a: -2x(x+3)+x(2x-1)=10

=>-2x^2-6x+2x^2-x=10

=>-7x=10

=>x=-10/7

b: Sửa đề: 2/3x(9/2x+1/4)-(3x^2+2)=3

=>3x^2+1/6x-3x^2-2=3

=>1/6x-2=3

=>x=30

sao sửa, đề nó vậy á

a) vì (3x - 2)(2y-3)=1

=> 3x-2 = 1 ; 2y-3 = 1

Ta có :+) 3x - 2 = 1

=> 3x = 3

=> x= 1

+) 2y-3 = 1

=> 2y = 4

=> y = 2

Vậy x=1; y = 2

b) Vì (x + 1)(2y-1) = 12

=> (x+1) và (2y-1) ϵ Ư(12) = {1 ; 2 ; 6 ; 3 ; 4 ; 12 }

Ta thấy : 2y - 1 là số lẻ

=> 2y-1 ϵ {1 ; 3 }

+ Nếu 2y - 1 = 1

=> 2y = 1 + 1

2y = 2

=> y = 1

=> x+1 = 12

=> x = 11

+ Nếu 2y - 1 = 3

=>2y = 4

=> y = 2

=> x+1 = 6

=> x = 5

Vậy x = 11 ; 5

y = 1 ; 2

a, \(3x=5y=7z=>\dfrac{3x}{105}=\dfrac{5y}{105}=\dfrac{7z}{105}=>\dfrac{x}{35}=\dfrac{y}{21}=\dfrac{z}{15}\)

áp dụng tính chất dãy tỉ số = nhau

\(=>\dfrac{x}{35}=\dfrac{y}{21}=\dfrac{z}{15}=\dfrac{x+y+z}{35+21+15}=\dfrac{10}{71}\)

\(=>\dfrac{x}{35}=\dfrac{10}{71}=>x=\dfrac{350}{71}\)

\(=>\dfrac{y}{21}=\dfrac{10}{71}=>y=\dfrac{210}{71}\)

\(=>\dfrac{z}{15}=\dfrac{10}{71}=>z=\dfrac{150}{71}\)

b, \(\)\(6x=5y=>\dfrac{x}{5}=\dfrac{y}{6}=>\dfrac{x}{20}=\dfrac{y}{24}\)

có \(7y=8z=>\dfrac{y}{8}=\dfrac{z}{7}=>\dfrac{y}{24}=\dfrac{z}{21}\)

\(=>\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=>\dfrac{3x}{60}=\dfrac{2y}{48}=\dfrac{4z}{84}\)

áp dụng t/c dãy tỉ số = nhau

\(=>\dfrac{3x}{60}=\dfrac{2y}{48}=\dfrac{4z}{84}=\dfrac{3x+2y+4z}{60+48+84}=\dfrac{12}{192}=\dfrac{1}{16}\)

\(=>\dfrac{3x}{60}=\dfrac{1}{16}=>x=1,25\)

\(=>\dfrac{2y}{48}=\dfrac{1}{16}=>y=1,5\)

\(=>\dfrac{4z}{84}=\dfrac{1}{16}=>z=1,3125\)

c, \(x:y:z=1:2:3=>\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}\)

\(=>x=\dfrac{y}{2},z=\dfrac{3y}{2}\)

thay x,z vào \(x^3+y^3+z^3=36=>\left(\dfrac{y}{2}\right)^3+y^3+\left(\dfrac{3y}{2}\right)^3=36\)

\(=>y=2\)

\(=>x=\dfrac{y}{2}=\dfrac{2}{2}=1,z=\dfrac{3y}{2}=\dfrac{3.2}{2}=3\)

d, \(\dfrac{x}{2}=\dfrac{y}{3}=>x=\dfrac{2y}{3}\)

thay x vào \(3x^3+y^3=51=>3.\left(\dfrac{2y}{3}\right)^3+y^3=51=>y=3\)

\(=>x=\dfrac{2.3}{3}=2\)

c, từ đoạn này á

\(\left(\dfrac{y}{2}\right)^3+y^3+\left(\dfrac{3y}{2}\right)^3=36\)

\(< =>\dfrac{y^3}{8}+\dfrac{8y^3}{8}+\dfrac{27y^3}{8}=36\)

\(=>\dfrac{36y^3}{8}=36=>36y^3=8.36=>y^3=8=>y=2\)

a: \(4x^3-36x\)

\(=4x\cdot x^2-4x\cdot9\)

\(=4x\left(x^2-9\right)=4x\left(x-3\right)\left(x+3\right)\)

b:Sửa đề: \(4x^3-y^3+4x^2y-xy^2\)

\(=4x^2\left(x+y\right)-y^2\left(x+y\right)\)

\(=\left(x+y\right)\left(4x^2-y^2\right)=\left(x+y\right)\left(2x-y\right)\left(2x+y\right)\)

c: \(a^2+2ab-5a-10b\)

=a(a+2b)-5(a+2b)

=(a+2b)(a-5)

d: \(\left(x+1\right)^3-27\)

\(=\left(x+1\right)^3-3^3\)

\(=\left(x+1-3\right)\left\lbrack\left(x+1\right)^2+3\left(x+1\right)+3^2\right\rbrack\)

\(=\left(x-2\right)\left(x^2+2x+1+3x+3+9\right)\)

\(=\left(x-2\right)\left(x^2-5x+13\right)\)

e: \(4xy^2-4x^2y-y^3\)

\(=y\cdot4xy-y\cdot4x^2-y\cdot y^2\)

\(=-y\left(4x^2-4xy+y^2\right)=-y\left(2x-y\right)^2\)

f: \(\left(5x-y\right)^2-\left(x-2y\right)^2\)

=(5x-y-x+2y)(5x-y+x-2y)

=(4x+y)(6x-3y)

=3(2x-y)(4x+y)

g: \(x^3+2x^2+x-16xy^2\)

\(=x\left(x^2+2x+1-16y^2\right)\)

\(=x\left\lbrack\left(x+1\right)^2-\left(4y\right)^2\right\rbrack\)

=x(x+1-4y)(x+1+4y)

a: \(\Leftrightarrow\left(x,y+1\right)\in\left\{\left(1;12\right);\left(12;1\right);\left(2;6\right);\left(6;2\right);\left(3;4\right);\left(4;3\right)\right\}\)

hay \(\left(x,y\right)\in\left\{\left(1;11\right);\left(12;0\right);\left(2;5\right);\left(6;1\right);\left(3;3\right);\left(4;2\right)\right\}\)

b: \(\Leftrightarrow\left(3x-2;2y-3\right)\in\left\{\left(1;1\right);\left(-1;-1\right)\right\}\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x-2=1\\2y-3=1\end{matrix}\right.\Leftrightarrow\left(x,y\right)=\left(1;2\right)\)

c: \(\Leftrightarrow\left(x+1,2y-1\right)\in\left\{\left(12;1\right);\left(4;3\right)\right\}\)

hay \(\left(x,y\right)\in\left\{\left(11;1\right);\left(3;2\right)\right\}\)