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a) 126+ (-29)+ 2004+ (-106)
= [126+ (-106)] +2004+ (-29)
= 20+ 2004- 29
= 2024-29
= 1995
b) (-199)+ (-200)+ (-201)
= -(199+200+201)
= -600
a) 126 + (-20) + 2004 + (-106)
= [126 + (-20) + (-106)] + 2004
= 0 + 2004
= 2004
b) (-199) + (-200) + (-201)
= [(-199) + (-201)] + (-200)
= (-400) + (-200)
= -600
Phần nào không hiểu bạn có thể nhắn hỏi mình nhe
Ta có : mẫu số 1 : 4 . 1
mẫu số hai : 4.7
... mẫu số thứ 96 = 100.103 = 10300
=> Số số hạng y là 100
Ta có :
\((y+..+y) + (\frac{3}{1.4} + \frac{3}{4.7} + ...+ \frac{3}{100.103})\)
\(= ( y+...+y) + [1. (\frac{1}{1.4} + \frac{1}{4.7} + ..+ \frac{1}{100.103})]\)
\(= (y+...y) + [1.(\frac{1}{1} - \frac{1}{4} + \frac{1}{4} - \frac{1}{7} + ...+ \frac{1}{100} - \frac{1}{103}) ]\)
\(= (y+...+y) + (1 - \frac{1}{103})\)
\(= (y+...+y) + \frac{102}{103}\)
\(=> (y+...+y) = \frac{308}{103} - \frac{102}{103} = \frac{206}{103}\)
\(=> y = \frac{206}{103} : 100 = \frac{206}{10300} = \frac{103}{5150}\) ( Chia 100 vì có 100 số hạng y)
Vậy \(y = \frac{103}{5150}\)
D = $\frac{2}{3}.\frac{5}{6}.\frac{9}{10}. ... .\frac{799}{780}$
= $\frac{2.2}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}. ... .\frac{38.41}{39.40}$
= $\frac{2.2}{2.3}.\frac{2.3. ... .38}{3.4. ... 39}.\frac{5.6. ... .41}{4.5. ... .40}$
= $\frac{2}{3}.\frac{2}{39}.\frac{41}{4}$
= $\frac{41}{3.39}$
D = \(\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}.....\frac{779}{780}\)
= \(\frac{2.2}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}.....\frac{38.41}{39.40}\)
= \(\frac{2}{3}.\frac{2.3.4....38}{3.4.5....39}.\frac{5.6.7.....41}{4.5.6.....40}\)
= \(\frac{2}{3}.\frac{2}{39}.\frac{41}{4}\)
= \(\frac{41}{117}\)
a) x(x+2) > 0
=> x2 + 2x > 0
Vì x2 luôn ≥ 0 với mọi x nên để x2 + 2x > 0 thì 2x > 0 => x>0
Vậy với x>0 thì x(x+2) > 0
b) ( x -1 )( x + 3) < 0
<=> x2 + 3x - x - 3 > 0
<=> x2 + 2x - 3 > 0
Vì x2 luôn ≥ 0 với mọi x nên để x2 + 2x - 3 < 0 thì 2x - 3 < 0 => 2x < 3 => x < 3/2
Vậy với x<3/2 thì ( x -1 )( x + 3) < 0
c) ( 1 - x )( y + 1 ) =-3
Ta có bảng:
1 - x | 1 | -1 | 3 | -3 |
y + 1 | 3 | -3 | 1 | -1 |
x | 0 | 2 | -2 | 4 |
y | 2 | -4 | 0 | -2 |
Vậy với x thuộc {…} và y thuộc {…} thì ( 1 - x )( y + 1 ) =-3
Làm mẫu câu a nha
a) \(x\left(x+2\right)>0\)
Th1 : \(\hept{\begin{cases}x>0\\x+2>0\end{cases}\Rightarrow\hept{\begin{cases}x>0\\x>-2\end{cases}\Rightarrow}x>0}\)
Th2 : \(\hept{\begin{cases}x< 0\\x+2< 0\end{cases}\Rightarrow\hept{\begin{cases}x< 0\\x< -2\end{cases}}\Rightarrow x< -2}\)
Vậy ta có : \(\orbr{\begin{cases}x>0\\x< -2\end{cases}}\)
a) Ta có: \(\left(x-3\right)\left(x-5\right)< 0\)
\(\Rightarrow\hept{\begin{cases}x-3< 0\\x-5>0\end{cases}}\)hoặc \(\hept{\begin{cases}x-3>0\\x-5< 0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x< 3\\x>5\end{cases}}\) (vô lý) hoặc \(\hept{\begin{cases}x>3\\x< 5\end{cases}}\)(thỏa mãn).
Vậy 3 < x < 5 thì (x-3)(x-5) <0.
b) \(-6x-\left(-7\right)=25\)
\(\Rightarrow-6x=25-7\)
\(\Rightarrow-6x=18\Rightarrow x=\frac{18}{-6}=-3\)
Vậy x = -3.
c) \(46-\left(x-11\right)=-48\)
\(\Rightarrow46-x+11=-48\)
\(\Rightarrow46+11+48=x\Rightarrow x=105\).
d) \(\left(x+15\right)\left(x-2\right)=0\)
\(\Rightarrow\)x + 15 = 0 hoặc x - 2 = 0
\(\Rightarrow x=-15\)hoặc \(x=2\).
e) \(3\left(4-x\right)-2\left(x-5\right)=12\)
\(\Rightarrow12-3x-2x+10=12\)
\(\Rightarrow-3x-2x=12-10-12\)
\(\Rightarrow-5x=-10\Rightarrow x=2\).
Chúc bn hc tốt!
a.\(\left(\dfrac{1}{2}+1\right).\left(\dfrac{1}{3}+1\right).\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{99}+1\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{100}{99}\)
\(=\dfrac{3.4.5...100}{2.3.4...99}\)
\(=\dfrac{100}{2}=50\)
a,
\(\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{99}+1\right)\\ =\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{100}{99}\\ =\dfrac{3\cdot4\cdot5\cdot...\cdot100}{2\cdot3\cdot4\cdot...\cdot99}\\ =\dfrac{100}{2}=50\)
b,
\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{100}-1\right)\\ =\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot\dfrac{-3}{4}\cdot...\cdot\dfrac{-99}{100}\\ =\dfrac{\left(-1\right)\left(-2\right)\left(-3\right)\cdot...\cdot\left(-99\right)}{2\cdot3\cdot4\cdot...\cdot100}\\ =\dfrac{\left(-1\right)\left(-1\right)\left(-1\right)\cdot...\left(-1\right)}{100}\left(\text{có }99\text{ số }-1\right)\\ =\dfrac{\left(-1\right)^{99}}{100}\\ =\dfrac{-1}{100}\)
c,
\(C=\dfrac{4}{30}+\dfrac{4}{70}+\dfrac{4}{126}+...+\dfrac{4}{798}\\ =\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{399}\\ =\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{19\cdot21}\\ =\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{19}-\dfrac{1}{21}\\ =\dfrac{1}{3}-\dfrac{1}{21}\\ =\dfrac{7}{21}-\dfrac{1}{21}\\ =\dfrac{6}{21}=\dfrac{2}{7}\)
a) -1 - 2 + 3 + 4 - 5 - 6 + 7 + 8 - 9 - 10 + 11 + 12 - ... - 2013 - 2014 + 2015 + 2016
= ( -1 - 2 + 3 + 4 ) - ( 5 + 6 - 7 - 8 ) - ( 9 + 10 - 11 - 12 ) - .......... - ( 2013 + 2014 - 2015 - 2016 )
= 4 - ( -4 ) - ( -4 ) - ......... - ( -4 )
= 4 + 4 + 4 +....... + 4
= { [ ( 2016 - 1 ) : 1 + 1 ] : 4 } . 4
= { [ 2015 : 1 + 1 ] : 4 } . 4
= { 2016 : 4 } . 4
= 504 . 4
= 2016
b) \(\left(\frac{1}{2}-1\right):\left(\frac{1}{3}-1\right):\left(\frac{1}{4}-1\right):\left(\frac{1}{5}-1\right):.........:\left(\frac{1}{100}-1\right)\)
\(=\frac{-1}{2}:\frac{-2}{3}:\frac{-3}{4}:\frac{-4}{5}:......:\frac{-99}{100}\)
\(=\frac{-1}{2}.\frac{3}{-2}.\frac{4}{-3}.\frac{5}{-4}.......\frac{100}{-99}\)
\(=\frac{-1.3.4........100}{2.2.3.4......99}\)
\(=\frac{-1.100}{2.2}\)
\(=\frac{-100}{4}\)
\(=-25\)
a) -1-2+3+4-5-6+7+8+...+2016=-3+3-7+7-...-2016+2016=0
b) \(\left(\frac{1}{2}-1\right):...:\left(\frac{1}{100}-1\right)=\frac{-1}{2}:\frac{-2}{3}:\frac{-3}{4}:...:\frac{-99}{100}\)
\(=\)\(\frac{-1}{2}.\frac{-3}{2}.....\frac{-100}{99}=\frac{-1}{2}.\left(-50\right)=25\)
a) \(126+\left(-20\right)+2004+\left(-106\right)\)
\(=106+2004+\left(-106\right)\)
\(=0+2004=2004\)
b) \(70+69+68+...+1+0+\left(-1\right)+\left(-2\right)+...+\left(-70\right)\)
\(=\left[70+\left(-70\right)\right]+\left[69+\left(-69\right)\right]\)\(+...+\left[2+\left(-2\right)\right]+\left[1+\left(-1\right)\right]+0\)
\(=0+0+...+0+0+0=0\)