a, \(2x\left(x+2\right)-\left(x+2\right)\left(x-2\right)=\left(x+2\right)^2=x^2+4x+4\)

b, \(\left(x-3\right)\left(x^2+3x+9\right)-\left(x^2-27x\right)=x^3-27-x^2+27x\)

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

2𝑥(𝑥+2)−(𝑥+2)(𝑥−2)

2𝑥^2+4𝑥−(𝑥+2)(𝑥−2)

2𝑥^2+4𝑥−(𝑥(𝑥−2)+2(𝑥−2))

2𝑥^2+4𝑥−(𝑥^2−2𝑥+2(𝑥−2))

2𝑥^2+4𝑥−(𝑥^2−2𝑥+2𝑥−4)

2𝑥^2+4𝑥−(𝑥^2−4)

2𝑥^2+4𝑥−𝑥^2+4

2𝑥^2−𝑥^2+4𝑥+4

k cho mk nhé

cảm ơn bn nhiều

chúc bn hok tốt

\(a,\left(x-3\right)\left(x^2+3x+9\right)-\left(x^2-1\right)\left(x+27\right)\)

\(=\left(x^3-27\right)-x^3-27x^2+x+27=x-27x^2\)

\(b,\left(3-x\right)^3-\left(x+3\right)\left(x^2-3x+9\right)\)

\(=27-9x+3x^2-x^3-\left(x^3+27\right)=3x^2-9x-2x^3\)

\(c,\left(x-2\right)\left(x^2+2x+4\right)-x\left(x-3\right)\left(x+3\right)\)

\(=\left(x^3-8\right)-x\left(x^2-9\right)=x^3-8-x^3+9x=9x-8\)

a) (x-3)(x²+3x+9)-(x²-1)(x+27)

=(x³-27)-(x³+27x²-x-27)

=x³-27-x³-27x²+x+27

=-27x²+x

=x(-27x+1)

b) (3-x)³-(x+3)(x²-3x+9)

=27-27x+9x²-x³-x³-27

=-2x³+9x²-27x

=x(-2x+9x-27)

c) (x-2)(x²+2x+4)-x(x-3)(x+3)

=x³-8-x(x²-9)

=x³-8-x³+9x

=9x-8

#H

\(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)

\(ĐKXĐ:x\ge0,x\ne1\)

\(\Leftrightarrow\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(\Leftrightarrow\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

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

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

\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

Vậy kết quả rút gọn của P là \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

a )

\(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)

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

\(=-3x^3-3x\)

b )

\(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)

\(=3x^2-6x-5x+5x^2-8x^2+24\)

\(=-11x+24\)

\(\left(x^2-2\right)\left(x^2-2x+2\right)\left(x^2+2x+2\right)\left(x^2+2\right)\)

\(=\left[\left(x^2-2\right)\left(x^2+2\right)\right].\left\{\left[\left(x^2+2\right)-2x\right].\left[\left(x^2+2\right)+2x\right]\right\}\)

\(=\left(x^4-4\right).\left[\left(x^2+2\right)^2-4x^2\right]\)

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

\(=x^8-16\)

\(1.A=\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}=\sqrt{5-2.2\sqrt{5}+4}-\sqrt{5+2.2\sqrt{5}+4}=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}=\text{|}\sqrt{5}-2\text{|}-\text{|}\sqrt{5}+2\text{|}=-4\)\(2.\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1-2\sqrt{x-1}+1}=\sqrt{\left(\sqrt{x-1}-1\right)^2}=\text{|}\sqrt{x-1}-1\text{|}=\sqrt{x-1}-1\)

\(\dfrac{17}{6}+\dfrac{26}{-12}-\dfrac{-13}{39}\)
\(=\dfrac{17}{6}-\dfrac{13}{6}+\dfrac{1}{3}\)
\(=\dfrac{4}{6}+\dfrac{1}{3}\)
\(=\dfrac{2}{3}+\dfrac{1}{3}\)
\(=1\)

Qui tắc rút gọn một phân thức đại số.

- Phân tích tử và mẫu thành nhân tử (nếu cần) để tìm nhân tử chung.

- Chia cả tử và mẫu cho nhân tử chung đó.

Rút gọn:

Đơn giản biểu thức tức là rút gọn hết biểu thức lại bạn nhé !

Ví dụ :

1 ) Đơn giản biểu thức 

    a x 13 + a x 87 + b x 70 + b x 30

       Bài làm

   a x 13 + a x 87 + b x 70 + b x 30

= a x ( 13 + 87 ) + b x ( 70 + 30 )

= a x 100 + b x 100

= ( a + b ) x 100

bạn rút gọn biểu thức

thank you so much <3 <3 <3 <3