em lớp 5 chưa học đến chị ạ

hoi kho lm giup mik

2 thg 11, 2018 — a, x^2-y^2-x+y= (x^2-y^2)-(x-y)=(x-y)(x+y)-(x-y)=(x-y)(x+y-1). b, x^4-x^3-x^2+1=(x^4-x^3)-(x^2-1)=x^3(x-1)-(x-1)(x+1)=(x-1)(x^3-x-1).

x² - y² + x-y

= (x-y)(x+y) + (x-y)

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

a) A = 4x²-6xy

= 2x(x - 3y)

Vậy A= 2x(x-3y) 

b) B= x(x-2) + 2(x-2)

B= (x-2)(x+2)

Vậy B= (x-2)(x+2)

c) C= 2x(x-5)+ 6(5-x)

C= 2x(x-5) - 6(x-5)

C= (2x-6)(x-5)

C= 2(x-3)(x-5)

Vậy C=2(x-3)(x-5)

\(a^3b^3+b^3c^3+c^3a^3=3a^2b^2c^2\)

\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)

Đặt \(\frac{1}{a}=x,\frac{1}{b}=y,\frac{1}{c}=z\)

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

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\)

mà \(a,b,c\)dương nên \(x=y=z\Rightarrow a=b=c\).

\(A=\left(2+\frac{a}{b}\right)\left(2+\frac{b}{c}\right)\left(2+\frac{c}{a}\right)=3^3=27\).

\(3a^2\)\(b^2\)\(c^2\)

\(=>ab+bc+ca=0\)

\(=>ab^2\)\(+bc^2\)\(+ca^2\)\(=0\)

\(TH1:ab+bc+ca=0\)

\(ab+bc=-ca\)

\(=>a+c=-\frac{ac}{b}\)

\(=>a+b=-\frac{ab}{c}\)

\(b+c=-\frac{bc}{a}\)

\(Thay\)\(A\)

\(=>A=-3\)

\(\left(ab-bc\right)^2\)\(+\left(bc-ca\right)^2\)\(+\left(ca-ab\right)^2\)\(=0\)

\(=>ab-bc=0\)

\(bc-ca=0\)

\(ca-ab=0\)

\(=>ab=bc=ca\)

\(=>a=b=c\)

\(Thay\)\(A\)

\(=>A=-24\)

\(=>A=\left(-3;-24\right)\)

Em làm sai mong anh thông cảm cho ạ

(x+1)(x+2)-(x-3)(x+4)=6

<=> (x² + 2x + x+2) -(x² + 4x - 3x  - 12)=6

<=> x² + 3x + 2 - x² - x +12=6

<=> (x² - x²) + (3x-x)+  (2+12) =6

<=> 2x + 14=6

<=> 2x = -8

<=> x=-4

Vậy x = -4 

$M=x^2+y^2-xy-x+y+1$

$4M=4x^2+4y^2-4xy-4x+4y+4$

$4M={\left(2x-y-1\right)}^2+3y^2+2y+3$

$12M=3{\left(2x-y-1\right)}^2+9y^2+6y+9$

$12M={\left(2x-y-1\right)}^2+{\left(3y+1\right)}^2+8$

$M\ge \frac{1}{3}$

$\{\begin{array}{c}3y+1=0;y=-\frac{1}{3}\\ 2x+\frac{1}{3}-1=0;x=-\frac{1}{3}\end{array}$