Toán mà lại đăng chủ đề Mĩ Thuật hả bạn

Bấm nhầm hay cố ý????????

Ơ ??? Tại sao toán mà lại đăng trong chủ đề Mĩ thuật ?

3x²-3xy-y-5x=-20

=>3x(x-y)-y-5x=-20

=>3x(x-y)+x-y-6x=-20

=>3x(x-y)+(x-y)-6x=-20

=>(x-y)(3x+1)-6x=-20

=>(x-y)(3x+1)-6x-2=-22

=>(x-y)(3x+1)-(6x+2)=-22

=>(x-y)(3x+1)-2(3x+1)=-22

=>(3x+1)(x-y-2)=-22

Ta có bảng sau

Vậy ta có 2 bộ (x,y) là (0;-20) và (7;6)

thank you ket ban ko

3x(x-y-2)+(x-y-2)+2=-20

(3x+1)(x-y-2)=-22

----->   3x(x-y)+x-6x-y=-20

------>   3x(x-y)+(x-y)-6x=-20

------>  ( x-y)(3x+1) - 6x= -20

------>( x-y)(3x+1)-6x-2= -20-2

------>  (x-y)(3x+1)-2(3x+2)

-------> (x-y-2)(3x+1)= -22

-----> TA CÓ BẢNG SAU:

            Vậy ta có x={   0 ; -20  ;   7  ;  6}

Đúng thì cho mik nha

CHÚC BN HỌC TỐT

Bài làm

3x² - 3xy - 5x - y = -20

<=> 3x² + x - 6x - 3xy - y = -20

<=> 3x² + x - 6x - 3xy - y = -22 + 2

<=>3x² + x - 6x - 3xy - y - 2= -22

<=> ( 3x² + x ) - ( 6x + 2 ) - ( 3xy + y ) = -22

<=> x( 3x + 1 ) - 2( 3x + 1 ) - y( 3x + 1 ) = -22

<=> ( 3x + 1 )( x - 2 - y ) = -22

Ta được bảng sau:

Vậy ta có cặp x, y  là: ( x = -4; y = -8 ); ( x = -1; y = -14 ); ( x = 0; y = 20 ); ( x = 7; y = 6 )

# Học tốt #

Phân tích đa thức thành nhân tử:(em làm luôn đấy,ko ghi lại đề)

\(\left(x^3+y^3\right)-\left(x+y\right)+3xy\left(x+y\right)\)

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

\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)\(=\left(x+y\right)\left[\left(x+y\right)^2-1^2\right]\)

\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)

\(8x^3+12x^2+6x+1=0.\)

\(\Leftrightarrow\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3=0\)

\(\Leftrightarrow\left(2x+1\right)^3=0\)

\(\Leftrightarrow2x+1=0\)

\(\Leftrightarrow x=-\frac{1}{2}\)

\(2x^2+5x-3=0\Leftrightarrow\left(2x^2+6x\right)+\left(-x-3\right)=0\)

\(\Leftrightarrow2x\left(x+3\right)-\left(x+3\right)=0\Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\)

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

\(x^2-2x-3=0\Leftrightarrow\left(x^2-3x\right)+\left(x-3\right)=0\)

\(\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)

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

\(\left(5x-1\right)+2\left(1-5x\right)\left(4+5x\right)+\left(5x+4\right)^2\)

\(=5x-1+2\left(4+5x-20x-25x^2\right)+25x^2+40x+16\)

\(=25x^2+45x+15+8+10x-40x-50x^2\)

\(=-25x^2+15x+23\)

\(\left(x-y\right)^3+\left(y+x\right)^3+\left(y-x\right)^3-3xy\left(x+y\right)\)

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

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

\(=x^3+3x^2y+3xy^2+y^3-3xy^2-3x^2y\)

\(=x^3+y^3\)

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

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

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

....

a)  x² - x - 12 

= x² - 4x + 3x - 12

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

= (x - 4)(x + 3)

b) x³ - y³ - 3x² + 3x - 1

= (x³ - 3x² + 3x - 1) - y³

= (x - 1)³ - y³

= (x - 1 - y) [ (x - 1)² + (x - 1)y + y² ]

= (x - y - 1)(x² - 2x + 1 + xy - y + y² )

d) 4x³ - 5x² - 16x + 20

= (4x³ - 8x²) + (3x² - 6x) - (10x - 20)

= 4x² (x - 2) + 3x(x - 2) - 10(x - 2)

= (x - 2)(4x² + 3x - 10)

= (x - 2)(4x² + 8x - 5x - 10)

= (x - 2)(x + 2)(4x - 5)

\(5x\left(3x^2y-2xy^2+1\right)-3xy\left(5x^2-3xy\right)+x^2y^2-10=0\)

\(\Leftrightarrow15x^3y-10x^2y^2+5x-15x^3y+9x^2y^2+x^2y^2-10=0\)

\(\Leftrightarrow5x=10\Leftrightarrow x=2\)

(5x-2)(3x+1)+(7-15x)(x+3)=-20

<=> 15x²+5x-6x-2+7x+21-15x²-45x+20=0

<=>39-39x=0

<=>39(1-x)=0

<=>1-x=0

=>x=1

(5x-2)(3x+1)+(7-15x)(x+3)=-20

=>\(15x^2-6x+5x-2+7x-15^2+21-45x=-20\)

=>\(-39x+19=-20\)

=>\(-39x=-39\)

=>\(x=1\)

vậy x=1

(5x-2)(3x+1)+(7-15x)(x+3)=-20

<=>\(15x^2+5x-6x-2+7x-15x^2+21-45x=-20\)

<=>\(-39x+19=0\)

<=>\(-39x=-19\)

<=>\(x=\dfrac{19}{39}\)

Vậy \(x=\dfrac{19}{39}\)

a) 5x² ( 3x² -7x+2)-15x(x-3)

=15x⁴-35x³+10x²-15x²+45x

=15x⁴-35x³-5x²+45x

c) (x+3)(x-3)(x-2)(x+1)

=(x²-9)(x²+x-2x-2)

=(x²-9)(x²-x-2)

=x⁴-x³-2x²-9x²+9x+18

=x⁴-x³-11x²+9x+18

d)(2x+1)²+(4x-1)²+2(2x+1)(4x+1)

=2x²+4x+1-16x²-8x+1

=2x²+4x+1-16x²-8x+1+16x²-4x+8x-2

=2x²+7

e) (2x²-3x)(5x²-2x+1)-10x²(x+3)

=10x⁴ -4x³+2x²-15x³+6x²-3 -10x²-30x

=10x⁴-19x³-2x²-30x-3

thanks bn nka

câu c) bn thíu dấu

đề là gì