Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
I 2x-3 I = I x+1 I
2x-3 = x+1
x+1 - 2x+3=0
x (1-2) +1+3=0
-1x +4 =0
-1x = 0-4
-1x =-4
x = -4 : -1
x =4
Trả lời:
\(\left|2x-3\right|=\left|x+1\right|\)
\(\Rightarrow2x-3=x+1\) hoặc \(2x-3=-\left(x+1\right)\)
TH1: \(2x-3=x+1\)
\(2x-x=1+3\)
\(x=4\)
TH2: \(2x-3=-\left(x+1\right)\)
\(2x-3=-x-1\)
\(2x+x=-1+3\)
\(3x=2\)
\(x=\frac{2}{3}\)
Vậy \(x=4;x=\frac{2}{3}\)
a) \(\left(\frac{1}{7}x-\frac{2}{3}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{1}{7}x-\frac{2}{3}=0\\-\frac{1}{5}x+\frac{3}{5}=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\frac{1}{7}x=\frac{2}{3}\\-\frac{1}{5}x=-\frac{3}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{14}{3}\\x=3\end{cases}}\)
b)\(\frac{1}{10}x-\frac{4}{5}x+1=0\)
\(\Leftrightarrow x.\left(\frac{1}{10}-\frac{4}{5}\right)+1=0\)
\(\Rightarrow-\frac{7}{10}x=-1\)
\(\Rightarrow x=\frac{10}{7}\)
c)\(\left(2x-\frac{1}{3}\right).\left(5x+\frac{2}{7}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{3}=0\\5x+\frac{2}{7}=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=\frac{1}{3}\\5x=-\frac{2}{7}\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{6}\\x=-\frac{2}{35}\end{cases}}\)
a, (1/7 . x - 2/3) . (-1/5 . x + 3/5) = 0
Suy ra : 1/7 .x -2/3 = 0 hoặc -1/5 .x + 3/5 =0
Vậy : 1/7 .x = 2/3 hoặc -1/5 .x = 3/5
x =2/3 : 1/7 hoặc x = 3/5 : (-1/5)
x = 14/3 hoặc x = -3
b, 1/10 .x - 4/5 .x + 1 =0
x . (1/10 - 4/5) + 1 = 0
x . (-7/10) + 1 = 0
x . -7/10 =0 +1 = 1
x = 1 : (-7/10)
x = -10/7
c, (2x - 1/3 ) . (5x +2/7) = 0
Suy ra : 2x - 1/3 = 0 hoặc 5x + 2/7 = 0
Vậy : 2x = 1/3 hoặc 5x = 2/7
x = 1/3 : 2 hoặc x = 2/7 : 5
x = 1/6 hoặc x = 2/35
a) 5x.(x+3/4) = 0
=> x = 0
x+3/4 = 0 => x = -3/4
b) \(\frac{x+7}{2010}+\frac{x+6}{2011}=\frac{x+5}{2012}+\frac{x+4}{2013}.\)
\(\Rightarrow\frac{x+7}{2010}+\frac{x+6}{2011}-\frac{x+5}{2012}-\frac{x+4}{2013}=0\)
\(\frac{x+7}{2010}+1+\frac{x+6}{2011}+1-\frac{x+5}{2012}-1-\frac{x+4}{2013}-1=0\)
\(\left(\frac{x+7}{2010}+1\right)+\left(\frac{x+6}{2011}+1\right)-\left(\frac{x+5}{2012}+1\right)-\left(\frac{x+4}{2013}+1\right)=0\)
\(\frac{x+2017}{2010}+\frac{x+2017}{2011}-\frac{x+2017}{2012}-\frac{x+2017}{2013}=0\)
\(\left(x+2017\right).\left(\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}\right)=0\)
=> x + 2017 = 0
x = -2017
a) để 2x - 3 > 0
=> 2x > 3
x > 3/2
b) 13-5x < 0
=> 5x < 13
x < 13/5
c) \(\frac{x+3}{2x-1}>0\)
=> x + 3 > 0
x > -3
d) \(\frac{x+7}{x+3}=\frac{x+3+4}{x+3}=1+\frac{4}{x+3}\)
Để x+7/x+3 < 1
=> 1 + 4/x+3 < 1
=> 4/x+3 < 0
=> không tìm được x thỏa mãn điều kiện
Bài làm:
a) \(\left|\frac{1}{2}x-\frac{5}{2}\right|-1=-\frac{1}{2}\)
\(\Leftrightarrow\left|\frac{1}{2}x-\frac{5}{2}\right|=\frac{1}{2}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x-\frac{5}{2}=\frac{1}{2}\\\frac{1}{2}x-\frac{5}{2}=-\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=3\\\frac{1}{2}x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)
+ Nếu x = 6
\(\left|12-\frac{1}{3}y\right|=\frac{5}{6}\)
\(\Leftrightarrow\orbr{\begin{cases}12-\frac{1}{3}y=\frac{5}{6}\\12-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{67}{6}\\\frac{1}{3}y=\frac{77}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{67}{2}\\y=\frac{77}{2}\end{cases}}\)
+ Nếu x = 4
\(\left|8-\frac{1}{3}y\right|=\frac{5}{6}\)
\(\Leftrightarrow\orbr{\begin{cases}8-\frac{1}{3}y=\frac{5}{6}\\8-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{43}{6}\\\frac{1}{3}y=\frac{53}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{43}{2}\\y=\frac{53}{2}\end{cases}}\)
Vậy ta có 4 cặp số (x;y) thỏa mãn: \(\left(6;\frac{67}{2}\right);\left(6;\frac{77}{2}\right);\left(4;\frac{43}{2}\right);\left(4;\frac{53}{2}\right)\)
b) \(\frac{3}{2}x-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{5}{3}\)
\(\Leftrightarrow\frac{3}{2}x-\frac{1}{2}x+\frac{1}{3}=\frac{5}{3}\)
\(\Leftrightarrow x=\frac{4}{3}\)
Thay vào ta được:
\(\frac{2.\frac{4}{3}+y}{\frac{4}{3}-2y}=\frac{5}{4}\)
\(\Leftrightarrow\frac{32}{3}+4y=\frac{20}{3}-10y\)
\(\Leftrightarrow14y=-4\)
\(\Rightarrow y=-\frac{2}{7}\)
Vậy ta có 1 cặp số (x;y) thỏa mãn: \(\left(\frac{4}{3};-\frac{2}{7}\right)\)
Câu 1: (2x-3)-(x-5)=(x+2)-(x-1)
2x -3 -x+5 = x+2 -x +1
2x -x -x +x = 2+1 +3 -5
x= 1
Câu 2: 2(x-1)-5(x+2)=10
2x -2 -5x -10 =10
2x -5x = 10 +2 +10
(2-5) x = 22
-3x= 22
x= 22/-3
Câu 1: ( 2x - 3 ) - ( x - 5 ) = ( x + 2 ) - ( x - 1 )
=> ( 2x - x ) - ( 3 - 5 ) = ( x - x ) + ( 2 + 1 )
=> x + 2 = 3
=> x = 1
Thử lại: ( 2 - 3 ) - ( 1 - 5 ) = ( 1 + 2 ) - ( 1 - 1 )
=> -1 + 4 = 3 - 0
=> 3 = 3 ( thoả mãn )
Câu 2: 2 ( x - 1 ) - 5 ( x + 2 ) = 10
=> ( 2x - 2 ) - ( 5x + 10 ) = 10
=> ( 2x - 5x ) - ( 2 + 10 ) = 10
=> -3x - 12 = 10
=> -3x = 22
=> x = -22/3
Thử lại: 2 ( -22/3 - 1 ) - 5 ( -22/3 + 2 ) = 10
=> 2 * -25/3 - 5 * -16/3 = 10
=> -50/3 - -80/3 = 10
=> (-50) - (-80)/3 = 10
=> 30 / 3 = 10 ( thoả mãn )