Tìm x, biết:
x + 15 = 41
x – 23 = 39
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\(\frac{70}{3}\left(\frac{39}{30}+\frac{39}{42}\right)-\frac{246}{7}\div\left(\frac{41}{56}+\frac{41}{72}\right)\)
\(=\frac{70}{3}\left(\frac{13}{10}+\frac{13}{14}\right)-\frac{246}{7}\div\left(\frac{41}{7\cdot8}+\frac{41}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(1+\frac{3}{10}+1-\frac{1}{14}\right)-\frac{246}{7}\div\left(\frac{40+1}{7\cdot8}+\frac{40+1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left[\left(1+1\right)+\left(\frac{3}{10}-\frac{1}{14}\right)\right]-\frac{246}{7}\div\left(\frac{5}{7}+\frac{1}{7\cdot8}+\frac{5}{9}+\frac{1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(2+\frac{8}{35}\right)-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)\right]\)
\(=\frac{70}{3}\cdot\frac{78}{35}-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\right]\)
\(=\frac{35\cdot2\cdot26\cdot3}{3\cdot35}-\frac{246}{7}\div\left(\frac{5}{7}+\frac{5}{9}+\frac{1}{7}-\frac{1}{9}\right)\)
\(=52-\frac{246}{7}\div\left[\left(\frac{5}{7}+\frac{1}{7}\right)+\left(\frac{5}{9}-\frac{1}{9}\right)\right]\)
\(=52-\frac{246}{7}\div\left(\frac{6}{7}+\frac{4}{9}\right)\)
\(=52-\frac{246}{7}\div\frac{82}{63}\)
\(=52-\frac{82\cdot3\cdot9\cdot7}{7\cdot82}\)
\(=52-27=25\)
\(\frac{57}{20}-\frac{26}{15}+\frac{139}{20}\div3\)
\(=\frac{57}{20}-\frac{26}{15}+\frac{139}{60}\)
\(=\frac{171}{60}-\frac{104}{60}+\frac{139}{60}=\frac{103}{30}\)
\(\frac{39}{4}+\frac{2}{3}\left(11-\frac{23}{4}\right)\)
\(=\frac{39}{4}+11\cdot\frac{2}{3}-\frac{23}{4}\cdot\frac{2}{3}\)
\(=\frac{39}{4}+\frac{22}{3}-\frac{56}{12}\)
\(=\frac{119}{12}+\frac{88}{12}-\frac{56}{12}=\frac{151}{12}\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2002}\right)\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2001}{2002}\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot2001\cdot2002\cdot2003}{2\cdot3\cdot4\cdot...\cdot2002\cdot2003\cdot2004}=\frac{1}{2004}\)
b, 92 : 4 - 27 = \(\dfrac{x+350}{x}\) + 315
23 - 27 = 1 + \(\dfrac{350}{x}\) + 315
316 + \(\dfrac{350}{x}\) = -4
\(\dfrac{350}{x}\) = - 316 - 4
\(\dfrac{350}{x}\) = -320
-320 \(x\) = 350
\(x\) = 350: (-320)
\(x\) = - \(\dfrac{35}{32}\) (loại)
Vậy \(x\) \(\in\) \(\varnothing\)
c, 720 : [ 41 - (2\(x\) - 5)] = 23.5
41 - (2\(x\) - 5) = 720 : (23.5)
41 - 2\(x\) + 5 = 18
46 - 2\(x\) =18
2\(x\) = 46 - 18
2\(x\) = 28
\(x\) = 28: 2
\(x\) = 14
Vậy \(x\) = 14
\(|x|-|27|=|46|+|-23|-|41| \\ |x|-27=46+23-41 \\ |x|-27=28 \\ |x|=28+27 \\ |x|=55 \\ x=\pm 55\)
Vậy \(x=55\) hoặc \(x=-55\).
\(\left|x\right|-\left|27\right|=\left|46\right|+\left|-23\right|-\left|41\right|\)
\(\left|x\right|-27=46+23-41\)
\(\left|x\right|-27=28\)
\(\left|x\right|\)=28+27=55
=>\(x\)=-55 hoặc 55
\(4^{x+1}-41=23\Leftrightarrow4^{x+1}=64=2^6=\left(2^2\right)^4=4^4\)
Suy ra \(x+1=4\Leftrightarrow x=3\)
số x đó là
x+23=116
x=116-23
x=93
mình chỉ biết thế
này thôi
(x+21)-75=39
x+21 = 39 + 75
x+21 = 114
x = 114 - 21
x = 93
vậy........
11(x-9)=781
x-9 = 781 : 11
x-9 = 71
x = 71 + 9
x = 80
vậy.........
280:(x-8)=14
x-8 = 280 : 14
x-8 = 20
x = 20 + 8
x = 28
vậy......
6(x+41)=600
x+41 = 600 : 6
x+41 = 100
x = 100 - 41
x = 59
vậy.............
15 ( x-9)=1050
x-9 = 1050 : 15
x-9 = 70
x = 70 + 9
x = 79
vậy..........
1 , ( x + 21 ) - 75 = 39
x + 21 =39+75
x + 21 =114
x =114 - 21
x = 93 Vậy x=93
2, 11(x-9)=781
x-9=781/11
x-9=71
x=71+9
x =80
Vậy x =80
3) 280/ (x-8)=14
x-8 =280/14
x-8 =20
x=20+8
x=28
Vậy x =28