\frac{1}{8}\right|+\left|x+\frac{2}{8}\right|+\left|x+\frac{5}{8}\right|\ge0\forall x\)
=> VP \(\ge0\) => \(4x\ge0\) => \(x\ge0\)
Do đó: \(x+\frac{1}{8}+x+\frac{2}{8}+x+\frac{5}{8}=4x\)
<=> \(3x+1=4x\) <=> \(x=1\left(tm\right)\)
Vậy x = 1
h) \(\left|x-2\right|-\left|2x+3\right|-x=-2\)
<=> \(\left|x-2\right|-\left|2x+3\right|=x-2\)(*)
Lập bảng xét dấu:
x -3/2 2
x - 2 2 - x | 2 - x 0 x - 2
2x + 3 -2x - 3 0 2x + 3 | 2x + 3
Xét x < -3/2 => pt (*) trở thành: 2 - x + 2x + 3 = x - 2
<=> x + 5 = x - 2 <=> 0x = -7 (vô lí)
Xét -3/2 \(\le\) x < 2 => pt (*) trở thành: 2 - x - 2x - 3 = x - 2
<=> 4x = 1 <=> x = 1/4 ((tm)
Xét x \(\ge\) 2 => pt (*) trở thành x - 2 - 2x - 3 = x - 2
<=> 2x = -3 <=> x = -3/2 (ktm)
Vậy x = 1/4
i) |2x - 3| - x = |2 - x|
<=> |2x - 3| - |2 - x| = x (*)
Lập bảng xét dấu
x 3/2 2
2x - 3 3 - 2x 0 2x - 3 | 2x - 3
2 - x 2 - x | 2 - x 0 x - 2
Xét x < 3/2 => pt (*) trở thành: 3 - 2x - 2 + x = x
<=> 2x = 1 <=> x = 1//2 ((tm)
Xét \(\frac{3}{2}\le x< 2\)=> pt (*) trở thành: 2x - 3 - 2 + x = x
<=> 2x = 5 <=> x = 5/2 (ktm)
Xét x \(\ge\)2 ==> pt (*) trở thành: 2x - 3 - x + 2 = x
<=> 0x = -5 (vô lí)
Vậy x = 1/2
k) 2|x - 3| - |4x - 1| = 0
<=> 2|x - 3| = |4x - 1|
<=> \(\orbr{\begin{cases}2\left(x-3\right)=4x-1\\2\left(x-3\right)=1-4x\end{cases}}\)
<=> \(\orbr{\begin{cases}2x-6=4x-1\\2x-6=1-4x\end{cases}}\)
<=> \(\orbr{\begin{cases}2x=-5\\6x=7\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-\frac{5}{2}\\x=\frac{7}{6}\end{cases}}\) Vậy ...
Ta có: 4 - |x - 5| = 0
=> | x - 5 | = 4
<=> x - 5 = 4
x - 5 = -4
<=> x = 4 + 5
x = -4 + 5
<=> x = 9
x = 1
giúp mik nha
\(4x=20\Leftrightarrow x=5\)
\(-3x=9\Leftrightarrow x=-3\)
\(x+2=0\Leftrightarrow x=-2\)
\(\left[{}\begin{matrix}8+x=0\Leftrightarrow x=-8\\6-x=0\Leftrightarrow x=6\end{matrix}\right.\)