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15-2n:n+1
2(n+1):n+1
15-2n-2(n+1):n+1
15-2n-2n-2:n+1
15-2:n+1
13:n+1
→n+1={1;13}
→n={9;12}
y+z+1x=x+z+2y=x+y−3z=1x+y+zy+z+1x=x+z+2y=x+y−3z=1x+y+z(đk x+y+z≠0≠0
⇒y+z+1x=x+z+2y=x+y−3z=y+z+1+x+z+2+x+y−3x+y+z=2⇒y+z+1x=x+z+2y=x+y−3z=y+z+1+x+z+2+x+y−3x+y+z=2
⇒1x+y+z=2⇒x+y+z=0,5⇒1x+y+z=2⇒x+y+z=0,5
⇒y+z=0,5−x,x+z=0,5−y,x+y=0,5−z⇒y+z=0,5−x,x+z=0,5−y,x+y=0,5−z
⇒0,5−x+1x=2⇒1,5−xx=2⇒1,5−x=2x⇒3x=1,5⇒x=12⇒0,5−x+1x=2⇒1,5−xx=2⇒1,5−x=2x⇒3x=1,5⇒x=12
⇒0,5−y+2y=2⇒2,5−yy=2⇒2,5−y=2y⇒3y=2,5⇒y=56⇒0,5−y+2y=2⇒2,5−yy=2⇒2,5−y=2y⇒3y=2,5⇒y=56
⇒z=0,5−12−56=−56⇒z=0,5−12−56=−56
Vậy x=12,y=56,z=−56
\(6n+9⋮4n-1\)
\(\Rightarrow2.\left(6n+9\right)⋮4n-1\)
\(\Rightarrow12n+18⋮4n-1\)
\(\Rightarrow12n-3+21⋮4n-1\)
\(\Rightarrow3.\left(4n-1\right)+21⋮4n-1\)
Vì \(3.\left(4n-1\right)⋮4n-1\Rightarrow21⋮4n-1\)
Mà 4n - 1 chia 4 dư 3; \(4n-1\ge-1\) do \(n\in N\)
\(\Rightarrow4n-1\in\left\{-1;3;7\right\}\)
\(\Rightarrow4n\in\left\{0;4;8\right\}\)
\(\Rightarrow n\in\left\{0;1;2\right\}\)
a: \(=\dfrac{1}{3}-\dfrac{37}{100}+\dfrac{1}{8}-\dfrac{32}{25}+\dfrac{-5}{2}+\dfrac{3}{2}\)
\(=\dfrac{1}{3}+\dfrac{1}{8}-1-\dfrac{37}{100}-\dfrac{128}{100}\)
\(=\dfrac{8+3-24}{24}-\dfrac{165}{100}\)
\(=\dfrac{-263}{120}\)
b: \(=\dfrac{5}{22}+\dfrac{3}{13}-\dfrac{13}{8}-\dfrac{2}{11}+\dfrac{3}{2}\)
\(=\dfrac{5}{22}-\dfrac{4}{22}-\dfrac{13}{8}+\dfrac{12}{8}+\dfrac{3}{13}\)
\(=\dfrac{1}{22}-\dfrac{1}{8}+\dfrac{3}{13}=\dfrac{173}{1144}\)
a: \(=\dfrac{1}{3}-\dfrac{37}{100}+\dfrac{1}{8}-\dfrac{32}{25}-\dfrac{5}{2}+\dfrac{3}{2}\)
\(=\dfrac{11}{24}+\left(-\dfrac{37}{100}+\dfrac{1}{8}-1\right)\)
\(=\dfrac{11}{24}-\dfrac{249}{200}=-\dfrac{59}{75}\)
b: \(=\dfrac{5}{22}+\dfrac{3}{13}-\dfrac{13}{8}-\dfrac{2}{11}+\dfrac{3}{2}\)
\(=\dfrac{5}{22}-\dfrac{4}{22}+\dfrac{3}{13}-\dfrac{13}{8}+\dfrac{12}{8}\)
\(=\dfrac{1}{22}+\dfrac{3}{13}-\dfrac{1}{8}=\dfrac{173}{1144}\)
\(B=\left(1-\frac{1}{12}\right)\left(1-\frac{1}{13}\right)\left(1-\frac{1}{14}\right)\)
\(=>B=\frac{11}{12}\cdot\frac{12}{13}\cdot\frac{13}{14}\)
\(=>B=\frac{11\cdot12\cdot13}{12\cdot13\cdot14}=\frac{11}{14}\)
Bài 1 :
a) 16.(38−2)−38(16−1)
= 16 * 38 - 2* 16 - 16 *38 + 38
= - 32 +38
= 6
b) (−41).(59+2)+59(41−2)
= (-41) * 59 - 41 *2 + 41 * 59 - 59 *2
= 2(-41 - 59)
= 2* (-100)
= -200
Bài 2 :
Tìm các số x ; y ; x biết rằng :
x + y = 2 ; y + z = 3 ; z + x = -5
=> 2(x+ y + z)= 2+3+(-5)= 0
=> x+y +z =0
=> z= (x+y+z) -(x+y) = 0-2 =-2
x= (x+y+z) -(y+z) = 0-3 =-3
y = (x+y+z) - (x+z) = 0 -(-5) =0+5 =5
Vậy x= -2 ; y=-3 ; z=5
Bài 3 : Tìm x ; y ∈ Z biết rằng :
( y + 1 ) . xy - 1 ) = 3
(sao có 3 dấu ngoặc z?)
Bài 3 : Tìm x ; y ∈∈ Z biết rằng :
( y + 1 ) . ( xy - 1 ) = 3
Th1: 3= -3 * (-1)
y+1 = -3 => y= -4
xy-1 = -1 => xy = 1
x(-4) =1
x= -1/4
Mà x,y thuộc Z => loại
Th2: 3= (-1) * (-3)
y+1 = -1 => y= -2
xy -1 = -3 => xy = 3
x(-2) =3
x= -3/2
Mà x,y thuộc Z => loại
Th3: 3= 1*3
y+ 1 = 1 => y= 0
xy -1 =3 => xy = 4
x*0 =4
ko tồn tại x
Th4: 3= 3*1
y+1 = 3 => y =2
xy -1 = 1 => xy = 2
x*2 =2
x= 1
Vậy x=1 và y=2
Bài 1:
a)16.(38-2)-38.(16-1)
=16.38-16.2-38.16+38.1
=16.38-38.16+38-16.2
=0+38-32
=6
b) (-41).(59+2)+59.(41-2)
= -41.59-41.2+59.41-59.2
=59.41-41.59-2.(59+41)
= -2.100
=-200
Bài 2 :
vì : x+y=2 ; y+z=3 ; z+x=-5
=>2.(x+y+z)=2(2+3-5) =0
=>x+y+z=0:2=0
=>y=(x+y+z)-(z+x)=0-(-5)=5
=>x=(x+y+z)-(y+z)=0-3=-3
=>z=(x+y+z)-(y+x)=0-2=-2
Vậy x=-3
z=-2
y=5
phần a (1)bn ko lm à ?
Bài 1 :
a)16.(38-2)-38.(16-1)
=16.38-16.2-38.16+38.1
=16.38-38.16+38-16.2
=38-32
=6
b)(-41).(59+2)+59.(41-2)
= -41.59-41.2+59.41-59.2
=59.41-41.59-2(41+59)
=-2.100
=-200
Bài 2:
Vì : x+y=2 ; y+z=3 ; z+x=-5
=>2.(x+y+z)=2.(2+3-5)=0
=>x+y+z=0:2=0
=>y=(x+y+z)-(z+x)=0+5=5
=>z=(x+y+z)-(x+y)=0-2=-2
=>x=(x+y+z)-(y+z)=0-3=-3
Vậy x=-3
y=5
z=-2
Bài 3 :
vì (y+1).(xy-1)=3
=>y+1;xy-1ϵƯ(3)=\(\left\{-3;-1;1;3\right\}\)
Vậy (x,y)=(1;2)
sao xy-1=3 mà xy=3
Xin lỗi ,mk nhầm , Bài 3 : (x,y)=(1;-2),(1;2)
Nhắm mắt chọn bừa 1 đáp án mà cũng chọn đúng