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a , 3x - 4 = 57 : 56
3x - 4 = 57-6
3x -4 = 5
3x = 9
3x = 32
suy ra x = 2
Vậy x =2
b , 3x-1 = 27
3x-1 = 33
suy ra x - 1 =3 suy ra x = 3-1 =2
Vậy x =2
c , (x-1)3 = 4(x-1)
( x - 1)3 : ( x-1 ) = 4
(x-1)3-1 = 4
(x-1)2 = 22
suy ra x-1 =2 suy ra x = 2+1 =3
Vậy x =3
d , 5x . 5x-2 = 130
5x . 5x : 52 = 130
52x = 130 x 25
52s = 3250
suy ra x thuộc rỗng
Vậy x thuộc rỗng
a) 3^x -4 =5^7 : 5^6
3^x -4 =5
3^x=9
x=2
b ) 3^x-1=26
3^x=27
x=3
đây bạn
\(a,x-7\frac{5}{8}=1\frac{1}{4}\)
=> \(x-\frac{61}{8}=\frac{5}{4}\)
=> \(x=\frac{5}{4}+\frac{61}{8}\)
=> \(x=\frac{10}{8}+\frac{61}{8}=\frac{71}{8}=8\frac{7}{8}\)
\(b,x+7\frac{5}{8}=9\frac{1}{4}\)
=> \(x+\frac{43}{5}=\frac{37}{4}\)
=> \(x=\frac{37}{4}-\frac{43}{5}=\frac{13}{20}\)
\(c,\left[x-7\frac{5}{8}\right]:\frac{1}{2}=3\)
=> \(\left[x-\frac{61}{8}\right]=3\cdot\frac{1}{2}\)
=> \(\left[x-\frac{61}{8}\right]=\frac{3}{2}\)
=> \(x-\frac{61}{8}=\frac{3}{2}\)
=> \(x=\frac{3}{2}+\frac{61}{8}=\frac{12}{8}+\frac{61}{8}=\frac{73}{8}=9\frac{1}{8}\)
d, \(\frac{x}{1\cdot3}+\frac{x}{3\cdot5}+\frac{x}{5\cdot7}+...+\frac{x}{97\cdot99}=99\)
=> \(\frac{x}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right]=99\)
=> \(\frac{x}{2}\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=99\)
=> \(\frac{x}{2}\left[1-\frac{1}{99}\right]=99\)
=> \(\frac{x}{2}\cdot\frac{98}{99}=99\)
=> \(\frac{98x}{198}=99\)
=> 98x = 99 . 198
=> 98x = 19602
=> x = 19602 : 98 = 9801/49
a) \(x-7\frac{5}{8}=1\frac{1}{4}\)
=> \(x=\frac{5}{4}+\frac{61}{8}\)
=> \(x=\frac{71}{8}\)
b) \(x+7\frac{5}{8}=9\frac{1}{4}\)
=> \(x=\frac{37}{4}-\frac{61}{8}\)
=> \(x=\frac{13}{8}\)
c) \(\left(x-7\frac{5}{8}\right):\frac{1}{2}=3\)
=> \(x-\frac{61}{8}=3.\frac{1}{2}\)
=> \(x-\frac{61}{8}=\frac{3}{2}\)
=> \(x=\frac{3}{2}+\frac{61}{8}\)
=> \(x=\frac{73}{8}\)
d) \(\frac{x}{1.3}+\frac{x}{3.5}+...+\frac{x}{97.99}=99\)
=> \(x.\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)=99\)
=> \(\frac{1}{2}x\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)=99\)
=> \(x\left(1-\frac{1}{99}\right)=99:\frac{1}{2}\)
=> \(x.\frac{98}{99}=198\)
=> \(x=198:\frac{98}{99}=\frac{9801}{49}\)
1.n—3 chia hết cho n—1
==> n—1–2 chia hết chi n—1
Vì n—1 chia hết cho n—1
Nên 2 chia hết cho n—1
==> n—1 € Ư(2)
n—1 € {1;—1;2;—2}
Ta có:
TH1: n—1=1
n=1+1
n=2
TH2: n—1=—1
n=—1+1
n=0
TH3: n—1=2
n=2+1
n=3
TH 4: n—1=—2
n=—2+1
n=—1
Vậy n€{2;0;3;—1}
Nếu bạn chưa học số âm thì không cần viết đâu
a) x + \(\frac{1}{6}\)=\(-\frac{3}{8}\)
x=\(\frac{-3}{8}-\frac{1}{6}\)
x=\(\frac{-13}{24}\)
b)\(\frac{1}{2}\)x + \(\frac{1}{8}\)x = \(\frac{3}{4}\)
\(\left(\frac{1}{2}+\frac{1}{8}\right)\)x = \(\frac{3}{4}\)
\(\frac{5}{8}\)x = \(\frac{3}{4}\)
x = \(\frac{3}{4}:\frac{5}{8}\)
x =\(\frac{6}{5}\)
c) 2 - l3/4-xl=7/12
l3/4-xl=17/12
nếu 3/4 - x > 0
thì 3/4 - x = 17/12
x= -2/3
nếu 3/4 - x <0
thì 3/4 - x = -17/12
x = 13/6
d) \(\left(2x-4.5\right):\frac{3}{4}-\frac{1}{3}=1\)
\(\left(2x-4.5\right):\frac{3}{4}=\frac{4}{3}\)
\(2x-4.5=1\)
2x = 5.5
x = 2.75
a, x + 1/6 = -3/8
x = -3/8 - 1/6
x = -13/24
b, 1/2x + 1/8x = 3/4
x * (1/2+1/8) = 3/4
x * 5/8 = 3/4
x = 3/4 : 5/8
x = 6/5
c, 2 - |3/4-x| = 7/12
|3/4-x| = 2 - 7/12
|3/4-x| = 17/12
=> 3/4 - x = 17/12 hoặc -17/12
# 3/4 - x = 17/12
=> x = -2/3
# 3/4 - x = -17/12
=> x = 13/6
=> x = { -2/3 ; 13/6}
d, (2x - 4,5) : 3/4 -1/3 = 1
(2x - 4,5) : 3/4 = 1 + 1/3
(2x - 4,5 ) : 3/4 = 4/3
2x - 4,5 = 4/3 * 3/4
2x - 4,5 = 1
2x = 1 + 4,5
2x = 5,5
x = 5,5 : 2
x = 2,75
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Dấu chấm là nhân
a) \(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\) \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
b) \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\) \(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}=1-\frac{1}{99}=\frac{98}{99}\)
c) Đặt \(C=\frac{4}{5.7}+\frac{4}{7.9}+....+\frac{4}{59.61}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{59}-\frac{1}{61}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{5}-\frac{1}{61}=\frac{56}{305}\)
\(\Rightarrow C=\frac{56}{305}:\frac{1}{2}=\frac{112}{305}\)
CHÚC BẠN HỌC TỐT NHA! ĐÚNG THÌ NHA!
\(a,x-1⋮x-3\)
\(\Rightarrow x-3+2⋮x-3\)
\(\Rightarrow2⋮x-3\)
\(x-3=\left\{-2;-1;1;2\right\}\)
\(x=\left\{1;2;4;5\right\}\)
\(b,x+6⋮x-1\)
\(\Rightarrow x-1+7⋮x-1\)
\(\Rightarrow7⋮x-1\)
\(x=\left\{-6;0;2;8\right\}\)
\(c,x⋮x-5\)
\(x-5+5⋮x-5\)
\(5⋮x-5\)
\(x=\left\{0;4;6;11\right\}\)
bài 1:x.y=-15 => x=3;y=-5
x=-3;y=5
x=5;y=-3
x=-5;y=3
x=-1;y=15
x=1;y=-15
Bài 1 đơn giản rồi nha, chỉ cần liệt kê các gặp số ra là xong
BÀi 2:
ta có:
\(\frac{n-3}{n-1}=\frac{n-1-2}{n-1}=1-\frac{2}{n-1}\)
Để n-3 chia hết cho n-1 <=> \(\frac{2}{n-1}\inℤ\Rightarrow2⋮n-1\)
\(\Rightarrow n-1\inƯ\left(2\right)\)
\(\Rightarrow n-1\in\left\{\pm1;\pm2\right\}\)
ta có bảng sau:
| n-1 | -2 | -1 | 1 | 2 |
| n | -1 | 0 | 2 | 3 |
\(n\in\left\{-1;0;2;3\right\}\)
2,
Q=\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)x\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
= \(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)x\left(\frac{3+\left(-2\right)+\left(-1\right)}{6}\right)\)
=\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)x0\)
Vậy Q=0