( x * 0,25 + 2012 ) * 2013 = ( 50 + 2012 ) * 2013
ai giải đước tớ sẽ tích 3 cái
Tìm x
(x×0,25+2012)×2013=(50+2012)×2013
x:0,25+x×6=40
tim x nha mn
l,[12 nhan 15 - x] nhan 1/4= 120 nhan 1/4
i, [x nhan 0,25 cong 2012 ] nhan 2013 =[50 cong 2012] nhan 2013
k,[ x - 1/2] nhan 5/3 = 7/4 - 1/2
\(\left[12\cdot15-x\right]\cdot\frac{1}{4}=120\cdot\frac{1}{4}\)
\(\Leftrightarrow\left[180-x\right]\cdot\frac{1}{4}=30\)
\(\Leftrightarrow180-x=30:\frac{1}{4}\)
\(\Leftrightarrow180-x=120\)
\(\Leftrightarrow x=60\)
\(\left[x\cdot0,25+2012\right]\cdot2013=\left[50+2012\right]\cdot2013\)
\(\Leftrightarrow x\cdot0,25\cdot2013+2012\cdot2013=2062\cdot2013\)
\(\Leftrightarrow x\cdot\frac{2013}{4}+4050156=4150806\)
\(\Leftrightarrow x\cdot\frac{2013}{4}=100650\)
\(\Leftrightarrow x=100650:\frac{2013}{4}=100650\cdot\frac{4}{2013}=200\)
mong các bạn giúp mik rồi mik tik cho
e) 17/5 : X = 34/5 : 4/3
f) X : 4/5= 25/8 : 5/4
g) (X x 0,25 + 2012) x 2013 = (50 + 2012) x 2013
h) (X - 1/2) x 5/3 = 7/4 - 1/
e: \(\dfrac{17}{5}:x=\dfrac{34}{5}:\dfrac{4}{3}=\dfrac{34}{5}\cdot\dfrac{3}{4}=\dfrac{102}{20}=\dfrac{51}{10}\)
\(\Leftrightarrow x=\dfrac{17}{5}:\dfrac{51}{10}=\dfrac{17}{5}\cdot\dfrac{10}{51}=\dfrac{10}{5}\cdot\dfrac{17}{51}=2\cdot\dfrac{1}{3}=\dfrac{2}{3}\)
f: \(x:\dfrac{4}{5}=\dfrac{25}{8}:\dfrac{5}{4}=\dfrac{25}{8}\cdot\dfrac{4}{5}=\dfrac{100}{40}=\dfrac{5}{2}\)
\(\Leftrightarrow x=\dfrac{5}{2}\cdot\dfrac{4}{5}=\dfrac{4}{2}=2\)
g: \(\left(0.25x+2012\right)\cdot2013=\left(50+2012\right)\cdot2013\)
\(\Leftrightarrow0.25x+2012=50+2012\)
\(\Leftrightarrow0.25x=50\)
hay x=200
h: \(\left(x-\dfrac{1}{2}\right)\cdot\dfrac{5}{3}=\dfrac{7}{4}-1=\dfrac{3}{4}\)
\(\Leftrightarrow x-\dfrac{1}{2}=\dfrac{3}{4}:\dfrac{5}{3}=\dfrac{9}{20}\)
\(\Leftrightarrow x=\dfrac{9}{20}+\dfrac{1}{2}=\dfrac{9}{20}+\dfrac{10}{20}=\dfrac{19}{20}\)
các bạn giúp mik được ko mik xin đó mik đang cần gấp nhé
e) 17/5 : X = 34/5 : 4/3
f) X : 4/5= 25/8 : 5/4
g) (X x 0,25 + 2012) x 2013 = (50 + 2012) x 2013
h) (X - 1/2) x 5/3 = 7/4 - 1/
e)175:x=345:43
3,4:x=6,8:43
3,4:x=5,1
x=3,4:5,1
x=23
f)x:45 =258:54
x:45=52
x=52×45
x=2
g)(x×0,25+2012)×2013=(50+2012)×2013g)
(x×0,25+2012)×2013=2062×2013
x×0,25+2012=(2062×2013):2013
x×0,25+2012=2062
x×0,25=2062−2012
x×0,25=50
x=50:0,25
x=200
h)(x−12)×53=74−12h)
(x−12)×53=54
x−12=54:53
x−12=34
x=34+12
x=54
Cho A = 2012 x 2012 x 2012 x ... x 2012 x 2012 ( 2013 thừa số 2012 )
B = 2013 x 2013 x 2013 x ... x2013 x 2013 ( 2012 thừa số 2013 )
Hỏi ( A + B ) chia cho 5 sẽ dư mấy ?
dư 0
vì ta thấy:
2010+2013=2015 chia hết cho 5
nên A+B chia 5 dư 0
tíc mình nha
Ta có :
2012 x 2012 x 2012 x ... x 2012 x 2012 ( 2013 chữ số 2012 ) = 2012 ^ 2013 => chữ số tận cùng là 8 (1)
2013 x 2013 x 2013 x ... x 2013 x 2013 ( 2012 chữ số 2013 ) = 2013 ^ 2012 => chữ số tận cùng là 9 (2)
Từ (1) và (2) => ( 8 + 9 ) : 5 = 17 : 5 dư 2
(X*0.25+2012)*2013=(50+2012)*2013
(0,25X + 2012) x 2013 = (50 + 2012) x 2013
0,25X + 2012 = 50 + 2012
0,25X = 50
X = \(\frac{50}{0,25}=200\)
Giải phương trình
X -3/2012+X-2/2013=X-2013/2+X-2012/3
Giải ra dùm mình luôn nha
Bạn tham khảo nhé :
Ta có :
\(\frac{x-3}{2012}+\frac{x-2}{2013}=\frac{x-2013}{2}+\frac{x-2012}{3}\)
\(\Leftrightarrow\)\(\left(\frac{x-3}{2012}-1\right)+\left(\frac{x-2}{2013}-1\right)=\left(\frac{x-2013}{2}-1\right)+\left(\frac{x-2012}{3}-1\right)\)
\(\Leftrightarrow\)\(\frac{x-2015}{2012}+\frac{x-2015}{2013}=\frac{x-2015}{2}+\frac{x-2015}{3}\)
\(\Leftrightarrow\)\(\frac{x-2015}{2012}+\frac{x-2015}{2013}-\frac{x-2015}{2}-\frac{x-2015}{3}=0\)
\(\Leftrightarrow\)\(\left(x-2015\right)\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2}-\frac{1}{3}\right)=0\)
Mà \(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2}-\frac{1}{3}\ne0\)
\(\Rightarrow\)\(x-2015=0\)
\(\Rightarrow\)\(x=2015\)
Vậy \(x=2015\)
Chsuc bạn học tốt ~
chuyển vế t cs
x+x-x-x=....
0x=.....
Suy ra vô nghiệm
(=) [(x-3/2012)-1]+[(x-2/2013)-1]= [(x-2013/2)-1]+[(x-2012/3)-1]
(=) x-2015/2012 . +. x-2015/2013. = x-2015/2. +. x-2015/3
(=)( x-2015 ) × (1/2012 + . 1/2013. -. 1/2 . - 1/3 .) =0
Mà 1/2012+1/2013-1/2-1/3≠0
=) x-2015 =0
(=) x=2015
Giải phương trình:
\(\dfrac{\sqrt{x-2012}-1}{x-2012}+\dfrac{\sqrt{y-2013}-1}{y-2013}+\dfrac{\sqrt{z-2014}-1}{z-2014}=\dfrac{3}{4}\)
Điều kiện: \(x\ge2012;y\ge2013;z\ge2014\)
Áp dụng bất đẳng thức Cauchy, ta có:
\(\left\{{}\begin{matrix}\dfrac{\sqrt{x-2012}-1}{x-2012}=\dfrac{\sqrt{4\left(x-2012\right)}-2}{2\left(x-2012\right)}\le\dfrac{\dfrac{4+x-2012}{2}-2}{2\left(x-2012\right)}=\dfrac{1}{4}\\\dfrac{\sqrt{y-2013}-1}{y-2013}=\dfrac{\sqrt{4\left(y-2013\right)}-2}{2\left(y-2013\right)}\le\dfrac{\dfrac{4+y-2013}{2}-2}{2\left(y-2013\right)}=\dfrac{1}{4}\\\dfrac{\sqrt{z-2014}-1}{z-2014}=\dfrac{\sqrt{4\left(z-2014\right)}-2}{2\left(z-2014\right)}\le\dfrac{\dfrac{4+z-2014}{2}-2}{2\left(z-2014\right)}=\dfrac{1}{4}\end{matrix}\right.\)
Cộng vế theo vế, ta được:
\(\dfrac{\sqrt{x-2012}-1}{x-2012}+\dfrac{\sqrt{y-2013}-1}{y-2013}+\dfrac{\sqrt{z-2014}-1}{z-2014}\le\dfrac{3}{4}\)
Đẳng thức xảy ra khi \(x=2016;y=2017;z=2018\)
Vậy....
\(\frac{x-1}{2013}\)+\(\frac{x-2}{2012}\)+\(\frac{x+3}{2013}\)+...+\(\frac{x-2012}{2}\)= 2012 ;
giải phương trình