a, \(7\left(x-1\right)+2x\left(1-x\right)=0\)
\(\Rightarrow\left(7-2x\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-2x=0\\x-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=1\end{cases}}\)
b, \(0,25-\left|3,5-x\right|=0\)
\(\Rightarrow\left|3,5-x\right|=2,5\)
\(\Rightarrow\orbr{\begin{cases}3,5-x=2,5\\3,5-x=-2,5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=6\end{cases}}\)
c, \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
\(\Rightarrow\left|x+\frac{3}{4}\right|=\frac{1}{2}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=\frac{-1}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{4}\\x=\frac{-5}{4}\end{cases}}\)
a) 7.(x-1) + 2x.(1-x) = 0
7.(x-1) - 2x.(x-1) = 0
(x-1).(7-2x) = 0
=> (x-1) = 0 => x = 1
7-2x = 0 => 2x = 7 => x = 7/2
KL:...
b) 0,25 - | 3,5-x| = 0
=> |3,5 - x| = 0,25
TH1: 3,5 - x = 0,25
x = 3,25
TH2: 3,5 - x = -0,25
x = 3,75
phần c bn dựa vào phần b mak lm nha
a) 7.(x - 1) + 2x.(1 - x) = 0
<=> -2x2 + 9x - 7 = 0
<=> \(\orbr{\begin{cases}x=1\\x=\frac{7}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=1\\x=\frac{7}{2}\end{cases}}\)
b) 0,25 - |3,5 - x| = 0
<=> - |3,5 - x| = 0 - 0,25
<=> - |3,5 - x| = -0,25
<=> |3,5 - x| = 0,25
Ta xét 2 th:
Th1: |x| = 3,5 - 0,25
<=> |x| = \(\frac{13}{4}\)
Th2: |x| = 3,5 + 0,25
<=> x = \(\frac{15}{4}\)
=> \(\orbr{\begin{cases}x=\frac{13}{4}\\\frac{15}{4}\end{cases}}\)
c) \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
<=> \(\left|x+\frac{3}{4}\right|=0+\frac{1}{2}\)
<=> \(\left|x+\frac{3}{4}\right|=\frac{1}{2}\)
Ta xét 2 th:
Th1: \(x+\frac{3}{4}=\frac{1}{2}\)
<=> \(x=\frac{1}{2}-\frac{5}{4}\)
<=> \(x=-\frac{3}{4}\)
Th2: \(x+\frac{3}{4}=\frac{1}{2}\)
<=> \(x=\frac{1}{2}-\frac{3}{4}\)
<=> \(x=-\frac{1}{4}\)
=> \(\orbr{\begin{cases}x=-\frac{3}{4}\\x=-\frac{1}{4}\end{cases}}\)
a) \(7.(x-1)+2x.(1-x) =0\)
\(=> 7.(x-1)-2x.(x-1)=0\)
\(=> (x-1)(7-2x)=0\)
=>\(\orbr{\begin{cases}x-1=0\\7-2\text{x}=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=1\\x=\frac{7}{2}\end{cases}}\)
Vậy ...
b) 0,25 - \(|3,5-x|\)= 0
=> \(|3,5-x|\) = 0,25
TH1: 3,5 - x = 0,25
=> x = 3,5 - 0,25 = 3,25
TH2: 3,5 - x = -0,25
=> x = 3,5 + 0,25 = 3,75
Vậy ...
c) \(|x+\frac{3}{4}|-\frac{1}{2}\)= 0
=> \(|x+\frac{3}{4}|\)= \(\frac{1}{2}\)
TH1: x + \(\frac{3}{4}\)= \(\frac{1}{2}\)
=> x =\(\frac{1}{2}\) - \(\frac{3}{4}\)
=> x = \(-\frac{1}{4}\)
TH2: x + \(\frac{3}{4}\) = - \(\frac{1}{2}\)
=> x = -\(\frac{1}{2}\) - \(\frac{3}{4}\)
=> x = \(-\frac{5}{4}\)
Vậy...