a) \(16^{^3}:8^2=\left(8.2\right)^3:8^2=8^3.2^3:8^2=\left(8^3:8^2\right).2^3=8.8=64\)

b)\(8^3.\left(0,125\right)^3=\left(8.0,125\right)^3=1^3=1\)

c)\(7^{^{200}}.\left(\frac{1}{7}\right)^{200}=\left(7.\frac{1}{7}\right)^{200}=1^{200}=1\)

d)\(4.\left(0,25\right)^3.64=4.\left(0,25\right)^3.4^3=4.\left(0,25.4\right)^3=4.1=4\)

e)....

cậu có thể tham khảo bài làm trên đây ạ, chúc cậu hok tốt ^^

a)

\(P=a\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}+\frac{a}{b}=a\sqrt{\frac{a^2\left(a+1\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}}+\frac{a}{a+1}\)

      =\(a\sqrt{\frac{a^2\left(a+1\right)^2+2a\left(a+1\right)+1}{a^2\left(a+1\right)^2}}+\frac{a}{a+1}=a\sqrt{\frac{\left[a\left(a+1\right)+1\right]^2}{\left[a\left(a+1\right)\right]^2}}+\frac{a}{a+1}\)

      \(=a.\frac{a\left(a+1\right)+1}{a\left(a+1\right)}+\frac{a}{a+1}=a+\frac{1}{a+1}+\frac{a}{a+1}=a+1\)

Vay P=a+1

phan b,c ap dung phan a la ra

CM bài toán phụ: \(x+y+z=0\) 

CM: \(I=\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\) với x,y,z dương

Ta có: \(I=\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}=\sqrt{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2-2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)}\)

\(=\sqrt{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2-2\cdot\frac{x+y+z}{xyz}}=\sqrt{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\)

\(=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

Áp dụng vào ta được: \(Q=1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{2020}-\frac{1}{2021}\)

\(Q=2021-\frac{1}{2021}=...\)

1, C . To watch

2,  B . Is

1)A                                                2)B

Ta có : xy + x + y = -1

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

=> (x + 1)(y + 1) = 0

=> \(\orbr{\begin{cases}x+1=0\\y+1=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-1\\y=-1\end{cases}}\)(đpcm) 

Vậy nếu xy + x + y = - 1 thì có ít nhất 1 số bằng - 1

xy + x + y = -1

<=> xy + x + y + 1 = 0

<=> x( y + 1 ) + 1( y + 1 ) = 0

<=> ( x + 1 )( y + 1 ) = 0

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\y+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\y=-1\end{cases}}\) ( đpcm )

thì phân tích thành nhân tử là oke

\(x^2+x+1>0\)

\(\Leftrightarrow x^2+x+\frac{1}{4}+\frac{3}{4}>0\)

\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)*đúng*

Ta có:\(x^2+x+1=\left(x^2+x+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)

Vì\(\left(x+\frac{1}{2}\right)^2\ge0\forall x\in R\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\left(đpcm\right)\)

Dãy số trên có số hạng là:

(2016-2):2=1007 (số hạng)

là 2026 mà bn.

\(\frac{\left(81,624:4\frac{4}{3}-4,505\right)+125\frac{3}{4}}{\left[\left(\left(\frac{11}{25}\right)^2:0,08+3,53\right)^2-\left(2,75\right)^2\right]:\frac{13}{25}}\)

\(=\frac{\left(\frac{10203}{125}.\frac{3}{16}-\frac{901}{200}\right)+\frac{503}{4}}{\left[\left(\frac{121}{625}.\frac{25}{2}+\frac{353}{100}\right)^2-\frac{121}{16}\right].\frac{25}{13}}\)

\(=\frac{\left(15,3045-\frac{901}{200}\right)+\frac{503}{4}}{\left(\frac{14161}{400}-\frac{121}{16}\right).\frac{25}{13}}\)

\(=\frac{136,5495}{\frac{696}{13}}\)

\(=2,550493534\)

1)\(\left(-8\right)^2=64\)

2)\(\left(-1,25\right)^2=1,56\)

3) \(3^5=243\)

4) \(2^5:2^3\Leftrightarrow2^{5-3}=2^2\)

5) \(\left(-4\right)^2\times\left(-4\right)=\left(-4\right)^{2+1}=\left(-4\right)^3\)

6) \(\left(\frac{2}{3}\right)^3\times\left(\frac{2}{3}\right)^2=\left(\frac{2}{3}\right)^{3+2}=\left(\frac{2}{3}\right)^5\)

6,1. = 64

2 . = 1,56

3 . =243

4 , = 2² = 4

5  , (-4³) = -12

6, = 2/3 ⁵ = 22 / 213