Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(\Leftrightarrow x^2=\dfrac{-5}{2}\cdot\dfrac{-10}{9}=\dfrac{50}{18}=\dfrac{25}{9}\)
=>x=5/3hoặc x=-5/3
c: \(\Leftrightarrow4\left(x-\dfrac{5}{8}\right)=\dfrac{1}{4}+\dfrac{3}{4}=1\)
=>x-5/8=1/4
hay x=2/8+5/8=7/8
d: \(\Leftrightarrow\left|x-3\right|=\dfrac{2}{5}+\dfrac{3}{5}=1\)
=>x-3=1 hoặc x-3=-1
=>x=4 hoặc x=2
e: =>1-1/2x=-3
=>1/2x=4
hay x=8
bài này áp dụng quy tắc nhân chéo nha :vv
a) \(\dfrac{x}{5}=\dfrac{2}{5}\Leftrightarrow5x=2.5=10\Leftrightarrow x=\dfrac{10}{5}=2\)
b) \(\dfrac{3}{8}=\dfrac{6}{x}\Leftrightarrow3x=6.8=48\Leftrightarrow x=\dfrac{48}{3}=16\)
c)\(\dfrac{1}{9}=\dfrac{x}{27}\Leftrightarrow9x=27\Leftrightarrow x=\dfrac{27}{9}=3\)
d)\(\dfrac{4}{x}=\dfrac{8}{6}\Leftrightarrow8x=4.6=24\Leftrightarrow x=\dfrac{24}{8}=3\)
e) \(\dfrac{3}{x-5}=\dfrac{-4}{x+2}\Leftrightarrow3\left(x+2\right)=-4\left(x-5\right)\\ \Leftrightarrow3x+4x=-6+20\\ \Leftrightarrow7x=14\Leftrightarrow x=\dfrac{14}{7}=2\)
g) \(\dfrac{x}{-2}=\dfrac{-8}{x}\Leftrightarrow x^2=\left(-2\right)\left(-8\right)=16\\ \Rightarrow x=\pm4\)
a , \(\left(\dfrac{-2}{3}+1\dfrac{1}{4}-\dfrac{1}{6}\right):\dfrac{-24}{10}\)
=\(\left(\dfrac{-2}{3}+\dfrac{5}{4}-\dfrac{1}{6}\right):\dfrac{-12}{5}\)
=\(\left(\dfrac{-8}{12}+\dfrac{15}{12}-\dfrac{2}{12}\right)\cdot\dfrac{-5}{12}\)
=\(\dfrac{5}{12}\cdot\dfrac{-5}{12}=\dfrac{-25}{144}\)
b , \(\dfrac{13}{15}\cdot0,25\cdot3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right)1\dfrac{23}{24}\)
=\(\dfrac{13}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right)\cdot\dfrac{57}{24}\)
=\(\dfrac{13}{20}-\dfrac{47}{60}\cdot\dfrac{57}{24}\)
=\(\dfrac{13}{20}-\dfrac{893}{480}=\dfrac{312}{480}-\dfrac{893}{480}=\dfrac{-581}{480}\)
c , \(\left(\dfrac{12}{32}+\dfrac{5}{-20}-\dfrac{10}{24}\right):\dfrac{2}{3}\)
=\(\left(\dfrac{180}{480}-\dfrac{120}{480}-\dfrac{200}{480}\right)\cdot\dfrac{3}{2}\)
= \(\dfrac{-7}{24}\cdot\dfrac{3}{2}=\dfrac{-7}{16}\)
d , \(4\dfrac{1}{2}:\left(2,5-3\dfrac{3}{4}\right)+\left(-\dfrac{1}{2}\right)\)
=\(\dfrac{9}{2}:\left(\dfrac{5}{2}-\dfrac{15}{4}\right)-\dfrac{1}{2}\)
=\(\dfrac{9}{2}:\dfrac{-5}{4}-\dfrac{1}{2}=\dfrac{9}{2}\cdot\dfrac{-4}{5}-\dfrac{1}{2}=\dfrac{-18}{5}-\dfrac{1}{2}=\dfrac{-41}{10}\)
e , \(\dfrac{-5}{2}:\left(\dfrac{3}{4}-\dfrac{1}{2}\right)=\dfrac{-5}{2}\left(\dfrac{3}{4}-\dfrac{2}{4}\right)\)
=\(\dfrac{-5}{2}:\dfrac{1}{4}=\dfrac{-5}{2}\cdot4=-10\)
Bạn ơi ; tách từng bài ra cho dễ làm :
1.7C-C= 7^2016-7
C = ( 7^2016-7 ) :6
\(C=7+7^2+7^3+.....+7^{2016}\)
\(\Rightarrow7C=7^2+7^3+7^4+...+7^{2017}\)
\(\Rightarrow7C-C=\left(7^2+7^3+.....+7^{2017}\right)-\left(7+7^2+7^3+....+7^{2016}\right)\)
\(\Rightarrow6C=2^{2017}-7\)
\(\Rightarrow C=\frac{2^{2017}-7}{6}\)
Bài 1:
a, \(\left(x-2\right)^2=9\)
\(\Rightarrow x-2\in\left\{-3;3\right\}\Rightarrow x\in\left\{-1;5\right\}\)
b, \(\left(3x-1\right)^3=-8\)
\(\Rightarrow3x-1=-2\Rightarrow3x=-1\)
\(\Rightarrow x=-\dfrac{1}{3}\)
c, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow x+\dfrac{1}{2}\in\left\{-\dfrac{1}{4};\dfrac{1}{4}\right\}\)
\(\Rightarrow x\in\left\{-\dfrac{3}{4};-\dfrac{1}{4}\right\}\)
d, \(\left(\dfrac{2}{3}\right)^x=\dfrac{4}{9}\)
\(\Rightarrow\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^2\)
Vì \(\dfrac{2}{3}\ne\pm1;\dfrac{2}{3}\ne0\) nên \(x=2\)
e, \(\left(\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{16}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{x-1}=\left(\dfrac{1}{2}\right)^4\)
Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(x-1=4\Rightarrow x=5\)
f, \(\left(\dfrac{1}{2}\right)^{2x-1}=8\) \(\Rightarrow\left(\dfrac{1}{2}\right)^{2x-1}=\left(\dfrac{1}{2}\right)^{-3}\) Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(2x-1=-3\) \(\Rightarrow2x=-2\Rightarrow x=-1\) Chúc bạn học tốt!!!
tớ sẽ ko làm cách chuyển vế đâu nha!
mong thông cảm
câu 1:
x = -3/4 - 25/100
x = -1
câu 2:
x = -25/10 . 13/3
x = -65/6
a;\(\dfrac{-6}{11}\) : \(\dfrac{12}{55}\) = \(\dfrac{-5}{2}\)
b;\(\dfrac{7}{12}\) + \(\dfrac{5}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{47}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{25}{72}\)
c;\(\dfrac{13}{10}\) : \(\dfrac{-5}{13}\) = \(\dfrac{-169}{50}\)
d; {\(\dfrac{5}{12}\) + \(\dfrac{5}{11}\) } : { \(\dfrac{5}{3}\) -\(\dfrac{7}{11}\) } = \(\dfrac{115}{132}\) : \(\dfrac{34}{33}\) = \(\dfrac{115}{136}\)
lưu ý mk ko chép đầu bài
mình cần gấp lắm đến chiều mai là phải nộp rùi
giúp mình nha thanks cá bạn trước 
ko có tâm trạng mà cười nữa
bài 1:
a) \(4\dfrac{1}{2}x:\dfrac{5}{12}=0,5\) ; b)\(1,5+1\dfrac{1}{4}x=\dfrac{2}{3}\)
\(\dfrac{9}{2}x:\dfrac{5}{12}=\dfrac{1}{2}\) \(\dfrac{3}{2}+\dfrac{5}{4}x=\dfrac{2}{3}\)
\(\dfrac{9}{2}x\) \(=\dfrac{1}{2}.\dfrac{5}{12}\) \(\dfrac{5}{4}x=\dfrac{2}{3}-\dfrac{3}{2}\)
\(\dfrac{9}{2}x\) \(=\dfrac{5}{24}\) \(\dfrac{5}{4}x=\dfrac{-5}{6}\)
\(x\) \(=\dfrac{5}{24}:\dfrac{9}{2}\) \(x=\dfrac{-5}{6}:\dfrac{5}{4}\)
\(x\) \(=\dfrac{5}{108}\) \(x=\dfrac{-2}{3}\)
c) Cho mình hỏi x ở đâu vậy ???
d)\(\left(x-5\right):\dfrac{1}{3}=\dfrac{2}{5}\) e)\(\left(4,5-2x\right):\dfrac{3}{4}=1\dfrac{1}{3}\)
\(\left(x-5\right)\) \(=\dfrac{2}{5}.\dfrac{1}{3}\) \(\left(\dfrac{9}{2}-2x\right):\dfrac{3}{4}=\dfrac{4}{3}\)
\(x-5\) \(=\dfrac{2}{15}\) \(\dfrac{9}{2}-2x\) =\(\dfrac{4}{3}.\dfrac{3}{4}\)
\(x\) \(=\dfrac{2}{15}+5\) \(\dfrac{9}{2}-2x=1\)
\(x\) \(=\dfrac{77}{15}\) \(2x=\dfrac{9}{2}-1\)
f) \(\left(2,7x-1\dfrac{1}{2}x\right):\dfrac{2}{7}=\dfrac{-21}{7}\) \(2x=\dfrac{7}{2}\)
\(\left(\dfrac{27}{10}x-\dfrac{3}{2}x\right):\dfrac{2}{7}=-3\) \(x=\dfrac{7}{2}:2\)
\(\left[x\left(\dfrac{27}{10}-\dfrac{3}{2}\right)\right]=-3.\dfrac{2}{7}\) \(x=\dfrac{7}{4}\)
\(x.\dfrac{6}{5}=\dfrac{-6}{7}\)
\(x=\dfrac{-6}{7}:\dfrac{6}{5}\)
\(x=\dfrac{-5}{7}\)
bài 2:
Theo bài ra ta có :\(\dfrac{a}{27}=\dfrac{-5}{9}=\dfrac{-45}{b}\)
\(\Rightarrow9a=27.\left(-5\right)\Rightarrow a=\dfrac{27.\left(-5\right)}{9}=-15\)
\(\Rightarrow\left(-5\right)b=\left(-45\right).9\Rightarrow b=\dfrac{\left(-45\right).9}{-5}=81\)
Vậy \(a=-15;b=81\)
a,5^x=125
=>5^x=5^3
=>x=3
b,3^2x=81
=>3^2x=3^4
=>2x=4
=>x=4:2=2
c,5^2x-3-2*5^2=5^2+3
5^2x-3-50=75
5^2x-3=75+50=125
5^2x-3=5^3
=>2x-3=3
=>2x=3+3=6
=>6:2=3
k cho mk nhé
\(a,125=5\cdot5\cdot5=5^3\Leftrightarrow x=3\)
\(b,81=3\cdot3\cdot3\cdot3=3^4\Leftrightarrow2x=4\Leftrightarrow x=4:2\Leftrightarrow x=2\)
\(c,5^{2x-3}-2\cdot5^2=5^2\cdot3\)
\(\Leftrightarrow5^{2x-3}=2\cdot5^2+5^2\cdot3\)
\(\Leftrightarrow5^{2x-3}=5^2\cdot\left(2+3\right)\)
\(\Leftrightarrow5^{2x-3}=5^2\cdot5\Leftrightarrow5^{2x-3}=5^3\)
\(\Leftrightarrow2x-3=3\Leftrightarrow2x=3+3\Leftrightarrow2x=6\Leftrightarrow x=6:2\Leftrightarrow x=3\)
`a)x^3=343=7^3`
`=>x=7`
Vậy `x=7`
`b)(x-2,5)^4=(x-2,5)^2`
`=>(x-2,5)^2[(x-2,5)^2-1]=0`
`+)(x-2,5)^2=0<=>x=2,5`
`+)(x-2,5)^2=1`
`TH1:x-2,5=1<=>x=3,5`
`th2:x-2,5=-1<=>x=1,5`
Vậy `x=0` hoặc `x=1,5` hoặc `x=3,5
`c)x^8/243=27`
`=>x^8=27.243`
`=>x^8=3^3*3^5=3^8`
`=>x=+-3`
siêu tốc
Giải:
a) \(x^3=343\)
\(\Rightarrow x^3=7^3\)
\(\Rightarrow x=7\)
b) \(\left(x-2,5\right)^4=\left(x-2,5\right)^2\)
Vì có cùng cơ số nhưng lại khác mũ thì \(\left(x-2,5\right)\in\left\{0;1\right\}\)
TH1:
\(\Rightarrow x-2,5=0\)
\(\Rightarrow x=2,5\)
TH2:
\(\Rightarrow\left[{}\begin{matrix}x-2,5=1\\x-2,5=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3,5\\x=1,5\end{matrix}\right.\)
c) \(\dfrac{x^8}{243}=27\)
\(\Rightarrow x^8=3^3.3^5\)
\(\Rightarrow x^8=3^8\)
\(\Rightarrow x=\pm3\)