Question

[Châu anh Phạm]

Answer 1

[Phong]

Bài 1:

`a)(2x+1):3+2=5`

`(2x+1):3=5-2`

`(2x+1):3=3`

`2x+1=3*3`

`2x+1=9`

`2x=9-1`

`2x=8`

`x=8/2`

`x=4`

Vậy: `...`

`b)(2x)/3=3x-1`

`2x=3(3x-1)`

`2x=9x-3`

`9x-2x=3`

`7x=3`

`x=3/7`

Vậy: `...`

`c)(x+5)/15+(x+6)/14=(x+7)/13+(x+8)/12`

`((x+5)/15+1)+((x+6)/14+1)=((x+7)/13+1)+((x+8)/12+1)`

`(x+20)/15+(x+20)/14=(x+20)/13+(x+20)/12`

`(x+20)/15+(x+20)/14-(x+20)/13-(x+20)/12=0`

`(x+20)(1/15+1/14-1/13-1/12)=0`

`(x+20)=0`

`x=-20`

Vậy: `...`

Answer 2

[Nguyễn Lê Phước Thịnh]

Bài 4:

\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\cdots+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)

=>\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\cdots+\frac{3}{x\left(x+3\right)}=\frac{101}{1540}\cdot3=\frac{303}{1540}\)

=>\(\frac15-\frac18+\frac18-\frac{1}{11}+\cdots+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)

=>\(\frac15-\frac{1}{x+3}=\frac{303}{1540}\)

=>\(\frac{1}{x+3}=\frac15-\frac{303}{1540}=\frac{1}{308}\)

=>x+3=308

=>x=305

Bài 3:

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

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

=>(x;y)∈{(3;2);(5;0);(1;-4);(-1;-2)}

b: xy+x+y+1=5

=>x(y+1)+(y+1)=5

=>(x+1)(y+1)=5

=>(x+1;y+1)∈{(1;5);(5;1);(-1;-5);(-5;-1)}

=>(x;y)∈{(0;4);(4;0);(-2;-6);(-6;-2)}

c: xy+3y+x=2

=>y(x+3)+x+3=2+3

=>(x+3)(y+1)=5

=>(x+3;y+1)∈{(1;5);(5;1);(-1;-5);(-5;-1)}

=>(x;y)∈{(-2;4);(2;0);(-4;-6);(-8;-2)}

Bài 2:

a: \(\frac{x-2}{-2}=\frac{3x+1}{3}\)

=>-2(3x+1)=3(x-2)

=>-6x-2=3x-6

=>-6x-3x=-6+2

=>-9x=-4

=>\(x=\frac49\)

b: \(\frac{x-5}{15}+\frac{x-7}{13}=\frac{x-9}{11}+\frac{x-11}{9}\)

=>\(\left(\frac{x-5}{15}-1\right)+\left(\frac{x-7}{13}-1\right)=\left(\frac{x-9}{11}-1\right)+\left(\frac{x-11}{9}-1\right)\)

=>\(\frac{x-20}{15}+\frac{x-20}{13}=\frac{x-20}{11}+\frac{x-20}{9}\)

=>x-20=0

=>x=20

c: \(\frac{x+9}{11}+\frac{x+12}{8}+\frac{x+10}{5}=-4\)

=>\(\left(\frac{x+9}{11}+1\right)+\left(\frac{x+12}{8}+1\right)+\left(\frac{x+10}{5}+2\right)=-4+4=0\)

=>\(\frac{x+20}{11}+\frac{x+20}{8}+\frac{x+20}{5}=0\)

=>x+20=0

=>x=-20

d: \(\left(2^{3x+2}-2\right):6+22=\frac{1}{\left(-3\right)^2}\cdot3^4\)

=>\(\left(2^{3x+2}-2\right):6+22=\frac{81}{9}=9\)

=>\(2^{3x+2}-2=\left(9-22\right)\cdot6=-13\cdot6=-78\)

=>\(2^{3x+2}=-78+2=-76\) (vô lý)

=>x∈∅