Tìm x ( Giải ra nhoa ) ♔ ♔ ♔
a) \(x\) x 2015 = 2015 x 2016
b( \(x\) x 0,25 +2013 ) x 2014 = ( 53 + 2013 ) x 2014
c) 71 + 65 x 4 = \(\frac{x+140}{x}\)+ 260
Giúp mình nhoa
Tìm x
a) \(x\) x 2015 = 2015 x 2016
b) (\(x\) x 0,25 +2013 ) x 2014 = ( 53 + 2013 ) x 2014
c) 71 + 65 x 4 = \(\frac{x+140}{x}\)+ 260
d) 583 - \(x\) = 237
e) \(x\)- 836 = 575
g) 142 x \(x\)= 3692
h) 8928 : \(x\) = 124
i) \(x\) : 34 = 168
g) 142 x \(x\) = 3692
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
1, TÌM SỐ CÓ 2 CHỮ SỐ , BIẾT RẰNG SỐ ĐÓ GẤP 8 LẦN TỔNG CÁC CHỮ SỐ CỦA NÓ ?
2, Tìm X, biết :
A, ( X . 0,25 + 2014 ) . 2015 = ( 53 + 2014 ) . 2016
B, 71 + 65 . 4 = X + 140 TRÊN X + 260
Tìm x:
a) (X x 0,25 + 1999) x 2000 = ( 53 + 1999 ) x 2000
b) 71 + 65 x 4 = \(\dfrac{x+140}{x}\) + 260
\(a.\left(x\times0,25+1999\right)\times2000=2052\times2000\\ \left(x\times0,25+1999\right)\times2000=4104000\\ x\times0,25+1999=4104000:2000\\ x\times0,25+1999=2052\\ x\times0,25=2052-1999\\ x\times0,25=53\\ x=53:0,25\\ x=212\)
tìm x
( x * 0,25 + 1999 ) * 2000 = ( 53 + 1999 ) * 2000
71 + 65 * 4 = \(\frac{x+140}{x}\) + 260
( 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
1 tìm x
a , ( x + 1 ) + ( x + 4 ) + ( x + 7 ) + ( x + 10) + ..........+ ( x + 28 ) = 155
b, ( x x 0,25 + 1999 ) x 2000 = ( 53 + 1999 ) x 2000
c, 71 + 65 x 4 = \(\frac{x+140}{x}\)x 260
( 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
Tìm x biết:
x-2014-2015/2013 + x-2013-2015/2014 + x-2014-2013/2015=3
\(x-2014-\frac{2015}{2013}+x-2013-\frac{2015}{2014}+x-2014-\frac{2013}{2015}=3\)
\(\Rightarrow\left(x+x+x\right)+\left(-2014-2014\right)-2013-\frac{2015}{2013}-\frac{2015}{2014}-\frac{2013}{2015}=3\)
\(3x-2013-\frac{2015}{2013}-\frac{2015}{2014}-\frac{2013}{2015}=3\)
\(3x=3+2013+\frac{2015}{2013}+\frac{2015}{2014}+\frac{2013}{2015}\)
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
tìm x
A) (\(\chi\)x 0,25 + 1999) x 2000= (53+ 1999) x 2000
b) 71 + 65 x 4=\(\frac{\chi+140}{\chi}\)+ 260.
Ai biết làm thì giúp mik nhé
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
a, x+1/2013+x+1/2014+x+1/2015=x+1/2016+x+1/2017
b,x-1/2013+x-2/2014+x-3/2015=x-4/2016-2
Giải phương trình:
\(\frac{\sqrt{x-2013}-1}{x-2013}+\frac{\sqrt{y-2014}-1}{y-2014}+\frac{\sqrt{z-2015}-1}{z-2015}=\frac{3}{4}\)
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 \(\left(x,y,z\right)\in\left\{\left(2017,2018,2019\right)\right\}\)