98 x 4 =....
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a, \(A=x\left(3x+1\right)+3x+1=\left(x+1\right)\left(3x+1\right)\)
Thay x = 33 ta được : \(32.100=3200\)
b, \(B=xy+2x+2y+4=x\left(y+2\right)+2\left(y+2\right)=\left(x+2\right)\left(y+2\right)\)
Thay x = 98 ; y = 98 ta được : \(100.100=10000\)
\(A=20\times21+21\times22+...+99\times100\)
\(3\times A=20\times21\times\left(22-19\right)+21\times22\times\left(23-20\right)+...+99\times100\times\left(101-98\right)\)
\(=20\times21\times22-19\times20\times21+...+99\times100\times101-98\times99\times100\)
\(=99\times100\times101-19\times20\times21\)
Suy ra \(A=\frac{99\times100\times101-19\times20\times21}{3}=360640\)
\(B=3\times4\times5+4\times5\times6+...+98\times99\times100\)
\(4\times B=3\times4\times5\times\left(6-2\right)+4\times5\times6\times\left(7-3\right)+...+98\times99\times100\times\left(101-97\right)\)
\(=3\times4\times5\times6-2\times3\times4\times5+...+98\times99\times100\times101-97\times98\times99\times100\)
\(=98\times99\times100\times101-2\times3\times4\times5\)
Suy ra \(B=\frac{98\times99\times100\times101-2\times3\times4\times5}{4}=24497520\)
`(x+1)/99+(x+2)/98+(x+3)/97+(x+4)/96=-4`
`=>(x+1)/99+1+(x+2)/98+1+(x+3)/97+1+(x+4)/96+1=-4+4`
`=>(x+100)/99+(x+100)/98+(x+100)/97+(x+100)/96=0`
`=>(x+100)(1/99+1/98+1/97+1/96)=0`
`=>x+100=0` (Vì `1/99+1/98+1/97+1/96\ne0`)
`=>x=-100`
Vậy ...
`#`𝐷𝑎𝑖𝑙𝑧𝑖𝑒𝑙
\(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\\ \dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}+4=0\\ \left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)+\left(\dfrac{x+4}{96}+1\right)=0\\ \dfrac{x+100}{99}+\dfrac{x+100}{98}+\dfrac{x+100}{97}+\dfrac{x+100}{96}=0\\ \left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{96}\right)=0\)
mà `1/99+1/98+1/97+1/96 \ne 0`
nên `x+100=0`
`x=-100`
Ta có\(M=\left[\left(1+\frac{1}{98}\right)+\left(\frac{1}{2}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{50}\right)\right].2.3...98\)
\(=\left[\frac{99}{1.98}+\frac{99}{2.97}+...+\frac{99}{49.50}\right].2.3...98=99\left(\frac{1}{1.98}+\frac{1}{2.97}+...+\frac{1}{49.50}\right).2.3...98\)
\(=99\left(\frac{k_1+k_2+...+k_{49}}{1.2.3...98}\right).2.3...98\left(k_1,k_2...k_{49}\varepsilonℕ^∗\right)=99\left(k_1+k_2+...+k_{49}\right)⋮99\Rightarrow M⋮99\left(đpcm\right)\)
1. \(x-8+32=68\)
\(x-8=68-32\)
\(x-8=36\)
\(x=36+8\)
\(x=44\)
_____
2. \(x+8+32=68\)
\(x+8=68-32\)
\(x+8=36\)
\(x=36-8\)
\(x=28\)
_____
3. \(98-x+34=43\)
\(98-x=43-34\)
\(98-x=9\)
\(x=98-9\)
\(x=89\)
_____
4. \(98+x-34=43\)
\(98+x=43+34\)
\(98+x=77\)
\(x=77-98\)
\(x=-21\)
_____
5. \(x:5:4=800\)
\(x:5=800.4\)
\(x:5=3200\)
\(x=3200.5\)
\(x=16000\)
_____
6. \(18+x=384:8\)
\(18+x=48\)
\(x=48-18\)
\(x=30\)
_____
7. \(\left(84,6-2\times x\right):3,02=5,1\)
\(84,6-2.x=5,1.3,02\)
\(84,6-2.x=15,402\)
\(2.x=84,6-15,402\)
\(2.x=69,198\)
\(x=69,198:2\)
\(x=34,599\)
_____
8. \(x\times5=120:6\)
\(x.5=20\)
\(x=20:5\)
\(x=4\)
_____
9. \(\left(15\times24-x\right):0,25=100:0,25\)
\(\left(15.24-x\right):0,25=400\)
\(15.24-x=400.0,25\)
\(15.24-x=100\)
\(360-x=100\)
\(x=360-100\)
\(x=260\)
1,x-8+32=68
=>x-8=68-32=36
=>x=36+8=42
2,x+8+32=68
=>x+8=68-32=36
=>x=36-8=28
3,98-x+34=43
=>98-x=43-34=9
=>x=98-9=89
4,98+x-34=43
=>98+x=43+34=77
=>x=77-98=-21
5,x:5:4=800
=>x:20=800
=>x=800x20=16000
6,18+x=384:8=48
=>x=48-18=30
7,(84,6-2x):3,02=5,1
=>84,6-2x=5,1x3,02=15,402
=>2x=84,6-15,402=69,198
=>x=69,198:2=34,599
8,5x=120:6=20
=>x=20:5=4
9,(15x24-x):0,25=100:0,25=400
=>360-x=400x0,25=100
=>x=360-100=260.
\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
\(\Rightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=0\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{99}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{98}\right)=0\)
\(\Rightarrow x+100=0\)
\(\Rightarrow x=-100\)
98 x 4 =392
392 nha bn
HT