BÀI 1: TÌM X:
a) I x - 1 I = 0
b) -11 . I 3x - 1I = -22
c) I x I <2
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\(\left|3x-1\right|=\left|2x+5\right|\)
\(\Rightarrow\orbr{\begin{cases}3x-1=2x+5\\3x-1+2x+5=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x-2x=5+1\\5x+4=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=6\\x=-\frac{4}{5}\end{cases}}\)
Ta có: \(\hept{\begin{cases}\left(x-1\right)^2\ge0\\\left|3y-1\right|\ge0\\\left|z+2\right|\ge0\end{cases}}\Rightarrow\left(x-1\right)^2+\left|3y-1\right|+\left|z+2\right|\ge0\)
Dấu "="\(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left|3y-1\right|=0\\\left|z+2\right|=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\3y-1=0\\x+2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{1}{3}\\z=-2\end{cases}}\)
Vậy x = 1, \(y=\frac{1}{3}\),z = -2
\(\left|3x-1\right|=\left|\dfrac{-1}{3}x+2\right|\)
<=> \(\left[{}\begin{matrix}3x-1=\dfrac{-1}{3}x+2\\-3x+1=\dfrac{-1}{3}x+2\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}3x-\dfrac{-1}{3}x=2+1\\-3x-\dfrac{-1}{3}x=2-1\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}\dfrac{10}{3}x=3\\\dfrac{-8}{3}x=1\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{9}{10}\\x=\dfrac{-3}{8}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=-\dfrac{1}{3}x+2\left(x\ge\dfrac{1}{3}\right)\\3x-1=\dfrac{1}{3}x-2\left(x< \dfrac{1}{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{10}{3}x=3\\\dfrac{8}{3}x=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{10}\left(tm\right)\\x=-\dfrac{3}{8}\left(tm\right)\end{matrix}\right.\)
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.
a: \(\Leftrightarrow\left\{{}\begin{matrix}3x-2>-4\\3x-2< 4\end{matrix}\right.\Leftrightarrow-\dfrac{2}{3}< x< 2\)
c: \(\Leftrightarrow\left[{}\begin{matrix}3x-1>5\\3x-1< -5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -\dfrac{4}{3}\end{matrix}\right.\)
d: \(\Leftrightarrow\left[{}\begin{matrix}3x+1>x-2\\3x+1< -x+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x>-3\\4x< 1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>-\dfrac{3}{2}\\x< \dfrac{1}{4}\end{matrix}\right.\)
Bài 1: a) min B=50 (vì |y-3|>=0) khi |y-3|=0=> y=3
b) tương tự min C=-1 khi x=100 và y=-200
a) \(\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\)
\(\frac{1}{3}:x=\frac{3}{5}-\frac{2}{3}=\frac{9}{15}-\frac{10}{15}=\frac{-1}{15}\)
\(x=\frac{-1}{15}.\frac{1}{3}\)
\(x=\frac{-1}{45}\)
Vậy x = \(\frac{-1}{45}\)
c) \(\left|2x-1\right|+1=4\)
\(\left|2x-1\right|=4-1=3\)
2x-1 = 3 ; -3
TH1: 2.x - 1 = 3
2.x = 3 + 1 = 4
x = 4 : 2 = 2
TH2: 2.x - 1 = -3
2.x = -3 + 1 = -2
x = -2 : 2 = -1
Vậy x \(\in\){ 2 ; -1 }
Ngại làm ấn máy ==
bài 1:
a. \((x+1)(x+3) - x(x+2)=7 \)
\(x^2+ 3x +x +3 - x^2 -2x =7\)
\(x^2+4x+3-x^2-2x=7\)
\(=> 2x+3=7\)
\(2x=4\)
\(x = 2\)
Bài 2:
a)
\((3x-5)(2x+11) -(2x+3)(3x+7) \)
\(= 6x^2 +33x-10x-55-6x^2-14x-9x-10\)
\(= (6x^2-6x^2)+(33x-10x-14x-9x)-(55+10)\)
\(=-65\)
\(\)
a, !x-1!=0
\(\Rightarrow x-1=0\)
\(\Rightarrow x=0+1\)
\(\Rightarrow x=1\)
Vậy x=1
b,-11.!3x-1!=-22
\(\Rightarrow!3x-1!=-22:\left(-11\right)\)
\(\Rightarrow!3x-1!=2\)
\(\Rightarrow\orbr{\begin{cases}3x-1=2\\3x-1=-2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=2+1\\3x=-2+1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=3\\3x=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3:3\\x=-1:3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-1}{3}\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=1\\x=-\frac{1}{3}\end{cases}}\)
c, !x!<2
\(\Rightarrow x\in\left\{-1;0;1\right\}\)
Vậy \(x\in\left\{-1;0;1\right\}\)
Hok tốt nhé!!
A )
| x - 1 | = 0
x = 0 + 1
x = 1
B )
-11 . | 3x - 1 | = -22
| 3x - 1 | = -22 : ( -11 )
| 3x - 1 | = 2
3x = 2 + 1
3x = 3
x = 3 : 3
x = 1