a: \(\Leftrightarrow2n-2+3⋮n-1\)

\(\Leftrightarrow n-1\in\left\{1;-1;3;-3\right\}\)

hay \(n\in\left\{2;0;4;-2\right\}\)

b \(\Leftrightarrow12n-9⋮3n+1\)

\(\Leftrightarrow12n+4-13⋮3n+1\)

\(\Leftrightarrow3n+1\in\left\{1;-1;13;-13\right\}\)

hay \(n\in\left\{0;-\dfrac{2}{3};4;-\dfrac{14}{3}\right\}\)

c: \(\Leftrightarrow n\left(n-1\right)+1⋮n-1\)

\(\Leftrightarrow n-1\in\left\{1;-1\right\}\)

hay \(n\in\left\{2;0\right\}\)

Nhìn cứ giống toán lớp 8,9 hơn lớp 6 đúng hok

c: \(=5+\dfrac{6}{7}-2-\dfrac{3}{8}-1-\dfrac{1}{8}=2+\dfrac{6}{7}-\dfrac{1}{2}=\dfrac{3}{2}+\dfrac{6}{7}=\dfrac{33}{14}\)

e: \(=\dfrac{-3}{5}\left(-\dfrac{1}{9}-\dfrac{5}{6}+\dfrac{5}{2}\right)=\dfrac{-3}{5}\cdot\dfrac{14}{9}=\dfrac{-14}{15}\)

d: =-76x10=-760

senpai trả lời đầy đủ ấy

c: =547x100=54700

d: =-76x10=-760

Bài 4:

a: (x+1)(y-2)=3

=>(x+1;y-2)∈{(1;3);(3;1);(-1;-3);(-3;-1)}

=>(x;y)∈{(0;5);(2;3);(-2;-1);(-4;1)}

b: (2x-1)(3y+1)=12

mà 2x-1 lẻ

nên (2x-1;3y+1)∈{(1;12);(-1;-12);(3;4);(-3;-4)}

=>(2x;3y)∈{(2;11);(0;-13);(4;3);(-2;-5)}

mà 3y⋮3

nên (2x;3y)∈{(4;3)}

=>(x;y)∈{(2;1)}

c: xy-x+y-2=0

=>x(y-1)+y-1-1=0

=>(x+1)(y-1)=1

=>(x+1;y-1)∈{(1;1);(-1;-1)}

=>(x;y)∈{(0;2);(-2;0)}

Bài 3:

a: 3n-1⋮n+2

=>3n+6-7⋮n+2

=>-7⋮n+2

=>n+2∈{1;-1;7;-7}

=>n∈{-1;-3;5;-9}

b:2n+3⋮3n-1

=>6n+9⋮3n-1

=>6n-2+11⋮3n-1

=>11⋮3n-1

=>3n-1∈{1;-1;11;-11}

=>3n∈{2;0;12;-10}

=>n∈{2/3;0;4;-10/3}

mà n là số nguyên

nên n∈{0;4}

Bài 4: 

c: \(\Leftrightarrow n+2\in\left\{1;-1;5;-5\right\}\)

hay \(n\in\left\{-1;-3;3;-7\right\}\)

Bài 4:

a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{x+y}{4+3}=\dfrac{14}{7}=2\)

Do đó: x=8; y=6

b: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{2x+3y}{2\cdot8+3\cdot12}=\dfrac{13}{52}=\dfrac{1}{4}\)

Do đó: x=2; y=3

a) \(\dfrac{x}{5}=\dfrac{2}{5}.\Rightarrow x=2.\)

b) \(\dfrac{3}{x-5}=\dfrac{-4}{x+2.}.\left(x\ne5;x\ne-2\right).\)

\(\Leftrightarrow\dfrac{3}{x-5}+\dfrac{4}{x+2}=0.\Leftrightarrow\dfrac{3x+6+4x-20}{\left(x-5\right)\left(x+2\right)}=0.\) 

\(\Rightarrow7x-14=0.\Leftrightarrow x=2\left(TM\right).\)

c) \(\dfrac{x}{-2}=\dfrac{-8}{x}.\left(x\ne0\right).\Leftrightarrow\dfrac{-x}{2}+\dfrac{8}{x}=0.\Leftrightarrow\dfrac{-x^2+16}{2x}=0.\Rightarrow-x^2+16=0.\Leftrightarrow x^2=16.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=4^2.\\x^2=\left(-4\right)^2.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4.\\x=-4.\end{matrix}\right.\)\(\left(TM\right).\)

d) \(\dfrac{x+2}{5}=\dfrac{45}{x+2}.\left(x\ne-2\right).\)

\(\Leftrightarrow\dfrac{x+2}{5}-\dfrac{45}{x+2}=0.\Leftrightarrow\dfrac{x^2+4x+4-225}{5x+10}=0.\Rightarrow x^2+4x-221=0.\)

\(\Leftrightarrow\left(x-13\right)\left(x+17\right)=0.\Leftrightarrow\left[{}\begin{matrix}x-13=0.\\x+17=0.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=13.\\x=-17.\end{matrix}\right.\) \(\left(TM\right).\)

đăng từng câu thì đầy đủ

Bài 3: 

a: =>1<x<=4

hay \(x\in\left\{2;3;4\right\}\)

b: =>-104<x<-99

hay \(x\in\left\{-103;-102;-101;-100\right\}\)

a: (326-43)+(174-57)

=326-43+174-57

=500-100

=400

b: (351-875)-(125-149)

=351-875-125+149

=500-1000

=-500

c: \(-415-\left\lbrace-215-\left\lbrack-115-\left(-315-2016\right)\right\rbrack\right\rbrace\)

=-415+215+[-115+315+2016]

=-200+[200+2016]

=-200+200+2016

=2016

d: [461+(-78)+40]+(-461)

=461-78+40-461

=-78+40

=-38

e: -323+[(-874)+564-241]
=-323-241+564-874

=-564+564-874

=-874

f: [(-59)+71]-[-83-(-95)]

=-59+71-(-83+95)

=12-12

=0