Tìm x thuộc z
6x+3 phần 3x-1
Giúp mình vs
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6x+3=2.(3x-1)+5 chia hết cho 3x-1=> 5 chia hết cho 3x-1=> 3x-1 thuộc Ư(5)={........}
+) A = \(\frac{3}{x-1}\)
=> x-1 \(\in\) Ư(3) = {-1,-3,1,3}
Ta có bảng :
x-1 | -1 | -3 | 1 | 3 |
x | 0 (loại) | -2 | 2 | 4 |
Vậy x = { -2,2,4 }
+) Bài B đề chưa rõ
+) C = \(\frac{11}{3x-1}\)
=> 3x-1 \(\in\) Ư(11) = { -1,-11,1,11 }
Ta có bảng :
3x-1 | -1 | -11 | 1 | 11 |
x | 0 (loại) | \(\frac{-10}{3}\) (loại) | \(\frac{2}{3}\) (loại) | 4 |
Vậy x = 4
+) M = \(\frac{x+2}{x-1}\)
Ta có: \(\frac{x+2}{x-1}=\frac{x-1+3}{x-1}=\frac{x-1}{x-1}+\frac{3}{x-1}=1+\frac{3}{x-1}\)
=> x-1 \(\in\) Ư(3) = {-1,-3,1,3}
Tiếp theo như bài A mình đã làm
E = \(\frac{x+7}{x+2}=\frac{x+2+5}{x+2}=\frac{x+2}{x+2}+\frac{5}{x+2}=1+\frac{5}{x+2}\)
=> x+2 \(\in\) Ư(5) = {-1,-5,1,5 }
Ta có bảng :
x+2 | -1 | -5 | 1 | 5 |
x | -3 | -7 | -1 | 3 |
Vậy x = { -7,-3,-1,3 }
\(\dfrac{1}{2}\left(6x-2y\right)\left(3x+y\right)=\dfrac{1}{2}.2\left(3x-y\right)\left(3x+y\right)=9x^2-y^2\)
\(\left(\dfrac{2}{3}z-\dfrac{2}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right).\dfrac{1}{2}=\left(\dfrac{1}{3}z-\dfrac{1}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}z\right).2.\dfrac{1}{2}=\dfrac{1}{9}z^2-\dfrac{1}{25}x^2\)
\(\left(5y-3x\right).\dfrac{1}{4}\left(12x+20y\right)=\left(5y-3x\right)\left(5y+3x\right).4.\dfrac{1}{4}=25y^2-9x^2\)
\(\left(\dfrac{3}{4}y-\dfrac{1}{2}x\right)\left(x+\dfrac{3}{2}y\right)=\left(\dfrac{3}{2}y-x\right)\left(\dfrac{3}{2}y+x\right)=\dfrac{9}{4}y^2-x^2\)
\(\left(a+b+c\right)\left(a+b+c\right)=\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
\(\left(x-y+z\right)\left(x+y-z\right)=x^2-\left(y-z\right)^2=x^2-y^2-z^2+2yz\)
\(A=2x^3+6x^2-3x+\dfrac{1}{2}=2\cdot\dfrac{1}{3}^3+6\cdot\dfrac{1}{3}^2-3\cdot\dfrac{1}{3}+\dfrac{1}{2}\)
=13/54
5)
để \(\frac{5x-3}{x+1}\)là số nguyên
\(5x-3⋮x+1\)
\(x+1⋮x+1\)
\(\Rightarrow5\left(x+1\right)⋮x+1\)
\(5x-3-\left(5x-5\right)⋮x+1\)
\(-2⋮x+1\)
\(\Rightarrow x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
x+1 | 1 | -1 | 2 | -2 |
x | 0 | -2 | 1 | -3 |
Vậy \(x\in\left\{0;-2;1;-3\right\}\)
PT <=> (3x - 1)(6x - 1)(4x - 1)(5x - 1) = 120
. <=> (18x² - 9x + 1)(20x² - 9x + 1) = 120
Đặt a = 19x² - 9x + 1 (Đk a > 0) ta có PT: (a - 1)(a + 1) = 120
<=> a² - 1 = 120
<=> a² = 121
<=> a = 11 (Vì a >0)
Với a = 11 ta có PT: 19x² - 9x - 10 = 0
<=> (10x + 19)(x - 1) = 0
<=> x = 1 (Vì x nguyên)
KL: x = 1
(x+1)+ (x+3) + (x+5)+.....+(x+99) = 0
x+1 + x+3 +x+5 +....+x+99 =0
Có số số hạng x là : (99-1):2+1= 50 số
Ta có: 50x + ( 1+3+5+...+99) = 0
Đặt A= 1+3+5+...+99
Tổng A là: (99+1).50:2= 2500
=> 50x + 2500 = 0
50x = 0-2500
50x= -2500
x= -2500 :50
x= -50
Vậy...
a) xy - 3x =-19
x(y-3) = -19
=> y-3 \(\in\)Ư(-19) ={ 1; 19; -19 ; -1}
=> y \(\in\){ 4; 22; -16; 2}
Sau bn lập bảng tìm x nha
b) 3x + 4y - xy = 16
3x + y(4-x) =16
12 - [ 3x+ y(4-x)] =12-16
12 - 3x - y(4-x)= -4
3(4-x)- y(4-x) = -4
(3-y) ( 4-x) =-4
Sau bn lập bảng tìm xy nha
Nguồn phần b là của bn Tài nha :>
Bài 1 :
\(\left(x+1\right)+\left(x+3\right)+\left(x+5\right)+...+\left(x+99\right)=0\)
Có tất cả các số số hạng là : \(\left(99-1\right)\div2+1=50\) ( số )
\(x+1+x+3+x+5+...+x+99=0\)
\(x+x+...+x+1+3+...+99=0\)
\(\left(x\times50\right)+\left[\left(99+1\right)\times50\div2\right]=0\)
\(\left(x\times50\right)+\left(100\times50\div2\right)=0\)
\(\left(x\times50\right)+\left(5000\div2\right)=0\)
\(\left(x\times50\right)+2500=0\)
\(x\times50=0-2500\)
\(x\times50=-2500\)
\(x=-2500\div50\)
\(x=-50\)
Bài 2 :
a ) \(xy-3x=-19\)
\(\Leftrightarrow\)\(x,y\inℤ\)và \(y-3\) \(\inƯ\)\(\left(-19\right)\)\(\in\)\(\left\{1;-1;19;-19\right\}\)
Ta có bảng sau
x | - 19 | 19 | - 1 | 1 |
y - 3 | 1 | - 1 | 19 | - 19 |
y | 4 | 2 | 22 | - 16 |
Vậy \(\left(x;y\right)\) \(\in\) \(\left\{\left(-19;4\right);\left(19;2\right);\left(-1;22\right);\left(1;-16\right)\right\}\)
b ) \(3x+4y-xy=16\)
\(\Leftrightarrow3x+4y-xy-12=16-12\)
\(\Leftrightarrow\left(3x-xy\right)+\left(4y-12\right)=4\)
\(\Leftrightarrow x\left(3-y\right)+4\left(-y\right)+3=4\)
\(\Leftrightarrow\left(3-y\right)\left(x+4\right)=4\)
\(\Leftrightarrow\)\(x;y\)\(\inℤ\)\(\Rightarrow\)\(3-y\) và \(x+4\)\(\in\)\(Ư\)\(\left(4\right)\)=
Ta có bảng sau :
x + 4 | 1 | - 1 | 2 | - 2 | 4 | - 4 |
x | - 3 | - 5 | - 2 | - 6 | 0 | - 8 |
y - 3 | 4 | - 4 | 2 | - 2 | 1 | - 1 |
y | 7 | - 1 | 5 | 1 | 4 | 2 |
Vậy \(\left(x;y\right)\)\(\in\)\(\left\{\left(-3;7\right);\left(-5;-1\right);\left(-2;5\right);\left(-6;1\right);\left(0;4\right);\left(-8;2\right)\right\}\)
27/12/2017 lúc 18:59
Ex1: Điền từ thích hợp vào chỗ trống
This is Ba. He(1)......... a student.Every morning he(2).........up at 5.30.He(3).............. his teeth and takes a(4)............... then has breakfast at 6.15. He goes to school(5)........six thirty.His house is(6).............his house so he walks.The classes(7)............at 7.15 and finish at 11.15.In the afternoon he plays sports with his friend,Nam. They play badminton but now they(8).................soccer.In the evening he (9)......his homework and goes to(10).........at 9.30
Ex2:Cho dạng đúng của động từ trong ngoặc
1.My sister(have)...........classes from Monday to Friday
2.She(read)................a book in her room now
3.He(get)........................up at 6.00 every day?
4.There(not be)..............a big yard behind his classroom
27/12/2017 lúc 18:59
Ex1: Điền từ thích hợp vào chỗ trống
This is Ba. He(1)......... a student.Every morning he(2).........up at 5.30.He(3).............. his teeth and takes a(4)............... then has breakfast at 6.15. He goes to school(5)........six thirty.His house is(6).............his house so he walks.The classes(7)............at 7.15 and finish at 11.15.In the afternoon he plays sports with his friend,Nam. They play badminton but now they(8).................soccer.In the evening he (9)......his homework and goes to(10).........at 9.30
Ex2:Cho dạng đúng của động từ trong ngoặc
1.My sister(have)...........classes from Monday to Friday
2.She(read)................a book in her room now
3.He(get)........................up at 6.00 every day?
4.There(not be)..............a big yard behind his classroom
27/12/2017 lúc 18:59
Ex1: Điền từ thích hợp vào chỗ trống
This is Ba. He(1)......... a student.Every morning he(2).........up at 5.30.He(3).............. his teeth and takes a(4)............... then has breakfast at 6.15. He goes to school(5)........six thirty.His house is(6).............his house so he walks.The classes(7)............at 7.15 and finish at 11.15.In the afternoon he plays sports with his friend,Nam. They play badminton but now they(8).................soccer.In the evening he (9)......his homework and goes to(10).........at 9.30
Ex2:Cho dạng đúng của động từ trong ngoặc
1.My sister(have)...........classes from Monday to Friday
2.She(read)................a book in her room now
3.He(get)........................up at 6.00 every day?
4.There(not be)..............a big yard behind his classroom