Tìm các cặp số nguyên dương x, y sao cho: 3x2y - 7y = 5x2 + 84
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x + x + x +....+ x + 1+4+7+...+28= 155
có baonhiêu sô trong dãy là có bấy nhiêu số x
(28-1) : 3 + 1 = 10 số
10*x + (tổng ) = 155
tổng = (28+1) x 10 : 2 = 145
10*x + 145 = 155
x =1
tick nhá
x + (1+4+7+10+.....+28)=155
x + (28+1) . 10 =155
x + 145 =155
suy ra x=155-145=10
Xét hiệu:
\(\frac{a}{b}-\frac{a+2007}{b+2007}=\frac{a.\left(b+2007\right)-b.\left(a+2007\right)}{b.\left(b+2007\right)}=\frac{ab+2007a-ab+2007b}{b.\left(b+2007\right)}=\frac{2007.\left(a-b\right)}{b.\left(b+2007\right)}\)
Xét 3 trường hợp:
TH1: a=b\(\Rightarrow\)a-b=0\(\Rightarrow\)\(\frac{2007.\left(a-b\right)}{b.\left(b+2007\right)}=\frac{2007.0}{b.\left(b+2007\right)}=0\)\(\Rightarrow\frac{a}{b}=\frac{a+2007}{b+2007}\)
TH2: a<b\(\Rightarrow\)a-b<0\(\Rightarrow\)\(2007.\left(a-b\right)< 0\Rightarrow\frac{2007.\left(a-b\right)}{b.\left(b+2007\right)}< 0\)\(\Rightarrow\frac{a}{b}< \frac{a+2007}{b+2007}\)
TH3: a>b\(\Rightarrow\)a-b>0\(\Rightarrow\)\(2007.\left(a-b\right)>0\Rightarrow\frac{2007.\left(a-b\right)}{b.\left(b+2007\right)}>0\)\(\Rightarrow\frac{a}{b}>\frac{a+2007}{b+2007}\)
Vậy với a=b thì \(\frac{a}{b}=\frac{a+2007}{b+2007}\)
a<b thì \(\frac{a}{b}< \frac{a+2007}{b+2007}\)
a>b thì \(\frac{a}{b}>\frac{a+2007}{b+2007}\)
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ta có 1 = 1 x 1
ta xét 1 Th
x + 1= 1
=> x= 0
và y - 2 = 1
=> y = 3
Th âm tự xét
1= (-1) x(-1)
(x + 1 ) ( y-2) = 1
=> x + 1 = 1 hoặc y - 2 =1
th1 : x + 1 =1
x = 1 -1
x = 0
th2 : y - 2 = 1
y = 1 + 2
y = 3
Vậy : x = 0 ; y = 3
tk mk nha! :)
GTNN:A=X2+2X+5
=>A=5
5=X2+2X+5
=>X2+2X=0
=>X=0
GTLN:M=4-/5x-2/-/3-y/
M=4-/5.0-2/-/ 3-3/
M=4-2-0=2
N=5-2x-x2
N =5-2*0-02
N=5
Mik nghĩ vậy còn bạn sao thì ko bít