Tìm x, biết: ( x - 1 ) + ( x - 2 ) + ( x - 3 ) + ... + ( x - 20 ) = 15
Giúp mình với, cảm ơn
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Giải:
a) \(2\dfrac{17}{20}-1\dfrac{15}{11}+6\dfrac{9}{20}:3\)
\(=\dfrac{57}{20}-\dfrac{26}{11}+\dfrac{129}{20}:3\)
\(=\dfrac{107}{220}+\dfrac{43}{20}\)
\(=\dfrac{29}{11}\)
b) \(4\dfrac{3}{7}:\left(\dfrac{7}{5}.4\dfrac{3}{7}\right)\)
\(=\dfrac{31}{7}:\left(\dfrac{7}{5}.\dfrac{31}{7}\right)\)
\(=\dfrac{31}{7}:\dfrac{31}{5}\)
\(=\dfrac{5}{7}\)
c) \(\left(3\dfrac{2}{9}.\dfrac{15}{23}.1\dfrac{7}{29}\right):\dfrac{5}{23}\)
\(=\left(\dfrac{29}{9}.\dfrac{15}{23}.\dfrac{36}{29}\right):\dfrac{5}{23}\)
\(=\dfrac{60}{23}:\dfrac{5}{23}\)
\(=12\)
ĐK : 6x \(\ge0\Rightarrow x\ge0\)
Khi đó |x + 1| = x + 1
|x + 2| = x +2
|x + 3| = x +3
|x + 4| = x + 4
|x + 5| = x +5
Khi đó |x + 1| + |x + 2| + |x + 3| + |x + 4| + |x + 5| = 6x
<=> x + 1 + x + 2 + x + 3 + x + 4 + x + 5 = 6x
<=> 5x + 15 = 6x
<=> x = 15 (tm)
Vậy x = 15
b) 3x + 2 - 3x + 1 - 3x = 15.340
=> 3x(32 - 3 - 1) = 15.340
<=> 3x . 5 = 15.340
<=> 3x = 341
<=> x = 41
Vậy x = 41
a,vì /x+1/,/x+2/,/x+3/,/x+4/,/x+5/\(\ge\)0 mà /x+1/+/x+2/+/x+3/+/x+4/+/x+5/=6x suy ra x>0
nên /x+1/+/x+2/+/x+3/+/x+4/+/x+5/=x+1+x+2+x+3+x+4+x+5=6x ( giải thích: /x/=x khi x \(\ge0\))
suy ra 5x+21=6x suy ra x=21
b, \(3^{x+2}-3^{x+1}-3^x=15.3^{40}\)
suy ra \(3^x\left(9-3-1\right)=5.3^{41}\)
suy ra \(3^x.5=5.3^{41}\Rightarrow x=41\)
60-[15*X+4]=15/2:1/2
60-[15*X+4]=15
15*X+4=60-15
15*X+4=45
15*X=45-4
15*X=41
X=41:15
X=41/15
ko ghi đề
\(60-\left(15.x+4\right)=\frac{15}{2}.\frac{2}{1}\)
\(60-\left(15.x+4\right)=15\)
\(15.x+4=60-15\)
\(15.x+4=45\)
\(15.x=45-4\)
\(15.x=41\)
\(x=41:15\)
\(x=\text{2.7333}\)
\(\dfrac{-2}{3}\left(x-\dfrac{1}{4}\right)=\dfrac{1}{3}\left(2x-1\right)\)
\(\Leftrightarrow\dfrac{-2}{3}x+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{-2}{3}x-\dfrac{2}{3}x=\dfrac{-1}{3}-\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{-4}{3}x=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{3}{8}\)
Vậy \(x=\dfrac{3}{8}\)
a, (3 - \(x\))(4y + 1) = 20
Ư(20) = { -20; -10; -5; -4; -2; -1; 1; 2; 4; 5; 10; 20}
Lập bảng ta có:
\(3-x\) | -20 | -10 | -5 | -4 | -2 | -1 | 1 | 2 | 4 | 5 | 10 | 20 |
\(x\) | 23 | 13 | 8 | 7 | 5 | 4 | 2 | 1 | -1 | -2 | -7 | -17 |
4\(y\) + 1 | -1 | -2 | -4 | -5 | -10 | -20 | 20 | 10 | 5 | 4 | 2 | 1 |
\(y\) | -1/2 | -3/4 | -5/4 | -6/4 | -11/4 | -21/4 | 19/4 | 9/4 | 1 | 3/4 | 1/4 | 0 |
Vậy các cặp \(x;y\) nguyên thỏa mãn đề bài là:
(\(x;y\)) =(-1; 1); (-17; 0)
b, \(x\left(y+2\right)\)+ 2\(y\) = 6
\(x\) = \(\dfrac{6-2y}{y+2}\)
\(x\in\) Z ⇔ 6 - \(2y⋮\) \(y\) + 2 ⇒-(2y + 4) +10 ⋮ \(y\) + 2 ⇒ -2(\(y\)+2) +10 ⋮ \(y\)+2
⇒ 10 ⋮ \(y\) + 2
Ư(10) = { -10; -5; -2; -1; 1; 2; 5; 10}
Lập bảng ta có:
\(y+2\) | -10 | -5 | -2 | -1 | 1 | 2 | 5 | 10 |
\(y\) | -12 | -7 | -4 | -3 | -1 | 0 | 3 | 8 |
\(x=\) \(\dfrac{6-2y}{y+2}\) | -3 | -4 | -7 | -12 | 8 | 3 | 0 | -1 |
Theo bảng trên ta có các cặp \(x;y\)
nguyên thỏa mãn đề bài lần lượt là:
(\(x;y\) ) =(-3; -12); (-4; -7); (-12; -3); (8; -1); (3; 0); (0;3 (-1; 8)
\(a,x+\dfrac{3}{7}=\dfrac{2}{5}+\dfrac{3}{10}\)
\(x+\dfrac{3}{7}=\dfrac{4}{10}+\dfrac{3}{10}\)
\(x+\dfrac{3}{7}=\dfrac{7}{10}\)
\(x=\dfrac{7}{10}-\dfrac{3}{7}\)
\(x=\dfrac{49}{70}-\dfrac{30}{70}\)
\(x=\dfrac{19}{70}\)
\(b,\dfrac{19}{20}-x=\dfrac{8}{5}-\dfrac{3}{4}\)
\(\dfrac{19}{20}-x=\dfrac{32}{20}-\dfrac{15}{20}\)
\(\dfrac{19}{20}-x=\dfrac{17}{20}\)
\(x=\dfrac{19}{20}-\dfrac{17}{20}\)
\(x=\dfrac{2}{20}\)
\(x=\dfrac{1}{10}\)
#Urushi☕
a) x - 1/2 = 3/5
x = 3/5 + 1/2
x = 11/10
b) x - 1/2 = -2/3
x = -2/3 + 1/2
x = -1/6
c) 2/5 - x = 0,25
x = 2/5 - 0,25
x = 2/5 - 1/4
x = 3/20
\(\left(x-1\right)+\left(x-2\right)+...+\left(x-20\right)=15\)
\(\left(x+x+...+x\right)-\left(1+2+...+20\right)=15\)
\(20x-\dfrac{20.21}{2}=15\)
\(20x-210=15\)
\(20x=210+15\)
\(20x=225\)
\(x=225:20\)
\(x=11,25\)
Đề thiếu 1 vế rồi em, phải = cái gì đó nữa chứ