(x^5)^2=x^18/x^7
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Mọi người tk mình đi mình đang bị âm nè!!!!!!
Ai tk mình mình tk lại nha !!!
a) 5 x 5 + 18
= 25 + 18
= 43
b) 5 x 7 + 23
= 35 + 23
= 58
c) 7 x 7 x 2
= 49 x 2
= 98
d) 15 – 6 x 2
= 15 – 12
= 3
(8 x 18 -5 x 18 -18 x3) x y + 2 x y =8 x 7 + 2
<=>144 x y -90 x y -54 x y + 2 x y = 58
<=>2 x y = 58
<=> y = 29
Vậy y = 29
Tìm y biết :
(8 x 18 - 5 x 18 - 18 x 3) x y + 2 x y = 8 x 7 + 2
[ 18 x ( 8 - 5 - 3 ) ] x y = 56 + 2
[ 18 x ( 8 - 5 - 3 ) ] x y = 58
( 18 x 0 ) x y = 58
0 x y + 2 x y = 58
2 x y = 58
y = 58 : 2
y = 29
a) x= 1/4
b) x= -7/144
c) x= 25/6
d) x=17/7
e) x= 80/63
f) x= -45/14
a, \(\dfrac{7}{18}x-\dfrac{2}{3}=\dfrac{5}{18}\)
\(\dfrac{7}{18}x=\dfrac{5}{18}+\dfrac{2}{3}\)
\(\dfrac{7}{18}x=\dfrac{17}{18}\)
\(x=\dfrac{17}{18}\div\dfrac{7}{18}\)
\(x=\dfrac{17}{7}\)
a) \(\dfrac{7}{18}\).x-\(\dfrac{2}{3}\)=\(\dfrac{5}{18}\)
\(\dfrac{7}{18}\).x =\(\dfrac{5}{18}\)+\(\dfrac{2}{3}\)
\(\dfrac{7}{18}\).x = \(\dfrac{17}{18}\)
x = \(\dfrac{17}{18}\) :\(\dfrac{7}{18}\)
x =\(\dfrac{17}{7}\)
b)\(\dfrac{4}{9}\) - \(\dfrac{7}{8}\).x =\(\dfrac{-2}{3}\)
\(\dfrac{7}{8}\).x =\(\dfrac{4}{9}\)-\(\dfrac{-2}{3}\)
\(\dfrac{7}{8}\).x =\(\dfrac{10}{9}\)
x =\(\dfrac{10}{9}\) : \(\dfrac{7}{8}\)
x =\(\dfrac{80}{63}\)
c)\(\dfrac{1}{6}\)+\(\dfrac{-5}{7}\): \(\dfrac{-7}{18}\)
\(\dfrac{1}{6}\)+\(\dfrac{90}{49}\)
\(\dfrac{589}{294}\)
(8 x 18 -5 x 18 -18 x3) x y + 2 x y =8 x 7 + 2
<=>144 x y -90 x y -54 x y + 2 x y = 58
<=>2 x y = 58
<=> y = 29
vậy y = 29
( mình ko chép lại đề bài đâu nha ,giải lun đó)
[18x(8-5-3)]xy+2xy=56+2
(18x0)xy+2xy=58
0xy+2xy=58
2xy=58
y=58:2=29
tick cho mình nha
(x5)2=x18x7(đk:x≠0)(x5)2=x18x7(đk:x≠0)
⇒x10=x11⇒x10=x11
⇒x11−x10=0⇒x11−x10=0
⇒x10(x−1)=0⇒x10(x−1)=0
⇒[x=0(ktm)x=1(tm)
(x4)3=x18x7(x4)3=x18x7
⇔ x12=x11x12=x11
⇔ x12−x11=0x12−x11=0
⇔ x11.(x−1)=0x11.(x−1)=0
⇔ [x11=0x−1=0[x11=0x−1=0
⇔ [x=0x=0+1=1[x=0x=0+1=1
Vậy xx ∈ {0;1}