K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
25 tháng 1
Bài 3:
Để A là số nguyên thì \(2n+7⋮n+3\)
=>\(2n+6+1⋮n+3\)
=>\(1⋮n+3\)
=>\(n+3\in\left\{1;-1\right\}\)
=>\(n\in\left\{-2;-4\right\}\)
Bài 4:
Để A là số nguyên thì \(2n+7⋮n+1\)
=>\(2n+2+5⋮n+1\)
=>\(5⋮n+1\)
=>\(n+1\in\left\{1;-1;5;-5\right\}\)
=>\(n\in\left\{0;-2;4;-6\right\}\)
Bài 5:
Để A là số nguyên thì \(6n-3⋮3n+1\)
=>\(6n+2-5⋮3n+1\)
=>\(-5⋮3n+1\)
=>\(3n+1\in\left\{1;-1;5;-5\right\}\)
=>\(3n\in\left\{0;-2;4;-6\right\}\)
=>\(n\in\left\{0;-\dfrac{2}{3};\dfrac{4}{3};-2\right\}\)
mà n nguyên
nên \(n\in\left\{0;-2\right\}\)
Bài 6:
Để A là số nguyên thì \(3n+4⋮n-1\)
=>\(3n-3+7⋮n-1\)
=>\(7⋮n-1\)
=>\(n-1\in\left\{1;-1;7;-7\right\}\)
=>\(n\in\left\{2;0;8;-6\right\}\)
H9
HT.Phong (9A5)
CTVHS
11 tháng 1
Bài 4:
a) 2x + 7 ⋮ x + 2
⇒ 2x + 4 + 3 ⋮ x + 2
⇒ 2(x + 2) + 3 ⋮ x + 2
⇒ 3 ⋮ x + 2
⇒ x + 2 ∈ Ư(3) = {1; -1; 3; -3}
⇒ x ∈ {-1; -3; 1; -5}
b) 2x + 7 ⋮ x - 3
⇒ 2x - 6 + 13 ⋮ x - 3
⇒ 2(x - 3) + 13 ⋮ x - 3
⇒ 13 ⋮ x - 3
⇒ x - 3 ∈ Ư(13) = {1; -1; 13; -13}
⇒ x ∈ {4; 2; 16; -10}
11 tháng 1
Bài 6:
a: \(3x-13⋮x+3\)
=>\(3x+9-22⋮x+3\)
=>\(-22⋮x+3\)
=>\(x+3\in\left\{1;-1;2;-2;11;-11;22;-22\right\}\)
=>\(x\in\left\{-2;-4;-1;-5;8;-14;19;-25\right\}\)
b: \(2x+24⋮x-4\)
=>\(2x-8+32⋮x-4\)
=>\(32⋮x-4\)
=>\(x-4\in\left\{1;-1;2;-2;4;-4;8;-8;16;-16;32;-32\right\}\)
=>\(x\in\left\{5;3;6;2;8;0;12;-4;20;-12;36;-28\right\}\)
Bài 5:
a: \(4x+3⋮x-2\)
=>\(4x-8+11⋮x-2\)
=>\(11⋮x-2\)
=>\(x-2\in\left\{1;-1;11;-11\right\}\)
=>\(x\in\left\{3;1;13;-9\right\}\)
b: \(2x+7⋮x-3\)
=>\(2x-6+13⋮x-3\)
=>\(13⋮x-3\)
=>\(x-3\in\left\{1;-1;13;-13\right\}\)
=>\(x\in\left\{4;2;16;-10\right\}\)
Bài 4:
a: \(2x+7⋮x+2\)
=>\(2x+4+3⋮x+2\)
=>\(3⋮x+2\)
=>\(x+2\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{-1;-3;1;-5\right\}\)
b: \(2x+7⋮x-3\)
=>\(2x-6+13⋮x-3\)
=>\(13⋮x-3\)
=>\(x-3\in\left\{1;-1;13;-13\right\}\)
=>\(x\in\left\{4;2;16;-10\right\}\)
NN
15 tháng 12 2022
92+(-13).36+(-13).34
=92+(-13).(36+34)
=92+(-13).70
=92+(-910)
=-818
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
6 tháng 1
Bài 2:
\(a,\dfrac{-1}{3}.\dfrac{5}{7}=\dfrac{-5}{21}\\ b,\dfrac{1}{2}.\dfrac{-3}{4}=\dfrac{-3}{8}\\ c,\dfrac{19}{7}.\dfrac{7}{15}=\dfrac{19.1}{1.15}=\dfrac{19}{15}\\ d,\dfrac{5}{11}:\left(-9\right)=\dfrac{5}{11}.\dfrac{-1}{9}=\dfrac{-5}{99}\\ e,-1:\dfrac{3}{5}=-1.\dfrac{5}{3}=-\dfrac{5}{3}\\ f,\dfrac{-2}{9}:\dfrac{2}{9}:\dfrac{-9}{14}=-\left(\dfrac{2}{9}:\dfrac{2}{9}\right).\dfrac{-14}{9}=-1.\dfrac{-14}{9}=\dfrac{14}{9}\\ k,\left(-\dfrac{2}{7}\right)^2=\left(-1\right)^2.\left(\dfrac{2}{7}\right)^2=1.\dfrac{2^2}{7^2}=\dfrac{4}{49}\\ l,\dfrac{1}{3}.\left(\dfrac{1}{3}\right)^3=\left(\dfrac{1}{3}\right)^4=\dfrac{1^4}{3^4}=\dfrac{1}{243}\\ m,\left(-\dfrac{1}{2}\right)^2.\left(\dfrac{1}{2}\right)^3=\left(-1\right)^2.\left(\dfrac{1}{2}\right)^{2+3}=1.\left(\dfrac{1}{2}\right)^5=1.\dfrac{1^5}{2^5}=\dfrac{1}{32}\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
6 tháng 1
Bài 1:
\(a,\dfrac{1}{-8}+\dfrac{-5}{8}=\dfrac{-1}{8}+\dfrac{-5}{8}=\dfrac{-\left(1+5\right)}{8}=-\dfrac{6}{8}=\dfrac{-6:2}{8:2}=-\dfrac{3}{4}\\ b,\dfrac{1}{7}+\dfrac{-3}{7}=\dfrac{1-3}{7}=-\dfrac{2}{7}\\ c,\dfrac{-12}{35}+\dfrac{-7}{35}=\dfrac{-\left(12+7\right)}{35}=\dfrac{-19}{35}\\ d,\dfrac{1}{6}+\dfrac{2}{5}=\dfrac{1.5+2.6}{6.5}=\dfrac{5+12}{30}=\dfrac{17}{30}\\ e,\dfrac{3}{5}+\dfrac{-7}{4}=\dfrac{3.4-7.5}{5.4}=\dfrac{12-35}{20}=\dfrac{-23}{20}\\ g,-2-\left(-\dfrac{1}{5}\right)=-2+\dfrac{1}{5}=\dfrac{-2.5+1}{5}=\dfrac{-9}{5}\\ h,\dfrac{2}{3}-\left(-1\right)=\dfrac{2}{3}+1=\dfrac{5}{3}\\ i,4-\dfrac{2}{3}=\dfrac{4.3-2}{3}=\dfrac{12-2}{3}=\dfrac{10}{3}\\ j,\dfrac{3}{4}-2=\dfrac{3-2.4}{4}=\dfrac{-5}{4}\\ k,-1-\left(-\dfrac{2}{3}\right)=-1+\dfrac{2}{3}=\dfrac{-1.3+2}{3}=\dfrac{-3+2}{3}=-\dfrac{1}{3}\)
28 tháng 11 2023
5. ( - 199 ) + ( - 200 ) + ( - 201 )
= [ ( - 199 ) + ( - 20 ) ] + ( - 200 )
= ( - 400 ) + ( - 200 )
= - 600
Câu 1:
\(M=132-\left\{100-\left[\left(78-73\right)^2:5+9\right]\right\}\)
\(=132-\left\{100-\left[5^2:5+9\right]\right\}\)
\(=132-100+5+9\)
=32+14
=46
Câu 2:
105-[(2x+7)-13]=25
=>(2x+7)-13=105-25=90
=>2x-6=90
=>2x=96
=>x=96/2=48
Câu 4:
\(N=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{99\cdot101}\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{1}{2}\cdot\dfrac{100}{101}=\dfrac{50}{101}\)
Câu 5:
\(xy-x+2y=3\)
\(\Rightarrow xy-x+2y-2=3-2\)
\(\Rightarrow\left(xy-x\right)+\left(2y-2\right)=1\)
\(\Rightarrow x\left(y-1\right)+2\left(y-1\right)=1\)
\(\Rightarrow\left(x+2\right)\left(y-1\right)=1\)
Ta có bảng sau:
Vậy các cặp (x;y) là (-1;2) ; (-3;0)
#YM