Question

Tìm x:

a)7.(x-1)+2x.(1-x)=0

b)0,25-|3,5-x|=0

c)|x+\(\frac{3}{4}\)| -\(\frac{1}{2}\)=0

Answer 1

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, \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)

\(\Rightarrow\left|x+\frac{3}{4}\right|=\frac{1}{2}\)

\(\Rightarrow\orbr{\begin{cases}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=\frac{-1}{2}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{4}\\x=\frac{-5}{4}\end{cases}}\)

Answer 2

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

Answer 3

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: \(x+\frac{3}{4}=\frac{1}{2}\)

<=> \(x=\frac{1}{2}-\frac{3}{4}\)

<=> \(x=-\frac{1}{4}\)

=> \(\orbr{\begin{cases}x=-\frac{3}{4}\\x=-\frac{1}{4}\end{cases}}\)

Answer 4

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...