|2x|-|-2,5|=|-7,5|
|2x-3|=1 |x-3,5|+|y-1,3|=0
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a,2,5-x=1,3
x=2,5-1,3
x=1,2
b,x-3,5=7,5
x=7.5+3,5
x=11
c,(x-1,5)-2,5=0
x-1,5=2,5
x=2,5+1,5
x=4
d,1,6-(x-0,2)=0
x-0,2=1,6
x=1,6+0,2
x=1,8
A=(3,7- -x)+2,5
vì 2,5>=0 với mọi x
=> x=-1,2
B= (x+1,5)-4,5
Vì -4,5<=0 với mọi x
=> x=-6
Học Tốt
a) \(\left|2x\right|-\left|-2,5\right|=\left|-7,5\right|\)
\(\Leftrightarrow\left|2x\right|-2,5=7,5\)
\(\Leftrightarrow\left|2x\right|=10\)
\(\Leftrightarrow\begin{cases}x\ge0\\2x=10\end{cases}\) hoặc \(\begin{cases}x< 0\\2x=-10\end{cases}\)
\(\Leftrightarrow\begin{cases}x\ge0\\x=5\left(tm\right)\end{cases}\) hoặc \(\begin{cases}x< 0\\x=-5\left(tm\right)\end{cases}\)
Vậy x={5;-5}
b)\(\left|3x\right|\cdot\left|-3,5\right|=\left|-2,8\right|\)
\(\Leftrightarrow\left|3x\right|\cdot3,5=2,8\)
\(\Leftrightarrow\left|3x\right|=\frac{4}{5}\)
\(\Leftrightarrow\begin{cases}x\ge0\\3x=\frac{4}{5}\end{cases}\) hoặc \(\begin{cases}x< 0\\3x=-\frac{4}{5}\end{cases}\)
\(\Leftrightarrow\begin{cases}x\ge0\\x=\frac{4}{15}\end{cases}\) hoặc \(\begin{cases}x< 0\\x=-\frac{4}{15}\end{cases}\)
Vậy x={4/15;-4/15}
c) \(\left(3x-5\right)\left(\frac{3}{2}x+2\right)\left(0,5x-10\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}3x-5=0\\\frac{3}{2}x+2=0\\0,5x-10=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{5}{3}\\x=-\frac{4}{3}\\x=20\end{array}\right.\)
a)|2x|-|-2,5|=|-7,5|
|2x|-2,5=7,5
|2x|=10
\(\Rightarrow\left[\begin{array}{nghiempt}2x=10\\2x=-10\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=5\\x=-5\end{array}\right.\)
Vậy x=5;-5
\(a,\Leftrightarrow\left|x+\dfrac{2}{5}\right|=\dfrac{7}{4}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{5}=\dfrac{7}{4}\left(x\ge-\dfrac{2}{5}\right)\\x+\dfrac{2}{5}=-\dfrac{7}{4}\left(x< -\dfrac{2}{5}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{27}{20}\left(tm\right)\\x=-\dfrac{43}{20}\left(tm\right)\end{matrix}\right.\)
\(b,\Leftrightarrow\left|x-\dfrac{13}{10}\right|=\dfrac{13}{10}\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{13}{10}=\dfrac{13}{10}\left(x\ge\dfrac{13}{10}\right)\\x-\dfrac{13}{10}=-\dfrac{13}{10}\left(x< \dfrac{13}{10}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
\(c,\Leftrightarrow\left|\dfrac{3}{4}-\dfrac{1}{2}x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}-\dfrac{1}{2}x=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\\\dfrac{1}{2}x-\dfrac{3}{4}=\dfrac{1}{2}\left(x>\dfrac{3}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{5}{2}\left(tm\right)\end{matrix}\right.\)
\(d,\Leftrightarrow\left|5-2x\right|=4\Leftrightarrow\left[{}\begin{matrix}5-2x=4\left(x\le\dfrac{5}{2}\right)\\2x-5=4\left(x>\dfrac{5}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{9}{2}\left(tm\right)\end{matrix}\right.\)
\(đ,\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\\x-1,3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\x=1,3\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
\(e,\Leftrightarrow\left\{{}\begin{matrix}x-2021=0\\x-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\x=2022\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
\(f,\Leftrightarrow\left|x\right|=\dfrac{1}{3}-x\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}-x\left(x\ge0\right)\\x=x-\dfrac{1}{3}\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\left(tm\right)\\0x=-\dfrac{1}{3}\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{6}\)
\(g,\Leftrightarrow\left[{}\begin{matrix}x-2=x\left(x\ge2\right)\\2-x=x\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=2\left(vô.lí\right)\\x=1\left(tm\right)\end{matrix}\right.\Leftrightarrow x=1\)
c: Ta có: \(\left|\dfrac{1}{2}x-2\right|-\left|x+3\right|=0\)
\(\Leftrightarrow\left|\dfrac{1}{2}x-2\right|=\left|x+3\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-2=x+3\\\dfrac{1}{2}x-2=-x-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{-1}{2}=5\\x\cdot\dfrac{3}{2}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-10\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Bài 3:
b: \(x^2+2x+1=\left(x+1\right)^2\)
c: \(x^2-16=\left(x-4\right)\left(x+4\right)\)
d: \(\left(2x-1\right)^2-\left(x+3\right)^2\)
\(=\left(2x-1-x-3\right)\left(2x-1+x+3\right)\)
\(=\left(x-4\right)\left(3x+2\right)\)
a) |4 (x-1)| = 12
=> 4(x-1) = 12 hoặc -12
Với 4(x-1) = 12
=> x-1 = 12:4
=> x-1 = 3
=> x= 3+1
=> x=4
Với 4(x-1) = -12
x-1 = (-12) : 4
x-1 = -3
x = -2
b) |2x +1| - 5 = 10
|2x +1| = 10 +5
|2x +1| = 15
=> |2x +1| = 15 hoặc -15
Với 2x +1 = 15
2x = 14
=> x= 14 :2
=> x=7
Với 2x+1 = -15
2x = (-15) -1
2x = -16
=> x= (-16) :2
=> x= -8
\dfrac{1}{2}
1) \(A=23+\left|2x-\frac{1}{3}\right|\)
Ta có: \(\left|2x-\frac{1}{3}\right|\ge0\forall x\)
\(\Rightarrow\left|2x-\frac{1}{3}\right|+23\ge23\forall x\)
\(A=23\Leftrightarrow\left|2x-\frac{1}{3}\right|=0\Leftrightarrow2x-\frac{1}{3}=0\Leftrightarrow2x=\frac{1}{3}\Leftrightarrow x=\frac{1}{6}\)
Vậy Amin=23 \(\Leftrightarrow x=\frac{1}{6}\)
Câu b, câu c tương tự
2) \(\left|x-3,5\right|+\left|y-1,3\right|=0\)
Ta có: \(\orbr{\begin{cases}\left|x-3,5\right|\ge0\forall x\\\left|y-1,3\right|\ge0\forall y\end{cases}}\Rightarrow\left|x-3,5\right|+\left|y-1,3\right|\ge0\forall x\)
Mà \(\left|x-3,5\right|+\left|y-1,3\right|=0\)
\(\Rightarrow\orbr{\begin{cases}\left|x-3,5\right|=0\\\left|y-1,3\right|=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x-3,5=0\\y-1,3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3,5\\y=1,3\end{cases}}}\)
Vậy x=3,5 ; y=1,3
c) Ta có: \(\left\{{}\begin{matrix}\left|x-1,5\right|\ge0\forall x\in Q\\\left|2,5-x\right|\ge0\forall x\in Q\end{matrix}\right.\)
\(\Rightarrow\left|x-1,5\right|+\left|2,5-x\right|\ge0\forall x\in Q\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\left|x-1,5\right|=0\\\left|2,5-x\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=1,5\\x=2,5\end{matrix}\right.\)
Vậy \(x=\left\{{}\begin{matrix}1,5\\2,5\end{matrix}\right.\).
e) \(\left(x-2\right)^2=1\)
\(\Rightarrow\left[{}\begin{matrix}x-2=\sqrt{1}\\x-2=-\sqrt{1}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\).
Mấy câu kia dễ rồi.
sửa lại ý c của N.Anh
Áp dụng bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) có:
\(\left|x-1,5\right|+\left|2,5-x\right|\ge\left|x-1,5+2,5-x\right|=1\)
\(\Rightarrow\left|x-1,5\right|+\left|2,5-x\right|\ge1>0\)
mà theo đề thì \(\left|x-1,5\right|+\left|2,5-x\right|=0\)
\(\Rightarrow\) k có gt \(x\) nào tm yêu cầu đề bài
a)\(\left|2x\right|-\left|-2,5\right|=\left|-7,5\right|\)
\(\Rightarrow\left|2x\right|-2,5=7,5\)
\(\Rightarrow\left|2x\right|=10\)
\(\Rightarrow\left[{}\begin{matrix}2x=10\Rightarrow x=5\\2x=-10\Rightarrow x=-5\end{matrix}\right.\)
b) \(\left|2x-3\right|=1\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=1\Rightarrow2x=4\Rightarrow x=2\\2x-3=-1\Rightarrow2x=2\Rightarrow x=1\end{matrix}\right.\)
c) \(\left|x-3,5\right|+\left|y-1,3\right|=0\)
Ta có: \(\left|x-3,5\right|\ge0\forall x\)
\(\left|y-1,3\right|\ge0\forall y\)
\(\Rightarrow\left|x-3,5\right|+\left|y-1,3\right|\ge0\forall x,y\)
Dấu "=" xảy ra
\(\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\Rightarrow x=3,5\\y-1,3=0\Rightarrow y=1,3\end{matrix}\right.\)
\(a)\left|2x\right|-\left|-2,5\right|=\left|-7,5\right|\)
\(\Rightarrow\left|2x\right|-2,5=7,5\)
\(\Rightarrow\left|2x\right|=10\)
\(\Rightarrow\left[{}\begin{matrix}2x=10\Rightarrow x=5\\2x=-10\Rightarrow x=-5\end{matrix}\right.\)
Vậy ...............
\(b)\left|2x-3\right|=1\)
\(\Rightarrow\left|2x\right|-3=1\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=1\Rightarrow2x=4\Rightarrow x=2\\2x-3=-1\Rightarrow2x=2\Rightarrow x=1\end{matrix}\right.\)
Vậy .........
\(c)\left|x-3,5\right|+\left|y-1,3\right|=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3,5=0\Rightarrow x=3,5\\y-1,3=0\Rightarrow y=1,3\end{matrix}\right.\)
Vậy ..............
Chúc bạn học tốt!