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

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

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

4: \(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)

5: \(\left(x+2\right)^3=x^3+6x^2+12x+8\)

\(\sqrt[3]{15\sqrt{3}-26}=\sqrt[3]{-\left(26-15\sqrt{3}\right)}\)

\(=-\sqrt[3]{8-3\cdot2^2\cdot\sqrt{3}+3\cdot2\cdot3-3\sqrt{3}}\)

\(=-\sqrt[3]{\left(2-\sqrt{3}\right)^3}=-\left(2-\sqrt{3}\right)=-2+\sqrt{3}\)

giúp mình với mình đang cần gấp

\(\left(3x-3\right).\left(5x-21x\right)+\left(7x+4\right).\left(9x-5\right)=44\)

\(=3x.\left(5x-21x\right)-3.\left(5x-21x\right)+7x.\left(9x-5\right)+4.\left(9x-5\right)=44\)

\(=3x.5x-3x.21x-3.5x+3.21x+7x.9x-7x.5+4.9x-4.5=44\)

\(=15x^2-63x^2-15x+63x^2+63x^2-35x+36x-20=44\)

\(=78x^2-14x-20=44\)

Sao cái đề sao sao ấy

 ( 3x-3 ) . ( 5x-21x) + ( 7x+4) . ( 9x-5) = 44

xem lại chỗ in đậm

hộ m câu này lun : 2x^2 + 3 ( x-1) ( x+1 ) = 5x ( x-1 )

c) C = \(\dfrac{4}{\sqrt{3}+1} - \dfrac{5}{\sqrt{3}-2} + \dfrac{6}{\sqrt{3}-3}\)
⇔ C = \(\dfrac{4(\sqrt{3}-1)}{2} - \dfrac{5(\sqrt{3}-2)}{-1} - \dfrac{6(\sqrt{3}+3)}{-6}\)
⇔ C = \(2\sqrt{3} -2 + 5\sqrt{3} + 10 - \sqrt{3} - 3\)
⇔ C = \(6\sqrt{3} + 5\)

1) ĐKXĐ: \(x\ge-5\)

\(pt\Leftrightarrow x+5=9\Leftrightarrow x=9-5=4\left(tm\right)\)

2) ĐKXĐ: \(x\ge3\)

\(pt\Leftrightarrow3\sqrt{x-3}-\sqrt{x-3}=6\)

\(\Leftrightarrow2\sqrt{x-3}=6\Leftrightarrow\sqrt{x-3}=3\)

\(\Leftrightarrow x-3=9\Leftrightarrow x=12\left(tm\right)\)

3) ĐKXĐ: \(x\ge-1\)

\(pt\Leftrightarrow\sqrt{\left(x+1\right)^2}-2\sqrt{x+1}=0\)

\(\Leftrightarrow x+1-2\sqrt{x+1}=0\)

\(\Leftrightarrow\sqrt{x+1}\left(\sqrt{x+1}-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+1=4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=3\left(tm\right)\end{matrix}\right.\)

tui uk.......u...a

`1)(a^[1/4]-b^[1/4])(a^[1/4]+b^[1/4])(a^[1/2]+b^[1/2])`

`=[(a^[1/4])^2-(b^[1/4])^2](a^[1/2]+b^[1/2])`

`=(a^[1/2]-b^[1/2])(a^[1/2]+b^[1/2])`

`=a-b`

`2)(a^[1/3]-b^[2/3])(a^[2/3]+a^[1/3]b^[2/3]+b^[4/3])`

`=(a^[1/3]-b^[2/3])[(a^[1/3])^2+a^[1/3]b^[2/3]+(b^[2/3])^2]`

`=(a^[1/3])^3-(b^[2/3])^3`

`=a-b^2`

\(x^2-x=x\left(x-1\right)\)

=\(\left(2-1\right)\left(2+1\right)\left(2^2-1\right)....\left(2^{20}-1\right)\) +1

=\(\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{20}+1\right)+1\)

=\(\left(2^4-1\right)\left(2^4+1\right)....\left(2^{20}+1\right)+1\)

=.....

=\(\left(2^{20}-1\right)\left(2^{20}+1\right)+1\)

=\(2^{40}-1+1\)

=\(2^{40}\)

Chuc ban hoc tot

Sai rồi, nếu mũ là 32 thì bài này làm thế đc chứ mũ 20 thì ko làm như này được