a) \(5\sqrt{48}-4\sqrt{27}-2\sqrt{57}+\sqrt{108}\)

\(=20\sqrt{3}-12\sqrt{3}-2\sqrt{57}+6\sqrt{3}\)

\(=\left(20-12+6\right)\sqrt{3}-2\sqrt{57}\)

\(=14\sqrt{3}-2\sqrt{57}\)

b) \(2\sqrt{24}-2\sqrt{54}+3\sqrt{6}-\sqrt{150}\)

\(=4\sqrt{6}-6\sqrt{6}+3\sqrt{6}-5\sqrt{6}\)

\(=\left(4-6+3-5\right)\sqrt{6}\)

\(=-4\sqrt{6}\)

1. \(\left(x+1\right)^3-125\)

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

\(=\left(x+1-5\right).\left[\left(x+1\right)^2+\left(x+1\right).5+5^2\right]\)

2. \(\left(x+4\right)^3-64\)

\(=\left(x+4\right)^3-4^3\)

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

3. \(x^3-\left(y-1\right)^3\)

\(=(x^3-y+1).\left[\left(x^2\right)+x.\left(y+1\right)+\left(y+1\right)^2\right]\)

\(\)4. \(\left(a+b\right)^3-c^3\)

\(=\left[\left(a+b\right)-c\right].\left[\left(a+b\right)^2+\left(a+b\right).c+c^2\right]\)

5. \(125-\left(x+2\right)^3\)

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

\(=\left(5-x-2\right).\left[5^2+5.\left(x+2\right)+\left(x+2\right)^2\right]\)

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

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

a) \(=y^4-81-y^4+4=-77\)

b) \(=a^2+b^2+c^2+2ab-2ac-2bc-a^2+2ac-b^2-2ab+2ac=c^2\)

c) Nhân 2 vào biểu thức

a)\(=\left(y^2-9\right)\left(y^2+9\right)-\left(y^4-4\right)=y^4-81-y^4+4=-77\)

b) \(\left(a+b-c\right)^2-\left(a-c\right)^2-2ab+2bc=b\left(2a-2c+b\right)-2ab+2bc=b^2\)

Sửa lại đề bạn nhé! 

c) \(P\left(3-1\right)=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(2P=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(2P=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

...

\(2P=3^{64}-1\Rightarrow P=\frac{3^{64}-1}{2}\)

a,

(y-3)(y+3)-(y²+2)(y²-2)

=y²-9-y⁴-4

=y²-y⁴-9-4

=y²-y⁴-13

b,

đề ghi thiếu chỗ mũ rồi

c,

(3+1)(3²+1)(3⁴+1)(3⁸+1)(3¹⁶+1)(3³²+1)

cái này mk k bt nữa

a) \(\dfrac{2}{5}-\dfrac{3}{15}\)

\(=\dfrac{2}{5}-\dfrac{3:3}{15:3}\)

\(=\dfrac{2}{5}-\dfrac{1}{5}\)

\(=\dfrac{1}{5}\)

b) \(\dfrac{9}{27}-\dfrac{2}{9}\)

\(=\dfrac{9:3}{27:3}-\dfrac{2}{9}\)

\(=\dfrac{3}{9}-\dfrac{2}{9}\)

\(=\dfrac{1}{9}\)

c) \(\dfrac{18}{24}-\dfrac{4}{8}\)

\(=\dfrac{18:6}{24:6}-\dfrac{4:2}{8:2}\)

\(=\dfrac{3}{4}-\dfrac{2}{4}\)

\(=\dfrac{1}{4}\)

d) \(\dfrac{6}{16}-\dfrac{10}{64}\)

\(=\dfrac{6\times2}{16\times2}-\dfrac{10:2}{64:2}\)

\(=\dfrac{12}{32}-\dfrac{5}{32}\)

\(=\dfrac{7}{32}\)

a: \(=\dfrac{1}{2}\cdot4\sqrt{3}+4\cdot3\sqrt{3}-2\cdot6\sqrt{3}\)

\(=2\sqrt{3}+12\sqrt{3}-12\sqrt{3}=2\sqrt{3}\)

b: \(=\left|2-\sqrt{5}\right|-\sqrt{\left(3+\sqrt{5}\right)^2}\)

\(=\sqrt{5}-2-3-\sqrt{5}\)

=-5

bạn viết có thánh đọc ra á :v

Bạn viết như vậy vẫn nhìn đc nhưng nhìn hơi khó

Thì các bạn vít ra giấy là hỉu nk mong giải giúp mk cái

ĐK:  \(x\ge0\)

\(x-6\sqrt{x}+9\)

\(=\left(\sqrt{x}\right)^2-2.\sqrt{x}.3+3^2\)

\(=\left(\sqrt{x}-3\right)^2\)

p/s: do không rõ đề với lại bạn có nhắc tới HĐT nên mk nghĩ đề là như z, sai thì bỏ qua nhé