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a: \(\Leftrightarrow x\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;9;-9;12;-12;18;-18;36;-36\right\}\)
mà -3<x<30
nên \(x\in\left\{-2;-1;1;2;3;4;6;9;12;18\right\}\)
b: \(\Leftrightarrow x\in\left\{0;4;-4;8;-8;12;-12;...\right\}\)
mà -16<=x<20
nên \(x\in\left\{-16;-12;-8;-4;0;4;8;12;16\right\}\)
c: \(\Leftrightarrow x-1+4⋮x-1\)
\(\Leftrightarrow x-1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{2;0;3;-1;5;-3\right\}\)
d: \(\Leftrightarrow2x+4-5⋮x+2\)
\(\Leftrightarrow x+2\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{-1;-3;3;-7\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 9:
Ta có: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{z}{-17}=\dfrac{-t}{-9}\)
\(\Leftrightarrow\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{-z}{17}=\dfrac{t}{9}=-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=-2\\\dfrac{-y}{3}=-2\\\dfrac{-z}{17}=-2\\\dfrac{t}{9}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\-y=-6\\-z=-34\\t=-18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\y=6\\z=34\\t=-18\end{matrix}\right.\)
Vậy: (x,y,z,t)=(-10;6;34;-18)
Bài 11:
Ta có: \(\dfrac{-7}{6}=\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}\)
\(\Leftrightarrow\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}=\dfrac{-7}{6}\)
Ta có: \(\dfrac{x}{18}=\dfrac{-7}{6}\)
\(\Leftrightarrow x=\dfrac{18\cdot\left(-7\right)}{6}=-21\)
Ta có: \(\dfrac{-98}{y}=\dfrac{-7}{6}\)
\(\Leftrightarrow y=\dfrac{-98\cdot6}{-7}=84\)
Ta có: \(\dfrac{-14}{z}=\dfrac{-7}{6}\)
\(\Leftrightarrow z=\dfrac{-14\cdot6}{-7}=12\)
Ta có: \(\dfrac{u}{-78}=\dfrac{-7}{6}\)
\(\Leftrightarrow u=\dfrac{-78\cdot\left(-7\right)}{6}=\dfrac{78\cdot7}{6}=91\)
Ta có: \(\dfrac{t}{102}=\dfrac{-7}{6}\)
\(\Leftrightarrow t=\dfrac{-7\cdot102}{6}=-7\cdot17=-119\)
Vậy: (x,y,z,t,u)=(-21;84;12;-119;91)
![](https://rs.olm.vn/images/avt/0.png?1311)
TA CÓ: \(\frac{3+x}{5+y}=\frac{3}{5}\)và \(x+y=16\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{3+x}{5+y}=\frac{3}{5}\Leftrightarrow\frac{3+x}{3}=\frac{5+y}{5}=\frac{3+x+y+5}{3+5}=\frac{24}{8}=3\)
\(\Rightarrow\hept{\begin{cases}\frac{3+x}{3}=3\Leftrightarrow3+x=9\Leftrightarrow x=6\\\frac{5+y}{5}=3\Leftrightarrow5+y=15\Leftrightarrow y=10\end{cases}}\)
Vậy \(x=6\)và \(y=10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Giải:
a) \(\dfrac{-5}{8}=\dfrac{x}{16}\)
\(\Rightarrow x=\dfrac{16.-5}{8}=-10\)
\(\dfrac{3x}{9}=\dfrac{2}{6}\)
\(\Rightarrow3x=\dfrac{2.9}{6}=3\)
\(\Rightarrow x=1\)
b) \(\dfrac{x+3}{15}=\dfrac{1}{3}\)
\(\Rightarrow x+3=\dfrac{1.15}{3}=5\)
\(\Rightarrow x=2\)
\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
\(\Rightarrow2x+1=\dfrac{6.7}{2}=21\)
\(\Rightarrow x=10\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\)
\(\Rightarrow x-6=\dfrac{18.4}{-12}=-6\)
\(\Rightarrow x=0\)
\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow y=\dfrac{-12.24}{18}=-16\)
\(\dfrac{3-x}{-12}=\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow\dfrac{3-x}{-12}=\dfrac{192}{-72}\)
\(\Rightarrow3-x=\dfrac{192.-12}{-72}=32\)
\(\Rightarrow x=-29\)
\(\Rightarrow\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow y+1=\dfrac{16.-72}{192}=-6\)
d) \(\dfrac{-2}{3}< \dfrac{x}{5}< \dfrac{-1}{6}\)
\(\Rightarrow\dfrac{-20}{30}< \dfrac{6x}{30}< \dfrac{-5}{30}\)
\(\Rightarrow6x\in\left\{-18;-12;-6\right\}\)
\(\Rightarrow x\in\left\{-3;-2;-1\right\}\)
\(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{-5;0;5;10\right\}\)
\(\Rightarrow x\in\left\{-1;0;1;2\right\}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=x+\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=\dfrac{5x+2}{5}\)
\(\Rightarrow5.\left(x+46\right)=20.\left(5x+2\right)\)
\(\Rightarrow5x+230=100x+40\)
\(\Rightarrow5x-100x=40-230\)
\(\Rightarrow-95x=-190\)
\(\Rightarrow x=-190:-95\)
\(\Rightarrow x=2\)
\(y\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y+\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow\dfrac{y^2+5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y^2+5=86\)
\(\Rightarrow y^2=86-5\)
\(\Rightarrow y^2=81\)
\(\Rightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\)
Chúc bạn học tốt!
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(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 \(\left(x;y\right)\)\(\in\)\(\left\{\left(-3;7\right);\left(-5;-1\right);\left(-2;5\right);\left(-6;1\right);\left(0;4\right);\left(-8;2\right)\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Do \(\frac{3+x}{7+y}=\frac{3}{7}\)=> x=3p, y=7q (p, q\(\in\)Z)
Ta có: x+y=3p+7q=20 hay 3(p+q)+4q=20 => 0<p+q<6
Do 20\(⋮\)4, 4q\(⋮\)4 => 3(p+q)\(⋮\)4 mà (3,4)=1 => p+q\(⋮\)4.
=> p+q=4 => q=(20-3.4):4=2 => y=2.7=14
=> p=4-2=2 => x=2.3=6
=>\(\frac{3+x}{7+y}=\)một phân số có thể rút gọn thành\(\frac{3}{7}\)
Giả sử x=3; y=7. Vì \(\frac{3+3}{7+7}=\frac{6}{14}=\frac{3}{7}\)Nhưng 3+7=10 (loại)
x=6; y=14. Vì\(\frac{3+6}{7+14}=\frac{9}{21}=\frac{3}{7}\)Và 6+14=20 (thỏa mãn)
Vậy x=6; y=14
![](https://rs.olm.vn/images/avt/0.png?1311)
Tớ mới học lớp 5
Tra loi: Vi x+3/y+5 = 3/5
=>(x+3).5 = (y+5) .3
=>5x+15=3y+15
=>5x=3y
=>x=3k;y=5k( k thuoc Z; k khac 0)
Ma x+y=16=> 3k+5k=16
=> k=2
=> x= 3 .2 =6 ;y=2.5=10
Vay x=6 va y+10