TH
4
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
NQ
0
XO
6 tháng 8 2020
Ta có :\(\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right).\left(2x-2\right)=\left(-\frac{3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)
=> \(\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right).\left(2x-2\right)=-\frac{1}{2}\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\)
=> \(2x-2=-\frac{1}{2}\)
=> \(2x=\frac{3}{2}\)
=> \(x=\frac{3}{4}\)
4 tháng 3 2020
Thay x =2 ; y= -1 vào biểu thức ta có:
16.2.(-1)^5 - 2.2^3.(-1)
=16.2.(-1) - 2.1/8
=-32 - 1/4
=-129/4
vậy...........................
học tốt!
1 tháng 10 2021
\(a,\Leftrightarrow\left|x+\dfrac{2}{5}\right|=\dfrac{7}{4}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{5}=\dfrac{7}{4}\left(x\ge-\dfrac{2}{5}\right)\\x+\dfrac{2}{5}=-\dfrac{7}{4}\left(x< -\dfrac{2}{5}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{27}{20}\left(tm\right)\\x=-\dfrac{43}{20}\left(tm\right)\end{matrix}\right.\)
\(b,\Leftrightarrow\left|x-\dfrac{13}{10}\right|=\dfrac{13}{10}\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{13}{10}=\dfrac{13}{10}\left(x\ge\dfrac{13}{10}\right)\\x-\dfrac{13}{10}=-\dfrac{13}{10}\left(x< \dfrac{13}{10}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
\(c,\Leftrightarrow\left|\dfrac{3}{4}-\dfrac{1}{2}x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}-\dfrac{1}{2}x=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\\\dfrac{1}{2}x-\dfrac{3}{4}=\dfrac{1}{2}\left(x>\dfrac{3}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{5}{2}\left(tm\right)\end{matrix}\right.\)
\(d,\Leftrightarrow\left|5-2x\right|=4\Leftrightarrow\left[{}\begin{matrix}5-2x=4\left(x\le\dfrac{5}{2}\right)\\2x-5=4\left(x>\dfrac{5}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{9}{2}\left(tm\right)\end{matrix}\right.\)
\(đ,\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\\x-1,3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\x=1,3\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
\(e,\Leftrightarrow\left\{{}\begin{matrix}x-2021=0\\x-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\x=2022\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
\(f,\Leftrightarrow\left|x\right|=\dfrac{1}{3}-x\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}-x\left(x\ge0\right)\\x=x-\dfrac{1}{3}\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\left(tm\right)\\0x=-\dfrac{1}{3}\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{6}\)
\(g,\Leftrightarrow\left[{}\begin{matrix}x-2=x\left(x\ge2\right)\\2-x=x\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=2\left(vô.lí\right)\\x=1\left(tm\right)\end{matrix}\right.\Leftrightarrow x=1\)
HL
0
14 tháng 7 2021
a, 1,5 +|2x - 2/3| = 3/2
|2x - 2/3| = 3/2 - 1,5
|2x - 2/3| = 0
<=> 2x - 2/3 = 0
<=> 2x = 0 + 2/3
<=> 2x = 2/3
<=> x = 2/3 : 2
<=> x = 1/3
Vậy x = 1/3
14 tháng 7 2021
b, 3/4 - |1/4 - x| = 5/8
|1/4 - x| = 3/4 - 5/8
|1/4 - x| = 1/8
<=> 1/4 - x = 1/8
1/4 - x = /1/8
<=> x = 1/4 - 1/8
x = 1/4 - ( -1/8)
<=> x = 1/8
x = 3/8
Vậy x thuộc { 1/8 ; 3/8 }
2H
5
25 tháng 4 2022
cho f(x) = 1/2x +4 =0
=> 1/2 x = 0-4
=> 1/2x = -4
=> x = -4 : 1/2
=> x= -8
vậy x=-8 là nghiệm của đa thức F(x)
Fudo lm thiếu 1 trường hợp r
Ta có \(\left(2x-5\right)^2=\left|2x-5\right|\)
\(\Leftrightarrow\orbr{\begin{cases}\left(2x-5\right)^2=2x-5\\\left(2x-5^2\right)=5-2x\end{cases}}\)
TH1: \(\left(2x-5\right)^2=2x-5\)
\(\Leftrightarrow\left(2x-5\right)^2-\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x-5-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\2x-5-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=5\\2x-6=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\2x=6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=3\end{cases}}\) (1)
TH2: \(\left(2x-5\right)^2=5-2x\)
\(\Leftrightarrow\left(2x-5\right)^2-\left(5-2x\right)=0\)
\(\Leftrightarrow\left(2x-5\right)^2+2x-5=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x-5+1\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\2x-4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=5\\2x=4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=2\end{cases}}\) (2)
Từ (1) và (2) \(\Leftrightarrow x\in\left\{\frac{5}{2};2;3\right\}\)
Vậy \(x\in\left\{\frac{5}{2};2;3\right\}\)
@@ Học tốt
|2x−5|+2x−5=0|2x−5|+2x−5=0
⇔|2x−5|=−2x+5⇔|2x−5|=−2x+5
⇔[2x−5=−2x+52x−5=2x−5⇔[2x−5=−2x+52x−5=2x−5
⇔[2x+2x=5+52x−2x=−5+5⇔[2x+2x=5+52x−2x=−5+5
⇔[4x=100x=0⇔[4x=100x=0
⇔⎡⎣x=52x=0