1)Sắp xếp theo thứ tự tăng dần
\(\left(\dfrac{-2}{5}\right)^3;\left(0,3\right)^2;\left(\dfrac{-3}{4}\right)^3;\left(-1,2\right)^2\)
2)Tìm x và y sao cho
\(\left(\dfrac{x}{y}\right)=\left(\dfrac{x}{y}\right)^2\)
3)Chứng minh rằng
a)\(16^5+2^{15}\) chia hết cho 33
b)\(333^{555}+555^{333}\) chia hết cho 37
4)Tìm x và y biết
a)\(\left(x-3\right)^2+\left(y+2\right)^2=0\)
b)\(\left(x-12+y\right)^{200}+\left(x-4-y\right)^{200}=0\)
5)Tính giá trị của...
Đọc tiếp
1)Sắp xếp theo thứ tự tăng dần
\(\left(\dfrac{-2}{5}\right)^3;\left(0,3\right)^2;\left(\dfrac{-3}{4}\right)^3;\left(-1,2\right)^2\)
2)Tìm x và y sao cho
\(\left(\dfrac{x}{y}\right)=\left(\dfrac{x}{y}\right)^2\)
3)Chứng minh rằng
a)\(16^5+2^{15}\) chia hết cho 33
b)\(333^{555}+555^{333}\) chia hết cho 37
4)Tìm x và y biết
a)\(\left(x-3\right)^2+\left(y+2\right)^2=0\)
b)\(\left(x-12+y\right)^{200}+\left(x-4-y\right)^{200}=0\)
5)Tính giá trị của x
a)\(8^x>16\)
b)\(\left(\dfrac{2}{3}\right)^x=\dfrac{4}{9}\)
c)\(\left(\dfrac{1}{2}\right)^{x-1}=8\)
d)\(3^x< 16\)
e)\(3^x< 27\)
Help me mai 5/8 6h mik đi học rồi
)
2) \(\dfrac{x}{y}=\left(\dfrac{x}{y}\right)^2\)
\(\Rightarrow\left(\dfrac{x}{y}\right)^2-\dfrac{x}{y}=0\)
\(\Rightarrow\dfrac{x}{y}\left(\dfrac{x}{y}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{y}=0\Rightarrow x=0;y\in R\\\dfrac{x}{y}-1=0\Rightarrow\dfrac{x}{y}=1\Rightarrow x=y\end{matrix}\right.\)
3) \(16^5+2^{15}=\left(2^4\right)^5+2^{15}=2^{20}+2^{15}=2^{15}.2^5+2^{15}.1=2^{15}.33⋮33\rightarrowđpcm\)
4)\(\left(x-3\right)^2+\left(y+2\right)^2=0\)
\(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0\\\left(y+2\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left(x-3\right)^2+\left(y+2\right)^2\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left(x-3\right)^2=0\Rightarrow x-3=0\Rightarrow x=3\\\left(y+2\right)^2=0\Rightarrow y+2=0\Rightarrow y=-2\end{matrix}\right.\)
\(\left(x-12+y\right)^{200}+\left(x-4-y\right)^{200}=0\)
\(\left\{{}\begin{matrix}\left(x-12+y\right)^{200}\ge0\\\left(x-4-y\right)^{200}\ge0\end{matrix}\right.\)
\(\Rightarrow\left(x-12+y\right)^{200}+\left(x-y-4\right)^{200}\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left(x-12+y\right)^{200}=0\\\left(x-y-4\right)^{200}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-12+y=0\Rightarrow x+y=12\\x-y-4=0\Rightarrow x-y=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x+y\right)+\left(x-y\right)=12+4\Rightarrow x+y+x-y=16\Rightarrow2x=16\Rightarrow x=8\\y=8-4=4\end{matrix}\right.\)