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10³ + 2¹⁵

= 1000 + 32768

= 33768

Mà 33768 : 33 = 1023 (dư 9)

Em xem lại đề

Trả lời:

( x + 2 )³ + ( x - 2 )³ - 2x ( x² + 10 )

= x³ + 6x² + 12x + 8 + x³ - 6x² + 12x - 8 - 2x³ - 20x 

= 4x

bài mình sai rồi, do mình hấp tấp quá, bài bạn @Quỳnh Anh đr bạn kham khảo bài bạn ấy nhé

\(\left(x+2\right)^3+\left(x-2\right)^3-2x\left(x^2+10\right)\)

\(=\left(x+2+x-2\right)\left[\left(x+2\right)^2+\left(x+2\right)\left(x-2\right)-\left(x-2\right)^2\right]-2x^3-20x\)

\(=2x\left(x^2+4x+4+x^2-4-x^2+4x-4\right)-2x^3-20x\)

\(=2x\left(x^2+8x-4\right)-2x^3-20x=2x^3+16x^2-8x-2x^3-20x\)

\(=16x^2-28x\)

\(a.\left(\dfrac{2}{5}.\dfrac{5}{2}\right).\dfrac{10}{21}=1.\dfrac{10}{21}=\dfrac{10}{21}\)

\(\dfrac{1}{2}.4+\dfrac{3}{4}.4=2+3=5\)

a tính bằng cách là

( 2/5 x 5/2 ) x 10/21 

= 1 x 10/21 

10/21

b tính bằng cách là

1/2 + ( 3/4 x 4)

=1/2 x 12/4 rút gọn là 1/2 x 3

= 3/2

i giúp em vớiiiiii

\(M=\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+...+\dfrac{3}{1+2+...+2022}\)

\(=\dfrac{3}{\dfrac{2\left(2+1\right)}{2}}+\dfrac{3}{\dfrac{3\left(3+1\right)}{2}}+...+\dfrac{3}{\dfrac{2022\left(2022+1\right)}{2}}\)

\(=\dfrac{6}{2\left(2+1\right)}+\dfrac{6}{3\left(3+1\right)}+...+\dfrac{6}{2022\cdot2023}\)

\(=\dfrac{6}{2\cdot3}+\dfrac{6}{3\cdot4}+...+\dfrac{6}{2022\cdot2023}\)

\(=6\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2022\cdot2023}\right)\)

\(=6\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)

\(=6\cdot\left(\dfrac{1}{2}-\dfrac{1}{2023}\right)=6\cdot\dfrac{2021}{4046}=\dfrac{12126}{4046}< 3\)

mà \(3< \dfrac{10}{3}\)

nên \(M< \dfrac{10}{3}\)

Lời giải:
\(K=\sqrt{4+|1-\sqrt{5}|}.(\sqrt{10}-\sqrt{2})=\sqrt{4+\sqrt{5}-1}.\sqrt{2}(\sqrt{5}-1)\)

\(=\sqrt{3+\sqrt{5}}.\sqrt{2}.(\sqrt{5}-1)=\sqrt{6+2\sqrt{5}}.(\sqrt{5}-1)\)

\(=\sqrt{(\sqrt{5}+1)^2}(\sqrt{5}-1)=(\sqrt{5}+1)(\sqrt{5}-1)=4\)

\(\Rightarrow x^2+2x+1-y^2-4y-4-7=0\\ \Leftrightarrow\left(x+1\right)^2-\left(y+2\right)^2=7\\ \Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^2=16\\\left(y+2\right)^2=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x+1=4\\y+2=3\end{matrix}\right.\\\left\{{}\begin{matrix}x+1=-4\\y+2=-3\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)