a: (2x+3)(y-4)=12

mà 2x+3 lẻ

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

=>(2x;y)∈{(-2;16);(-4;-8);(0;8);(-6;0)}

=>(x;y)∈{(-1;16);(-2;-8);(0;8);(-3;0)}

b; x(2y+1)-4y=3

=>x(2y+1)-4y-2=1

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

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

=>(x-2;2y+1)∈{(1;1);(-1;-1)}

=>(x;2y)∈{(3;0);(1;-2)}

=>(x;y)∈{(3;0);(1;-1)}

c: xy+2x+y+11=0

=>x(y+2)+y+2+9=0

=>(x+1)(y+2)=-9

=>(x+1;y+2)∈{(1;-9);(-9;1);(-1;9);(9;-1);(3;-3);(-3;3)}

=>(x;y)∈{(0;-11);(-10;-1);(-2;7);(8;-3);(2;-5);(-4;1)}

${Tham\; khảo:(}^{}$

${2x-3)}^2=9$

$\Rightarrow [\begin{array}{c}2x-3=3\\ 2x-3=-3\end{array}\Rightarrow [\begin{array}{c}x=3\\ x=0\end{array}$

Vậy x = 3 hoặc x = 0

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

\(\left(2x-3\right)^2=3^2\)

⇒\(2x-3=+-3\)

\(TH1:2x-3=3\text{⇒}x=3\)

\(TH2:2x-3=-3\text{⇒}x=0\)

\(x\left(2x-4\right)-2x\left(x+3\right)-3\left(x-1\right)-29=0\)

\(\Leftrightarrow2x^2-4x-2x^2-6x-3x+3-29=0\)

\(\Leftrightarrow-13x-26=0\)

\(\Leftrightarrow-13x=26\)

\(\Leftrightarrow x=26:-13\)

\(\Leftrightarrow x=-2\)

Vậy ...

\(x\left(2x-4\right)-2x\left(x+3\right)-3\left(x-1\right)-29=0\)

\(2x^2-4x-2x^2-6x-3x+3-29=0\)

\(2x^2-4x-2x^2-6x-3x=0+29-3\)

\(\left(2x^2-2x^2\right)+\left(-4x-6x-3x\right)=26\)

\(0+\left(-4-6-3\right)x=26\)

\(\Rightarrow-13x=26\rightarrow x=-2\)

a)\(\text{ 4x - 15 = -75 - x}\)

\(4x-15+75+x=0\)

\(5x+60=0\)

\(5x=-60\)

\(x=-14\)

Vậy....

Thêm dấu suy ra trc mỗi dòng nha

Học tốt

b)\(|2x-7|+2=13\)

\(|2x-7|=11\)

\(\Leftrightarrow\hept{\begin{cases}2x-7=11\\2x-7=-11\end{cases}\Leftrightarrow\hept{\begin{cases}2x=18\\2x=4\end{cases}\Leftrightarrow}\hept{\begin{cases}x=9\\x=2\end{cases}}}\)

vậy x=9 hoặc x=2

a)

(2x-1)²=9

<=> 2x-1=\(\sqrt{9}\)

<=> 2x-1=3

<=> 2x=4

<=> x=2

b)

2x(x-3)=0

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

Vậy x=0 hoặc x=3

Tìm số nguyên x,biết:

a,(2x-1)² = 9

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

\(\Rightarrow2x-1=3\)

\(\Rightarrow2x=4\Rightarrow x=2\)

b,2x(x-3) = 0

\(\Rightarrow\orbr{\begin{cases}2x=0\\x-3=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

giúp mik với!!

a. ( x - 2 ) ( 6 - 2x ) = 0

=> x - 2 = 0 hoặc 6 - 2x = 0 

     x = 2                     2x = 6

                                    x = 3

Vậy x = 2; x =3 

b. ( x + 1 )³ = -27

    ( x + 1 )³ = ( -3 )³ 

=> x + 1 = -3 

     x = -4

a) (x - 2)(6 - 2x) = 0

=> \(\orbr{\begin{cases}x-2=0\\6-2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\2x=6\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

b) (x + 1)³ = -27

=> (x + 1)³ = (-3)³

=> x + 1 = -3

=> x = -4

(2x-123)-(x+27)=0

2x-123-x-27=0

x-150=0

x=150

\(\left(-3x+2\right)-\left(5-3x\right)=-3\)

\(\Rightarrow-3x+2-5+3x=-3\)

\(\Rightarrow-3x+3x=-3+5-2\)

\(\Rightarrow0x=0\Rightarrow x\in Z\)

\(3+x-\left(3x-1\right)=6-2x\)

\(\Rightarrow3+x-3x+1=6-2x\)

\(\Rightarrow x-3x+2x=6-1-3\)

\(\Rightarrow0x=2\left(loại\right)\)

\(\left(x-5\right)\left(3x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\3x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-\frac{4}{3}\end{cases}}}\)

\(7x\left(2x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}7x=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)

\(\left(3x-1\right)2x=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=0\end{cases}}}\)

khó hiểu quá

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