Tính nhanh
A= 4 x 2 x 25 x 5 x 175
B=42-104:(50 x 273-50 x 73)
C=3 x 53 x 6+2 x 9 x 87-18 x 40)
D=99 - 97 + 95 - 93 + 91-...+ 7 - 5 + 3 - 1
Tính nhanh
A= 4 x 2 x 25 x 5 x 175
B=42-104:(50 x 273-50 x 73)
C=3 x 53 x 6+2 x 9 x 87-18 x 40)
D=99 - 97 + 95 - 93 + 91-...+ 7 - 5 + 3 - 1
bài 1 : 101 x 125 + 101 x 25 - 101 x 50
bài 2 : 76 x 115 + 56 x 24 + 59 x 24
bài 3 : thực hiện phép tính : a ) 90-84+ 8 - 72 +66-60+54-48
b ) 99-97+95-93+91-89+.........+7-5+3-1
bài 4 : tìm số tự nhiên x biết :
a) \(x\) x 16 -\(x\) x 9 = 56
bài 5 :tìm số tự nhiên x , biết
a) \(x\) + 2 x \(x\) +3 x \(x\)+4 x \(x\) +5 x \(x\) = 165
b ) 1+2+3+4+.....+\(x\)=55
GIẢI GIÚP E Ạ
Bài 1:
\(101\cdot125+101\cdot25-101\cdot50\)
\(=101\cdot\left(125+25-50\right)\)
\(=101\cdot100\)
\(=10100\)
Bài 2:
\(76\cdot115+56\cdot24+59\cdot24\)
\(=76\cdot115+24\cdot\left(56+59\right)\)
\(=76\cdot115+24\cdot115\)
\(=115\cdot\left(76+24\right)\)
\(=115\cdot100\)
\(=11500\)
5:
a: =>15x=165
=>x=11
b: =>x(x+1)/2=55
=>x^2+x=110
=>x=10
4: =>7x=56
=>x=8
Bài 1:
101•125+101•25+101•50
= 101•(125+25-50)
=101•100
=10100
Bài 2:
76•115+56•24+59•24
= 76•115+24•(56+59)
= 76•115+24•115
= 115•(76+24)
= 115•100
= 11500
[(x+1)/99]+[(x+3)/97]+[(x+5)/95]= [(x+7)/93]+[(x+9)/91]+[(x+11)/89]
các bạn giúp mình với a. Mình cảm ơn trước
\(\frac{x+1}{99}+\frac{x+3}{97}+\frac{x+5}{95}=\frac{x+7}{93}+\frac{x+9}{91}+\frac{x+11}{89}\)
\(\Rightarrow\frac{x+1}{99}+1+\frac{x+3}{97}+1+\frac{x+5}{95}+1\)\(=\frac{x+7}{93}+1+\frac{x+9}{91}+1+\frac{x+11}{89}+1\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}\)\(=\frac{x+100}{93}+\frac{x+100}{91}+\frac{x+100}{89}\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}\)\(-\frac{x+100}{93}-\frac{x+100}{91}-\frac{x+100}{89}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{93}-\frac{1}{91}-\frac{1}{89}\right)=0\)
Mà \(\left(\frac{1}{99}< \frac{1}{97}< \frac{1}{95}< \frac{1}{93}< \frac{1}{91}< \frac{1}{89}\right)\)nên \(\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{93}-\frac{1}{91}-\frac{1}{89}\right)< 0\)
\(\Rightarrow x+100=0\Leftrightarrow x=-100\)
Vậy x = -100
Giải pt sau:
1,x+2/2002 +x+5/1999 +x+201/1803=-3
2,x+1/99 +x+3/97 +x+5/95=x+9/91 +x+8/92 +x+7/93.
\(\frac{x+2}{2002}+\frac{x+5}{1999}+\frac{x+201}{1803}=-3\)
\(\Rightarrow\frac{x+2}{2002}+1+\frac{x+5}{1999}+1+\frac{x+201}{1803}+1=0\)
\(\Rightarrow\frac{x+2004}{2002}+\frac{x+2004}{1999}+\frac{x+2004}{1803}=0\)
\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)=0\)
Dễ thấy \(\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)>0\)nên x + 2004 = 0
Vậy x = -2004
Giải pt sau:
1,x+2/2002 +x+5/1999 +x+201/1803=-3
2,x+1/99 +x+3/97 +x+5/95=x+9/91 +x+8/92 +x+7/93.
\(\frac{x+2}{2002}+\frac{x+5}{1999}+\frac{x+201}{1803}=-3\)
\(\Leftrightarrow\frac{x+2}{2002}+1+\frac{x+5}{1999}+1+\frac{x+201}{1803}+1=-3+1+1+1\)
\(\Leftrightarrow\frac{x+2004}{2002}+\frac{x+2004}{1999}+\frac{x+2004}{1803}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)=0\)
\(\Leftrightarrow x+2004=0\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\ne0\right)\)
<=> x=-2004
a,\(\frac{x+2}{2002}+\frac{x+5}{1999}+\frac{x+201}{1803}=-3\)
\(< =>\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+5}{1999}+1\right)+\left(\frac{x+201}{1803}+1\right)=0\)
\(< =>\frac{x+2004}{2002}+\frac{x+2004}{1999}+\frac{x+2004}{1803}=0\)
\(< =>\left(x+2004\right).\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)=0\)
Do \(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\ne0\)
\(=>x+2004=0\)
\(=>x=-2004\)
\(\frac{x+2}{2002}+\frac{x+5}{1999}+\frac{x+201}{1803}=-3\)
\(\Leftrightarrow\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+5}{1999}+1\right)+\left(\frac{x+201}{1803}+1\right)=0\)
\(\Leftrightarrow\frac{x+2004}{2002}+\frac{x+2004}{1999}+\frac{x+2004}{1803}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)=0\)
\(\Leftrightarrow x=-2004\)
\(\frac{x+1}{99}+\frac{x+3}{97}+\frac{x+5}{95}=\frac{x+9}{91}+\frac{x+8}{92}+\frac{x+7}{93}\)
\(\Leftrightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)=\left(\frac{x+9}{91}+1\right)+\left(\frac{x+8}{92}+1\right)+\left(\frac{x+7}{93}+1\right)\)
\(\Leftrightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}=\frac{x+100}{91}+\frac{x+100}{92}+\frac{x+100}{93}\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{95}-\frac{1}{91}-\frac{1}{92}-\frac{1}{93}\right)=0\)
Để ý thấy cụm đằng sau < 0 nên x=-100
Giải các phương trình sau ;
a) x-4/96 + x-7/93 + x-8/92 + x-10/90 + x-15/85 = 5
b) x+3/97 + x+5/95 + x+9/91 = x+91/98 + x+92/93 + x+61/99
Tính nhanh:
a, 2 x 31 x 12 + 4 x 6 x 42 + 8 x 27 x3
b, 2 x 53 x 12 + 4 x 6 x 87 -3 x 8 x 40
c, 5 x 7 x 77 -7 x 60 + 49 x 25 - 15 x 42
a, 2 x 31 x 12 + 4 x 6 x 42 + 8 x 27 x 3
= 2 x 12 x 31 + 4 x 6 x 42 + 8 x 3 x 27
= 24 x 31 + 24 x 42 + 24 x 27
= 24 x ( 31 + 24 + 27 )
= 24 x 82
= 1968
b, 2 x 53 x 12 + 4 x 6 x 87 - 3 x 8 x 40
= 24 x 53 + 24 x 87 - 24 x 40
= 24 x ( 53 + 87 - 40 )
= 24 x 100
= 2400
c, Tương tự
a) 2 x 31 x 12 + 4 x 6 x 42 + 8 x 27 x 3
= 2 x 12 x 31 + 4 X 6 x 42 + 8 x 3 x27
=24 x 31 + 24 x 42 + 24 x 27
= 24 x ( 31 + 24 + 27 )
= 24 x 82
=1968
Giải các phương trình sau:
9) x-49/ 50 + x-50/ 49 = 49/ x-50 + 50/ x-49
7) x+25/ 75 + x+30/70 = x+35/65 + x+40/60
8) 99-x/101 + 97-x/103 + 95-x/105 + 93-x/107 = 4
10) x+14/86 + x+15/85 + x+16/84 + x+17/83 + x+116/4 = 0
9: \(\dfrac{x-49}{50}+\dfrac{x-50}{49}=\dfrac{49}{x-50}+\dfrac{50}{x-49}\)
=>x-99=0
hay x=99
7: \(\Leftrightarrow\left(\dfrac{x+25}{75}+1\right)+\left(\dfrac{x+30}{70}+1\right)=\left(\dfrac{x+35}{65}+1\right)+\left(\dfrac{x+40}{60}+1\right)\)
=>x+100=0
hay x=-100
8:
Sửa đề: \(\dfrac{99-x}{101}+\dfrac{97-x}{103}+\dfrac{95-x}{105}+\dfrac{93-x}{107}=-4\)
\(\Leftrightarrow\left(\dfrac{99-x}{101}+1\right)+\left(\dfrac{97-x}{103}+1\right)+\left(\dfrac{95-x}{105}+1\right)+\left(\dfrac{93-x}{107}+1\right)=0\)
=>200-x=0
hay x=200
(x-1/99+x-99)+(x-3/97+x-7/93)+(x-5/95+x-95/5)=6