tìm x
a) |3/2 -1/3 | *-6/7x-7/5=25/7*21/5
b) (x-3)^2 + -9/25= 2/5 * 8/5
c)( x-1/2)^2 – 9/16= 1
d) 3-x/ 5-x= ( -3/5)^2
tìm x a) |3/2 -1/3 | *-6/7x-7/5=25/7*21/5 b) (x-3)^2 + -9/25= 2/5 * 8/5 c)( x-1/2)^2 – 9/16= 1 d) 3-x/ 5-x= ( -3/5)^2
By Elliana
Đáp án:
Giải thích các bước giải:
a) | $\frac{3}{2}$ – $\frac{1}{3}$ | . $\frac{-6}{7x}$ -$\frac{7}{5}$ = $\frac{25}{7}$ . $\frac{21}{5}$
<=> | $\frac{7}{6}$ | . $\frac{-6}{7x}$ – $\frac{7}{5}$ = 15
<=> $\frac{7}{6}$ . $\frac{-6}{7x}$ = 15 + $\frac{7}{5}$
<=> $\frac{-1}{x}$ = $\frac{82}{5}$
<=> $\frac{-82}{82x}$ = $\frac{82}{5}$
=> -82x = 5
=> x = $\frac{-5}{82}$
b) $(x-3)^{2}$ +$\frac{-9}{25}$ =$\frac{2}{5}$ . $\frac{8}{5}$
<=> $(x-3)^{2}$ +$\frac{-9}{25}$ = $\frac{16}{25}$
<=> $(x-3)^{2}$ = $\frac{16}{25}$ – $\frac{-9}{25}$ = $\frac{16}{25}$ + $\frac{9}{25}$
<=> $(x-3)^{2}$ = $\frac{25}{25}$ =1 = $1^{2}$ = $(-1)^{2}$
TH1 $(x-3)^{2}$ = $1^{2}$ => x-3 = 1 <=> x = 1 + 3 = 4
TH2 $(x-3)^{2}$ = $(-1)^{2}$ => x-3 = -1 <=> x = -1 + 3 = 2
Vậy x = { 2;4 }
c) $(x – \frac{1}{2}$)$^{2}$ – $\frac{9}{16}$ = 1
<=> $(x – \frac{1}{2}$)$^{2}$ = 1 + $\frac{9}{16}$ = $\frac{25}{16}$
<=> $(x – \frac{1}{2}$)$^{2}$ = $\frac{5^{2}}{4^{2}}$ = $\frac{(-5)^{2}}{(-4)^{2}}$
TH1 $(x – \frac{1}{2}$)$^{2}$ = $\frac{5^{2}}{4^{2}}$
<=> x – $\frac{1}{2}$ =$\frac{5}{4}$
<=> x = $\frac{5}{4}$ + $\frac{1}{2}$ = $\frac{7}{4}$
TH2 $(x – \frac{1}{2}$)$^{2}$ = $\frac{(-5)^{2}}{(-4)^{2}}$ = $\frac{5^{2}}{4^{2}}$
<=> x – $\frac{1}{2}$ =$\frac{5}{4}$
<=> x = $\frac{5}{4}$ + $\frac{1}{2}$ = $\frac{7}{4}$
Vậy x = $\frac{7}{4}$
d) $\frac{3-x}{5-x}$ = ($\frac{-3}{5}$) $^{2}$ = $\frac{9}{25}$
=> $\frac{25.(3-x)}{25.(5-x)}$ = $\frac{9.(5-x)}{25.(5-x)}$
=> 25.(3-x) = 9. (5-x)
<=> 75-25x = 45 – 9x
<=> 75-45 = -9x + 25x = 16x
<=> 30 = 16x
=> x = $\frac{30}{16}$ = $\frac{15}{8}$
Vậy..
1+1=2