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( 19x + 2.52 ) : 14 = ( 13 - 8 )2- 4 2
( 19x + 2.52 ) : 14 = 52- 4 2
( 19x + 2.52 ) : 14 = 25- 16
( 19x + 2.52 ) : 14 =9
( 19x + 50 ) : 14 =9
19x + 50=9 . 14
19x + 50=126
19x=126-50
19x=76
x=76:19
x=4
a) \(\frac{3}{7}-\frac{1}{7}x=\frac{2}{3}\)
=> \(\frac{1}{7}x=\frac{3}{7}-\frac{2}{3}=-\frac{5}{21}\)
=> \(x=-\frac{5}{21}:\frac{1}{7}=-\frac{5}{21}\cdot7=-\frac{5}{3}\)
b) \(3x^2-2=72\)=> 3x2 = 74 => x2 = 74/3 => x không thỏa mãn
c) \(\left(19x+2\cdot5^2\right):14=\left(13-8\right)^2-4^2\)
=> \(\left(19x+2\cdot25\right):14=5^2-4^2=9\)
=> \(\left(19x+50\right):14=9\)
=> \(19x+50=126\)
=> \(19x=76\)
=> x = 4
d) \(x:\frac{1}{2}+x:\frac{1}{4}+x:\frac{1}{8}+x:\frac{1}{16}+x:\frac{1}{32}=343\)
=> \(x\cdot2+x\cdot4+x\cdot8+x\cdot16+x\cdot32=343\)
=> \(x\left(2+4+8+16+32\right)=343\)
=> x . 62 = 343
=> x = 343/62
\(a,\Rightarrow12x-91=101\\ \Rightarrow12x=192\\ \Rightarrow x=16\\ b,\Rightarrow x:23+45=133\\ \Rightarrow x:23=88\\ \Rightarrow x=\dfrac{88}{23}\\ c,\Rightarrow\left(6x-39\right):7=3\\ \Rightarrow6x-39=21\\ \Rightarrow6x=60\\ \Rightarrow x=10\\ d,\Rightarrow3x-24=\dfrac{148}{73}\\ \Rightarrow3x=\dfrac{1900}{73}\\ \Rightarrow x=\dfrac{1900}{219}\\ e,\Rightarrow\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\\ f,\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\\ d,\left(9-x\right)^3=64=4^3\\ \Rightarrow9-x=4\\ \Rightarrow x=5\\ h,\Rightarrow x=27\\ i,\Rightarrow6x=312\cdot12=624\cdot6\\ \Rightarrow x=624\\ j,\Rightarrow\left(19x+104\right):14=25-42=-17\\ \Rightarrow19x+104=-238\\ \Rightarrow19x=-342\\ \Rightarrow x=-18\)
Ta có : \(\left(19x+2\cdot5^2\right):14=\left(13-8\right)^2-4\)
\(\Leftrightarrow\left(19x+2\cdot25\right):14=5^2-4\)
\(\Leftrightarrow\left(19x+50\right)=21\cdot14\)
\(\Leftrightarrow19x+50=294
\)
\(\Leftrightarrow19x=294-50\)
\(\Leftrightarrow x=\frac{244}{19}=12\frac{16}{19}\)
( 19x + 2 . 5\(^2\)) : 14 = ( 13 - 8 )\(^2\) - 4
( 19x + 2 . 25 ) : 14 = 5\(^2\)- 4
( 19x + 50 ) : 14 = 25 - 4
( 19x + 50 ) : 14 = 21
19x + 50 = 21 . 14
19x + 50 = 294
19x = 294 - 50
19x = \(\frac{244}{19}\)
( x + 2 ) chia hết cho ( x - 1 ) => \(\frac{x+2}{x-1}\in Z\)
\(\frac{x+2}{x-1}=\frac{x-1+3}{x-1}=1+\frac{3}{x-1}\)
\(\Rightarrow x-1\inƯ\left(3\right)\)
\(\Rightarrow x-1\in\left(-1;1-3;3\right)\)
Ta có các giá trị X thỏa mãn :
\(x-1=-3\Rightarrow x=-2\)
\(x-1=1\Rightarrow x=2\)
\(x-1=3\Rightarrow x=4\)
\(\Rightarrow x-1=-1\Rightarrow x=0\)
Điều kiện \(x\ne1\)
Vậy các giá trị của x là : \(=-2;0;2;4\)
a,vi (x-1)chia het cho (x-1) suy ra (x+2)-(x-1) chia het cho (x-1) Suy ra 3 chia het cho (x-19) suy ra x-1 thuoc {1;3} suy ra x thuoc {2;4}
a ) \(\left(19x+2,5^2\right)\div14=\left(13-8\right)^2-4^2\)
\(\Rightarrow\left(19x+50\right)\div14=5^2-16\)
\(\Rightarrow\left(19+50\right)\div14=25-16=9\)
\(\Rightarrow19x+50=9\times14\)
\(\Rightarrow19x+50=126\)
\(\Rightarrow19x=126-50\)
\(\Rightarrow19x=76\)
\(\Rightarrow x=4\)
b ) \(1440\div\left[41-\left(2x-5\right)\right]=2^4.3\)
\(\Leftrightarrow1440\div\left[41-\left(2x-5\right)\right]=48\)
\(\Leftrightarrow41-\left(2x-5\right)=30\)
\(\Leftrightarrow2x-5=11\)
\(\Leftrightarrow2x=16\)
\(\Leftrightarrow x=8\)
240-[23+(13+24.3-x)]=132
240-[23+(13+168-x)]=132
240-[23+(181-x)]=132
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x-8:4-(46-23.2+6.3)=0
\(x-8:4-\left(46-23.2+6.3\right)=0\)
\(x-2-\left(46-46+18\right)=0\)
\(x-2-18=0\)
\(x-2=0+18\)
\(x-2=18\)
\(x=18+2\)
\(x=20\)