Tìm x
2012.(2011-x)+25=55:53
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a, Mình nghĩ là đề sai .
b, Ta có : \(\frac{x-45}{55}+\frac{x-47}{45}=\frac{x-55}{45}+\frac{x-53}{47}\)
=> \(\frac{x-45}{55}-1+\frac{x-47}{45}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)
=> \(\frac{x-45}{55}-\frac{55}{55}+\frac{x-47}{53}-\frac{53}{53}=\frac{x-55}{45}-\frac{45}{45}+\frac{x-53}{47}-\frac{47}{47}\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
=> \(\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)
=> \(x-100=0\)
=> \(x=100\)
Vậy phương trình trên có tập nghiệm là \(S=\left\{100\right\}\)
c, Ta có : \(\frac{2-x}{2010}-1=\frac{1-x}{2011}-\frac{x}{2012}\)
=> \(\frac{2-x}{2010}-1=\frac{1-x}{2011}+\frac{-x}{2012}\)
=> \(\frac{2-x}{2010}+1=\frac{1-x}{2011}+1+\frac{-x}{2012}+1\)
=> \(\frac{2-x}{2010}+\frac{2010}{2010}=\frac{1-x}{2011}+\frac{2011}{2011}+\frac{-x}{2012}+\frac{2012}{2012}\)
=> \(\frac{2012-x}{2010}=\frac{2012-x}{2011}+\frac{2012-x}{2012}\)
=> \(\frac{2012-x}{2010}-\frac{2012-x}{2011}-\frac{2012-x}{2012}=0\)
=> \(\left(2012-x\right)\left(\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)=0\)
=> \(2012-x=0\)
=> \(x=2012\)
Vậy phương trình trên có tập nghiệm là \(S=\left\{2012\right\}\)
1: Tính
a: 27+ 55+ (-17) + (-55)
= (27-17) + (55-55)
= 10 + 0
= 10
b: (-92)+ (-251)+ (-8)+ 25
= -(92+8) - (251 - 25)
= -100 - 226
= -326
c: (235- 47) + (175- 53)
= 235 - 47 + 175 - 53
= (235+175) - (47+53)
= 410 - 100
= 310
d: (756- 217)- (183- 44)
= 756 - 217 - 183 + 44
= (756+44) - (217+183)
= 800 - 400
= 400
2: Tìm x, biết
\(\text{x. (x+ 3)= 0}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x+3=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)
vậy______
\(\text{(x-2). (5- x) = 0}\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\5-x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=5\end{cases}}\)
vậy________
(x- 3). (2x +1) = 7
=> x-3 và 2x + 1 thuộc Ư(7)
ta có bảng sau :
x-3 | 1 | -1 | 7 | -7 |
2x+1 | 7 | -7 | 1 | -1 |
x | 4 | 2 | 10 | -4 |
x | 3 | -3 | 0 | -1 |
vậy x thuộc {4;3;2;-3;10;0;-4;-1}
\(\dfrac{x-45}{55}+\dfrac{x-47}{53}=\dfrac{x-55}{45}+\dfrac{x-53}{47}\)
\(\Leftrightarrow\left(\dfrac{x-45}{55}-1\right)+\left(\dfrac{x-47}{53}-1\right)=\left(\dfrac{x-55}{45}-1\right)+\left(\dfrac{x-53}{47}-1\right)\)
\(\Leftrightarrow\dfrac{x-100}{55}+\dfrac{x-100}{53}=\dfrac{x-100}{45}+\dfrac{x-100}{47}\)
\(\Leftrightarrow\dfrac{x-100}{55}+\dfrac{x-100}{53}-\dfrac{x-100}{45}-\dfrac{x-100}{47}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\right)=0\)
Do \(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\ne0\) nên x - 100 = 0 <=> x = 100
a)(3/2-0,5)/x=7/2+1/4
(3/2-1/2)/x=14/4+1/4
1/x=15/4
x=1:15/4
x=4/15
b)(x*0,25+2010)*2011=(53+2010)*(2012-1)
(x*0,25+2010)*2011=2063*2011
=>0,25x+2010=2063
0,25x=2063-2010
0,25x=53
x=53/0,25
x=212
<=>x-45/55 -1 + x-47/53 -1=x-55/45 -1 + x-53/47-1
<=>x-100/55 + x-100/53 = x-100/45 + x-100/47
<=>(x-100)(1/55 + 1/53 - 1/45 - 1/47 )=0
vi (1/55 + 1/53 - 1/45 - 1/47 ) luon khac 0 nen x-100=0 <=>x=100
\(\dfrac{25}{27};\dfrac{51}{53};\dfrac{47}{49};\dfrac{103}{105};\dfrac{2009}{2011};\dfrac{1963}{1963};\)
\(1,=38-42+14-25+27+15=27\)
\(2,=12+21-23+21-10=21\)
\(3,=57-725-605+53=-1220\)
\(4,=55+45+15-15+55-45=110\)
A = 1/25x27 + 1/27x29+1/29x31 + ...+ 1/53x55
2A = 3/25x27+ 3/27x29+...+3/53x55
2A = 1/25-1/27+1/27-1/29+...+1/53-1/55
2A = 1/25-1/55
2A = 6/275
A = 6/275:2=3/275
k mk nha, mk lm vội nên coi lại chút giùm mk xem có đúng k
2012.(2011-x)+25=55:53
=>2012.(2011-x)+25=55-3
=>2012.(2011-x)+25=52
=>2012.(2011-x)+25=25
=>2012.(2011-x)=25-25
=>2012.(2011-x)=0
=>2011-x=0:2012
=>2011-x=0
=>x=2011-0
=>x=2011