Em đăng kí nhận gp

Em đăng kí nhận quà khảo sát ứng dụng

Ta có:

+) Số M được chia thành 3 phần \(p_1;p_2;p_3\)

 \(\Rightarrow M=p_1+p_2+p_3\)

+) \(p_3\) hơn \(p_1\) là 42

\(\Rightarrow p_3-p_1=42\)

+) \(p_1;p_2\) tỉ lệ với \(\text{3;4}\) và \(p_2;p_3\) tỉ lệ với \(5;9\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{p_1}{3}=\dfrac{p_2}{4}\\\dfrac{p_2}{5}=\dfrac{p_3}{9}\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{p_1}{15}=\dfrac{p_2}{20}\\\dfrac{p_2}{20}=\dfrac{p_3}{36}\end{matrix}\right.\\ \Rightarrow\dfrac{p_1}{15}=\dfrac{p_2}{20}=\dfrac{p_3}{36}\)

Áp dụng tính chất dãy tỉ số bằng nhau kết hợp \(p_3-p_1=42\) được:

\(\dfrac{p_1}{15}=\dfrac{p_2}{20}=\dfrac{p_3}{36}=\dfrac{p_3-p_1}{36-15}=\dfrac{42}{21}=2\)

Suy ra:

\(\left\{{}\begin{matrix}p_1=15\cdot2=30\\p_2=20\cdot2=40\\p_3=36\cdot2=72\end{matrix}\right.\\ \Rightarrow M=p_1+p_2+p_3=30+40+72=142\)

Vậy...

a) \(\left(x+2023\right)^2+\left(y-\dfrac{1}{2}\right)^2=0\)

Nhận xét:

\(\left\{{}\begin{matrix}\left(x+2023\right)^2\ge0,\forall x\\\left(y-\dfrac{1}{2}\right)^2\ge0,\forall y\end{matrix}\right.\\ \Rightarrow\left(x+2023\right)^2+\left(y-\dfrac{1}{2}\right)^2\ge0,\forall x,y\)

Do đó, \(\left(x+2023\right)^2+\left(y-\dfrac{1}{2}\right)^2=0\) khi và chỉ khi:

\(\left\{{}\begin{matrix}x+2023=0\\y-\dfrac{1}{2}=0\end{matrix}\right.\\ \Rightarrow x=-2023;y=\dfrac{1}{2}\)

Vậy...

b) \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-1\right)^{10}\le0\) \(\left(1\right)\)

Nhận xét:

\(\left\{{}\begin{matrix}\left(\dfrac{1}{2}x-5\right)^{20}\ge0,\forall x\\\left(y^2-1\right)^{10}\ge0,\forall y\end{matrix}\right.\)

\(\Rightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-1\right)^{10}\ge0,\forall x,y\) \(\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra:

\(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-1\right)^{10}=0\) khi và chỉ khi:

\(\left\{{}\begin{matrix}\dfrac{1}{2}x-5=0\\y^2-1=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=5\\y^2=1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=10\\\left[{}\begin{matrix}y=1\\y=-1\end{matrix}\right.\end{matrix}\right.\\ \Rightarrow\left(x;y\right)\in\left\{\left(10;1\right),\left(10;-1\right)\right\}\)

Vậy...

Đặt \(A=1+4+4^2+4^3+...+4^{2021}\)
\(4A=4+4^2+4^3+4^4+...+4^{2022}\\ 4A-A=\left(4+4^2+4^3+4^4+...+4^{2022}\right)-\left(1+4+4^2+4^3+...+4^{2021}\right)\\ 3A=4^{2022}-1\\ A=\dfrac{4^{2022}-1}{3}\)Vậy...

Điều phải chứng minh là gì vậy bạn?

Ta có:

\(a+b=4\\ \Rightarrow\left(a+b\right)^2=a^2+2ab+b^2=4^2=16\)
Thay \(ab=3\) được:
\(a^2+2\cdot3+b^2=16\\ \Rightarrow a^2+6+b^2=16\\ \Rightarrow a^2+b^2=10\\ \Rightarrow a^2-ab+b^2=10-3=7\\ \Rightarrow\left(a+b\right)\left(a^2-ab+b^2\right)=a^3+b^3=4\cdot7=28\)

Vậy...

Ta có:
+) \(a-b=4\)
\(\Rightarrow\left(a-b\right)^2=a^2-2ab+b^2=4^2=16\)
Thay \(ab=5\) được:
\(\Rightarrow a^2-2\cdot5+b^2=a^2-10+b^2=16\\ \Rightarrow a^2+b^2=26\\ \Rightarrow a^2+ab+b^2=26+5=31\\ \Rightarrow\left(a-b\right)\left(a^2+ab+b^2\right)=a^3-b^3=4\cdot31=124\)

Vậy...

\(5.x-176=3^4.3^2\\ \Rightarrow5.x-176=81.9\\ \Rightarrow5.x-176=\\ \Rightarrow5.x-176=729\\ \Rightarrow5.x=729+176\\ \Rightarrow5.x=905\\ \Rightarrow x=905:5\\ \Rightarrow x=181\)

Vậy \(x=181\)

Sửa đề:

 \(3^{2n+3}-3^{2n+2}=2\cdot3^{10}\\ \Rightarrow3^{2n+2}\cdot3-3^{2n+2}=2\cdot3^{10}\\ \Rightarrow3^{2n+2}\cdot\left(3-1\right)=2\cdot3^{10}\\ \Rightarrow2\cdot3^{2n+2}=2\cdot3^{10}\\ \Rightarrow3^{2n+2}=3^{10}\\ \Rightarrow2n+2=10\\ \Rightarrow2n=8\\ \Rightarrow n=4\)

Vậy...