a) x=0    y=0         hoặc   x=1    y=8

b) x=2    y=4         hoặc   x=-2   y=-8       

b) x = 4 ; y = -2

     x = 10 ; y = -4

      x = 2 ; y = 4

bó tay luôn

Bài 1:

a. 

$\frac{2}{3}\times \frac{x}{y}=\frac{8}{15}$

$\frac{x}{y}=\frac{8}{15}: \frac{2}{3}=\frac{4}{5}$

b.

$\frac{x}{y}: \frac{3}{4}=\frac{2}{5}$

$\frac{x}{y}=\frac{3}{4}\times \frac{2}{5}=\frac{3}{10}$

c.

$\frac{3}{5}: \frac{x}{y}=\frac{4}{7}$

$\frac{x}{y}=\frac{3}{5}: \frac{4}{7}=\frac{21}{20}$

Bài 2:

Chiều dài hình chữ nhật là:

$\frac{3}{5}: \frac{3}{4}=\frac{4}{5}$ (m)

Chu vi hình chữ nhật:

$2\times (\frac{3}{4}+\frac{4}{5})=\frac{31}{10}$ (m)

1234567

a. y⁴ - 14y² + 49

Gọi y² là t, ta có:

t² - 14t + 49

<=> t² - 14t + 7²

<=> (t - 7)²

Thay x² = t

<=> (x² - 7)²

b. \(\dfrac{1}{4}-x^2\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^2-x^2\)

\(\Leftrightarrow\left(\dfrac{1}{2}-x\right)\left(\dfrac{1}{2}+x\right)\)

c. x⁴ - 16

<=> (x²)² - 4²

<=> (x² - 4)(x² + 4)

d. x² - 9

<=> x² - 3²

<=> (x - 3)(x + 3)

Bài 1: 

a: \(y^2-14y^2+49=\left(y-7\right)^2\)

b: \(\dfrac{1}{4}-x^2=\left(\dfrac{1}{2}-x\right)\left(\dfrac{1}{2}+x\right)\)

c: \(x^4-16=\left(x-2\right)\left(x+2\right)\left(x^2+4\right)\)

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

_Mấy bác cứ thik đăng nhiều :v , nhìn mak ko muốn lm . E lm bài 1 thôi :v còn các bài còn lại bác tự lm ( nó cx dễ thôi mà ) _

Bài 1 :

\(a) 2x-13=25+6x\)

\(\Rightarrow2x-6x=25+13\)

\(\Rightarrow-4x=38\)

\(\Rightarrow x=-\dfrac{19}{2}\)

Vậy .......

\(b) 12-x=3x+6\)

\(\Rightarrow-x-3x=6-12\)

\(\Rightarrow-4x=-6\)

\(\Rightarrow x=\dfrac{3}{2}\)

Vậy .....

\(c) 40-(25-2x)=x\)

\(\Rightarrow40-25+2x=x\)

\(\Rightarrow15+2x=x\)

\(\Rightarrow2x-x=-15\)

\(\Rightarrow x=-15\)

Vậy ......

\(d) |x-3|=5\)

\(\Rightarrow\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

Vậy ....

e) \(|x-3|+(x+2)+(x+1)=12\)

\(\Rightarrow\left|x-3\right|+x+2+x+1=12\)

\(\Rightarrow\left|x-3\right|+2x+3=12\)

\(\Rightarrow\left|x-13\right|+2x=9\)

\(\Rightarrow\left[{}\begin{matrix}x-3+2x=9\\-\left(x-3\right)+2x=9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=6\end{matrix}\right.\) \(( x = 6 \) ko thỏa mãn điều kiện )

Vậy ....

Trả lời:

7, 5( x + y )² + 15( x + y )

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

9, 7x( y - 4 )² - ( 4 - y )³ 

= 7x ( 4 - y )² - ( 4 - y )

= ( 4 - y )² ( 7x - 4 + y )

11, ( x + 1 )( y - 2 ) - ( 2 - y )²

= ( x + 1 )( y - 2 ) - ( y - 2 )²

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

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

8, 9x ( x - y ) - 10 ( y - x )² 

= 9x ( x - y ) - 10 ( x - y )²

= ( x - y )[ ( 9x - 10 ( x - y ) ]

= ( x - y )( 9x - 10x + 10y )

= ( x - y )( 10y - x )

10, ( a - b )² - ( a + b )( b - a ) 

= ( b - a )² - ( a + b )( b - a )

= ( b - a )( b - a - a - b )

= - 2a( b - a )

= 2a ( a - b )

12, 2x ( x - 3 ) + y ( x - 3 ) + ( 3 - x )

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

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

d) \(\frac{4x^2-12x+9}{9-4x^2}=-\frac{\left(2x+3\right)^2}{\left(2x-3\right)\left(2x+3\right)}=\frac{2x+3}{2x-3}\)