a/ x = (2015 . 2016) : 2015 =2016

b/ 503.5x + 2014 . 2013 = 106742 + 2014 . 2013 => 503.5x = 106742 => x = 212

c/ 71x + 260x = x+ 140 + 260x => 70x = 140 => x= 2

d/ x = 583 - 237 = 346

e/ x = 575 + 836 = 1411

g/ x = 3692 : 142 = 26

h/ x = 8928 : 124 = 72

i/ x = 168 . 34 = 5712

g/ x = 3692 : 142 = 26

\(a.\left(x\times0,25+1999\right)\times2000=2052\times2000\\ \left(x\times0,25+1999\right)\times2000=4104000\\ x\times0,25+1999=4104000:2000\\ x\times0,25+1999=2052\\ x\times0,25=2052-1999\\ x\times0,25=53\\ x=53:0,25\\ x=212\)

( X x 0,25+ 1999) x2000= ( 53+1999) x2000

( X x 0,25+ 1999)          = ( 53+1999)x2000:2000

( X x 0,25+1999)            =( 53+1999)

( X x 0,25+1999)             = 2052

X+0,25                           = 2052 - 1999

X+ 0,25                             = 53

X                                       = 53:0,25

X                                      = 212

71+65 x4 = x+140/x  + 260

x+140/x + 260= 71+65x4

x+140/x + 260=71 + 260

x+140/x=71

x/x +140/x=71

1+ 140/x=71

140/x=71-1

140/x=70

      x=140/70

      x=2

k nha linh

( x + 1 ) + ( x + 4 ) + ( x + 7 ) + ( x + 10) + ..........+ ( x + 28 ) = 155

10x + ( 1 + 4 + 7 + ... + 28 )    = 155

10x + 145                              = 155

10x                                          = 155 - 145

10x                                          = 10

x                                              = 10 : 10

Vậy x = 1

\(a,\left[x+1\right]+\left[x+4\right]+\left[x+7\right]+\left[x+10\right]+...+\left[x+28\right]=155\)

\(\Leftrightarrow x+1+x+4+x+7+x+10+...+x+28=155\)

\(\Leftrightarrow(x+x+x+...+x)+(1+4+7+...+28)=155\)

\(\Leftrightarrow10x+145=155\)

\(\Leftrightarrow10x=155-145\)

\(\Leftrightarrow10x=10\)

\(\Leftrightarrow x=1\)

\(b)(x\times0,25+1999)\times2000=(53+1999)\times2000\)

\(\Leftrightarrow(x\times0,25+1999)\times2000=4104000\)

\(\Leftrightarrow(x\times0,25+1999)=4104000:2000\)

\(\Leftrightarrow(x\times0,25+1999)=2052\)

\(\Leftrightarrow x\times0,25=2052-1999\)

\(\Leftrightarrow x\times0,25=53\)

\(\Leftrightarrow x=53:0,25=212\)

Câu c tự làm

còn câu c nữa giúp mình với đi các bạn.

\(x-2014-\frac{2015}{2013}+x-2013-\frac{2015}{2014}+x-2014-\frac{2013}{2015}=3\)

\(\Rightarrow\left(x+x+x\right)+\left(-2014-2014\right)-2013-\frac{2015}{2013}-\frac{2015}{2014}-\frac{2013}{2015}=3\)

\(3x-2013-\frac{2015}{2013}-\frac{2015}{2014}-\frac{2013}{2015}=3\)

\(3x=3+2013+\frac{2015}{2013}+\frac{2015}{2014}+\frac{2013}{2015}\)

bạn ơi bài này số lớn quá bạn sử dungjmays tính rồi tự tính nhé

Đáp án của bạn Hoàng Đình Đại sai rùi nhưng dù sao cx cảm ơn nhiều

a)(x.0,25+1999)x2000=(53+1999)x2000

(x.0,25+1999)=53+1999

x.0,25=53

x=53x4

x=210

a)(x.0,25+1999)x2000=(53+1999)x2000

(x.0,25+1999)=53+1999

x.0,25=53

x=53x4

x=210

k mik đi

(x .0,25 + 1999) . 2000 = (53+1999) .2000

(x . 0,25 +1999 )= (53 +1999)

x. 0,25= 53

x         = 53 :0, 25

x         = 212

dau cham la dau nhan

Đặt \(\sqrt{x-2013}=a\left(a>0\right)\)

\(\sqrt{y-2014}=b\left(b>0\right)\)

\(\sqrt{z-2015}=c\left(c>0\right)\)

Có \(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)

<=> \(\frac{a-1}{a^2}-\frac{1}{4}+\frac{b-1}{b^2}-\frac{1}{4}+\frac{c-1}{c^2}-\frac{1}{4}=0\)

<=> \(\frac{4a-4-a^2}{4.a^2}+\frac{4b-4-b^2}{4b^2}+\frac{4c-4+c^2}{4c^2}=0\)

<=>\(\frac{-\left(a^2-4a+4\right)}{4a^2}-\frac{b^2-4b+4}{4b^2}-\frac{c^2-4c+4}{4c^2}=0\)

<=> \(\frac{\left(a-2\right)^2}{4a^2}+\frac{\left(b-2\right)^2}{4b^2}+\frac{\left(c-2\right)^2}{4c^2}=0\).

Có \(\frac{\left(a-2\right)^2}{4a^2}\ge0\forall a>0\)

\(\frac{\left(b-2\right)^2}{4b^2}\ge0\forall b>0\)

\(\frac{\left(c-2\right)^2}{4c^2}\ge0\forall c>0\)

=> \(\frac{\left(a-2\right)^2}{4a^2}+\frac{\left(b-2\right)^2}{4b^2}+\frac{\left(c-2\right)^2}{4c^2}\ge0\) với moi a,b,c >0

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}a-2=0\\b-2=0\\c-2=0\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}a=2\\b=2\\c=2\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}\sqrt{x-2013}=2\\\sqrt{y-2014}=2\\\sqrt{z-2015}=2\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x-2013=4\\y-2014=4\\z-2015=4\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x=2017\\y=2018\\z=2019\end{matrix}\right.\)(t/m)

Vậy \(\left(x,y,z\right)\in\left\{\left(2017,2018,2019\right)\right\}\)

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