Tìm x biết:
\(a\)) \(\left(19x+2.5^2\right):14=\left(13-8\right)^2-4^2\)
\(b\)) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\)
\(c\)) \(11-\left(-53+x\right)=97\)
\(\text{Tìm x, biết:}\)
\(a\)) \(\left(19x+2.5^2\right):14=\left(13-8\right)^2-4^2\)
\(b\)) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\)
\(c\)) \(11-\left(-53+x\right)=97\)
\(d\)) \(-\left(x+84\right)+213=-16\)
Tìm x,biết :
a)\(\left(19x+2.5^2\right):14=\left(13-8\right)^2-4^2\)
b)x+(x+1)+(x+2)+...+(x+30)=1240
c)11-(-53+x)=97
d)-(x+84)+213=-16
a) \(\left(19x+2\cdot5^2\right):14=\left(13-8\right)^2-4^2\)
\(\Leftrightarrow\left(19x+2\cdot25\right):14=5^2-4^2\)
\(\Leftrightarrow19x+50=\left(25-16\right)\cdot14\)
\(\Leftrightarrow19x+50=9\cdot14\)
\(\Leftrightarrow19x+50=126-50\)
\(\Leftrightarrow19x=76\)
\(\Leftrightarrow x=76:19\)
\(\Leftrightarrow x=4\)
b) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\)
\(\Leftrightarrow\left(x+x+x+x...+x\right)+\left(1+2+3+...+30\right)=1240\)
\(\Leftrightarrow31x+\frac{\left[\left(30-1\right):1+1\right]\cdot\left(30+1\right)}{2}=1240\)
\(\Leftrightarrow31x+465=1240\)
\(\Leftrightarrow31x=775\)
\(\Leftrightarrow x=25\)
c)\(11-\left(-53+x\right)=97\)
\(\Leftrightarrow-53+x=-86\)
\(\Leftrightarrow x=-33\)
d) \(-\left(x+84\right)+213=-16\)
\(\Leftrightarrow-\left(x+84\right)=-229\)
\(\Leftrightarrow x+84=229\)
\(\Leftrightarrow x=145\)
a) \(\left(19x+2.5^2\right):14=\left(13-8\right)^2-4^2\)
b) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\)
c) \(|x+7|=20+5.\left(-3\right)\)
a, (19x+2.52) : 14 = (13-8)2 - 42
(19x + 2.25) : 14 = 52 - 42
(19x + 50) : 14 = 25 - 16
(19x + 50) : 14 = 9
19 x + 50 = 9.14
19x + 50 = 126
19x = 126 - 50
19x = 76
x = 76 : 19
x = 4
vậy____
b) x + (x + 1) + (x + 2) + (x + 3)+.....+(x+30) = 1240
(x+x+x+...+x) + (1+2+3+...+30) = 1240
31x + 465 = 1240
31x = 1240 - 465
31x = 775
x = 775 : 31
x = 25
vậy____
c) |x + 7| = 20 + 5.(-3)
|x + 7| = 20 + (-15)
|x + 7| = 5
\(\Rightarrow\orbr{\begin{cases}x+7=5\\x+7=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5-7\\x=-5-7\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-2\\x=-12\end{cases}}\)
vậy_____
Bài 2 : Tìm x , biết :
a, \(\left(19x+2\cdot5^2\right)\text{ : }14=\left(13-8\right)^2-4^2\)
b, \(x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+30\right)=1240\)
c, \(11-\left(-53+x\right)=97\)
d, \(-\left(x+84\right)+213=-16\)
e, \(13-12+11+10-9+8-7-6+5-4+3+2-1=x\)
\(a,\left(19x+2.5^2\right):14=\left(13-8\right)^2-4^2\)
\(\Leftrightarrow\left(19x+50\right):14=5^2-4^2\)
\(\Leftrightarrow\left(19x+50\right):14=9\)
\(\Leftrightarrow19x+50=126\)
\(\Leftrightarrow19x=76\Leftrightarrow x=4\)
b) x + ( x + 1 ) + ( x + 2 ) + ... + ( x + 30 ) = 1240
x + x + 1 + x + 2 + ... + x + 30 = 1240
( x + x + ... + x ) + ( 1 + 2 + ... + 30 ) = 1240
Số số hạng là : ( 30 - 1 ) : 1 + 1 = 30 ( số )
Tổng là : ( 30 + 1 ) . 30 : 2 = 465
=> 31x + 465 = 1240
=> 31x = 775
=> x = 25
Vậy........
c) 11 - ( -53 + x ) = 97
11 + 53 - x = 97
64 - x = 97
x = 64 - 97
x = -33
Tìm x biết:\(\left(19x+2.5^2\right):14=\left(13-8\right)^2-4^2\)
\(\left(19x+2.5^2\right):14=\left(13-8\right)^2-4^2\)
\(\Rightarrow\left(19x+50\right):14=25-16\)
\(\Rightarrow19x+50=9.14=126\)
\(\Rightarrow19x=76\)
\(\Rightarrow x=76:19=4\)
(19x+2.5²):14 =(13-8)²-4²
(19x + 50):14 = 5²-4²
(19x + 50):14 = 9
19x + 50 = 9.14
19x+50 = 126
19x =76
x =76:19
x =4
Chúc các bn học tốt ah🥰🥰
a, \(\left(19x+2.5^2\right):14=\left(13-8\right)^2-4^2\) b, \(x+\left(x+1\right)+\left(x+2\right)+....+\left(x+30\right)=1240\)
tìm x,biết:
a)\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
b)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
c)\(\left(x+2\right)^2=\frac{38}{25}+\frac{9}{10}-\frac{11}{15}+\frac{13}{21}-\frac{15}{28}+\frac{17}{36}-...+\frac{197}{4851}-\frac{199}{4950}\)
giúp tớ với,huhu
Tìm x,
\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Leftrightarrow\frac{x}{\left(x+2\right)\left(x+14\right)}=\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}-\frac{1}{x+8}-\frac{1}{x+14}\)
\(\Leftrightarrow\frac{x}{\left(x+2\right)\left(x+14\right)}=\frac{1}{x+2}-\frac{1}{x+14}\)
\(\Leftrightarrow\frac{x}{\left(x+2\right)\left(x+14\right)}=\frac{\left(x+14\right)-\left(x+2\right)}{\left(x+2\right)\left(x+14\right)}\)
\(\Leftrightarrow x=\left(x+14\right)-\left(x+2\right)\)
\(\Leftrightarrow x=x+14-x-2\)
\(\Leftrightarrow x=\left(x-x\right)+\left(14-2\right)\)
\(\Leftrightarrow x=0+12\)
\(\Leftrightarrow x=12\)
Tìm x
\(\frac{2}{\left(x+2\right).\left(x+4\right)}+\frac{4}{\left(x+4\right).\left(x+8\right)}+\frac{6}{\left(x+8\right).\left(x+14\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\frac{12}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Rightarrow x=12\)