a: \(\Leftrightarrow\left(x+y;y-5\right)\in\left\{\left(1;16\right);\left(2;8\right);\left(4;4\right);\left(8;2\right);\left(16;1\right);\left(-4;-4\right);\left(-8;-2\right);\left(-16;-1\right)\right\}\)

\(\Leftrightarrow\left(x+y;y\right)\in\left\{\left(1;21\right);\left(2;13\right);\left(4;9\right);\left(8;7\right);\left(16;6\right);\left(-4;1\right);\left(-8;3\right);\left(-16;4\right)\right\}\)

\(\Leftrightarrow\left(x,y\right)\in\left\{\left(1;7\right);\left(10;6\right)\right\}\)

b: \(\Leftrightarrow\left(2x+3\right)\left(2y-1\right)=10\)

\(\Leftrightarrow\left(2x+3;2y-1\right)\in\left\{\left(1;10\right);\left(10;1\right);\left(2;5\right);\left(5;2\right);\left(-1;-10\right);\left(-10;-1\right);\left(-2;-5\right);\left(-5;-2\right)\right\}\)

mà 2x+3;2y-1 đều là các số lẻ

nên \(\left(x,y\right)\in\varnothing\)

a: \(\Leftrightarrow\left(x+y;y-5\right)\in\left\{\left(1;16\right);\left(2;8\right);\left(4;4\right);\left(8;2\right);\left(16;1\right);\left(-4;-4\right);\left(-8;-2\right);\left(-16;-1\right)\right\}\)

\(\Leftrightarrow\left(x+y;y\right)\in\left\{\left(1;21\right);\left(2;13\right);\left(4;9\right);\left(8;7\right);\left(16;6\right);\left(-4;1\right);\left(-8;3\right);\left(-16;4\right)\right\}\)

\(\Leftrightarrow\left(x,y\right)\in\left\{\left(1;7\right);\left(10;6\right)\right\}\)

b: \(\Leftrightarrow\left(2x+3\right)\left(2y-1\right)=10\)

\(\Leftrightarrow\left(2x+3;2y-1\right)\in\left\{\left(1;10\right);\left(10;1\right);\left(2;5\right);\left(5;2\right);\left(-1;-10\right);\left(-10;-1\right);\left(-2;-5\right);\left(-5;-2\right)\right\}\)

mà 2x+3;2y-1 đều là các số lẻ

nên \(\left(x,y\right)\in\varnothing\)

a: 3x=7y

=>x/7=y/3=(x-y)/(7-3)=-16/4=-4

=>x=-28; y=-12

b: x/6=y/5

=>x/6=2y/10=(x+2y)/(6+10)=20/16=5/4

=>x=30/4=15/2; y=25/4

c: Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{5}=\dfrac{2x+3y+5z}{2\cdot2+3\cdot\left(-3\right)+5\cdot5}=\dfrac{6}{20}=\dfrac{3}{10}\)

=>x=3/5; y=-9/10; z=3/2

d: x/2=y/3

=>x/8=y/12

y/4=z/5

=>y/12=z/15

=>x/8=y/12=z/15

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{10}{5}=2\)

=>x=16; y=24; z=30

Th1: 2x+3 ≥ 0
Khi đó: |2x+3| =x+2
 (2x+3)= x+2
- 2x+3= x+2
-2x-x= 2-3
 x= -1
Th2: 2x+3 < 0
Khi đó: |2x+3|=x+2
 -(2x+3) = x +2
 -2x-3 = x+2
 -3x = 5
 x=-5/3

Vậy x= -1

      x= -5/3

Lớp 6 cugx học dạng v nè

`x/2=y/3 <=> x/8=y/12;

`y/4=z/5 <=> y/12=z/15.`

`<=> x/8=y/12=z/15=(x^2-y^2)/(64-144)=16/80=1/5`.

`@ x/8=1/5 <=> x= 8/5`.

`@ y/12=1/5 <=> y=12/5`.

`@ z/15=1/5 <=> y=15/5`.

Vậy...

Lời giải:

a. Đặt $\frac{x}{2}=\frac{y}{3}=a\Rightarrow x=2a; y=3a$

$x^2-y^2=(2a)^2-(3a)^2=-16$

$\Rightarrow -5a^2=-16\Rightarrow a=\pm \frac{4}{\sqrt{5}}$

Nếu $a=\frac{-4}{\sqrt{5}}$ thì:

$x=2a=\frac{-8}{\sqrt{5}}; y=3a=\frac{-12}{\sqrt{5}}; z=\frac{5}{4}y=-3\sqrt{5}$

Nếu $a=\frac{4}{\sqrt{5}}$ thì:

$x=2a=\frac{8}{\sqrt{5}}; y=3a=\frac{12}{\sqrt{5}}; z=\frac{5}{4}y=3\sqrt{5}$

b.

Nếu $x\geq \frac{-3}{2}$ thì:

$2x+3=x+2$

$\Leftrightarrow x=-1$

Nếu $x< \frac{-3}{2}$ thì:

$-2x-3=x+2$

$\Leftrightarrow -5=3x\Leftrightarrow x=\frac{-5}{3}$

Thử lại thấy 2 giá trị $-1, \frac{-5}{3}$ đều tm

c.

$f(x)=-3x+6=0$

$\Leftrightarrow -3x=-6\Leftrightarrow x=2$

Vậy $x=2$ là nghiệm của đa thức.

khó + lười + nhiều = không làm

Hello

ko thích làm

a , |2x+4|+|y-6|=0

=> 2 x + 4 = 0 => x = 0 

=> y - 6 = 0 => y = 6

Vậy x = 0 và y = 6

a. 2x+4= 2.0+4=4
y-6=2-6=-4

=)) l4l;l-4l

Bài 2: 

a: \(3x^2-3xy=3x\left(x-y\right)\)

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

c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)

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

ỳtct7ct7c7c7t79tc9

\(\Leftrightarrow\left(y-2\right)^2+10=\dfrac{20}{\left(x+1\right)^2+2}\)

Ta có: \(\left(y-2\right)^2+10\ge10\)

\(\left(x+1\right)^2+2\ge2\Rightarrow\dfrac{20}{\left(x+1\right)^2+2}\le\dfrac{20}{2}=10\)

\(\Rightarrow\left(y-2\right)^2+10\ge\dfrac{20}{\left(x+1\right)^2+2}\)

Đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}y-2=0\\x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)