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: \(x+\frac{3}{4}=\frac{1}{2}\)

<=> \(x=\frac{1}{2}-\frac{3}{4}\)

<=> \(x=-\frac{1}{4}\)

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

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

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

=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)

b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c) TT

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

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

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

=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)

\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)

=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)

=> \(\left|50x-140\right|=\left|25x+24\right|\)

=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)

=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)

Bài 2 : a. |2x - 5| = x + 1

 TH1 : 2x - 5 = x + 1

    => 2x - 5 - x = 1

    => 2x - x - 5 = 1

    => 2x - x = 6

    => x = 6

TH2 : -2x + 5 = x + 1

   => -2x + 5 - x = 1

   => -2x - x + 5 = 1

   => -3x = -4

   => x = 4/3

Ba bài còn lại tương tự

cho mình

\(\Leftrightarrow\hept{\begin{cases}2x=0\\x-\frac{1}{7}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0:2\\x=0+\frac{1}{7}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=0\\x=\frac{1}{7}\end{cases}}\)

2x(x-1/7)=0

=> 2x=0 hoặc x-1/7=0

=> x=0 hoặc x=1/7

vậy x=0 hoặc x=1/7

tk mk nha

a/ Đặt \(\hept{\begin{cases}\frac{x+1}{x-2}=a\\\frac{x+1}{x-4}=b\end{cases}}\) thì có

\(a^2+b-\frac{12b^2}{a^2}=0\)

\(\Leftrightarrow\left(a^2-3b\right)\left(a^2+4b\right)=0\)

b/ \(2x^2+3xy-2y^2=7\)

\(\Leftrightarrow\left(2x-y\right)\left(x+2y\right)=7\)

\(2x\left(x-\frac{1}{7}\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x=0\\x-\frac{1}{7}=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{7}\end{array}\right.\)

Vậy \(x\in\left\{0;\frac{1}{7}\right\}\)

Chúc bạn học tốt .