Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
TH1: \(x+y+z+t=0\)
\(P=\left(1+\dfrac{x+y}{z+t}\right)^{2023}+\left(1+\dfrac{y+z}{x+t}\right)^{2023}+\left(1+\dfrac{z+t}{x+y}\right)^{2023}+\left(1+\dfrac{t+x}{y+z}\right)^{2023}\)
\(=\left(\dfrac{x+y+z+t}{z+t}\right)^{2023}+\left(\dfrac{x+y+z+t}{x+t}\right)^{2023}+\left(\dfrac{x+y+z+t}{x+y}\right)^{2023}+\left(\dfrac{x+y+z+t}{y+z}\right)^{2023}\)
\(=0+0+0+0=0\) là số nguyên (thỏa mãn)
TH2: \(x+y+z+t\ne0\), áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2023x+y+z+t}=\dfrac{y}{x+2023y+z+t}=\dfrac{z}{x+y+2023z+t}+\dfrac{t}{x+y+z+2023t}\)
\(=\dfrac{x+y+z+t}{\left(2023x+y+z+t\right)+\left(x+2023y+z+t\right)+\left(x+y+2023z+t\right)+\left(x+y+z+2023t\right)}\)
\(=\dfrac{x+y+z+t}{2026\left(x+y+z+t\right)}=\dfrac{1}{2026}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2023x+y+z+t}=\dfrac{1}{2026}\\\dfrac{y}{x+2023y+z+t}=\dfrac{1}{2026}\\\dfrac{z}{x+y+2023z+t}=\dfrac{1}{2026}\\\dfrac{t}{x+y+z+2023t}=\dfrac{1}{2026}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2026x=2023x+y+z+t\\2026y=x+2023y+z+t\\2026z=x+y+2023z+t\\2026t=x+y+z+2023t\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}4x=x+y+z+t\\4y=x+y+z+t\\4z=x+y+z+t\\4t=x+y+z+t\end{matrix}\right.\)
\(\Rightarrow4x=4y=4z=4t\) (vì đều bằng \(x+y+z+t\))
\(\Rightarrow x=y=z=t\)
Do đó:
\(P=\left(1+\dfrac{x+x}{x+x}\right)^{2023}+\left(1+\dfrac{x+x}{x+x}\right)^{2023}+\left(1+\dfrac{x+x}{x+x}\right)^{2023}+\left(1+\dfrac{x+x}{x+x}\right)^{2023}\)
\(=2^{2023}+2^{2023}+2^{2023}+2^{2023}\)
\(=4.2^{2023}=2^{2025}\in Z\)
Em kiểm tra lại đề, 2 ngoặc cuối bị giống nhau, chắc em ghi nhầm
\(\dfrac{x-2023}{6}+\dfrac{x-2023}{10}+\dfrac{x-2023}{15}+\dfrac{x-2023}{21}=\dfrac{8}{21}\)
\(\left(x-2023\right)\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)=\dfrac{8}{21}\)
\(\left(x-2023\right).\dfrac{8}{21}=\dfrac{8}{21}\)
\(x-2023=1\)
\(x=2024\)
Vậy..............
\(...\Rightarrow\left(x-2023\right)\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)=\dfrac{8}{21}\)
\(\Rightarrow\left(x-2023\right)\left(\dfrac{35+21+14+1}{210}\right)=\dfrac{8}{21}\)
\(\Rightarrow\left(x-2023\right).\dfrac{71}{210}=\dfrac{8}{21}\)
\(\Rightarrow\left(x-2023\right).\dfrac{71}{210}=\dfrac{8}{21}.\dfrac{210}{71}=\dfrac{80}{71}\)
\(\Rightarrow x-2023=\dfrac{80}{71}\Rightarrow x=\dfrac{80}{71}+2023=\dfrac{143713}{71}\)
a: \(\left(2x-y+7\right)^{2022}>=0\forall x,y\)
\(\left|x-1\right|^{2023}>=0\forall x\)
=>\(\left(2x-y+7\right)^{2022}+\left|x-1\right|^{2023}>=0\forall x,y\)
mà \(\left(2x-y+7\right)^{2022}+\left|x-1\right|^{2023}< =0\forall x,y\)
nên \(\left(2x-y+7\right)^{2022}+\left|x-1\right|^{2023}=0\)
=>\(\left\{{}\begin{matrix}2x-y+7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2x+7=9\end{matrix}\right.\)
\(P=x^{2023}+\left(y-10\right)^{2023}\)
\(=1^{2023}+\left(9-10\right)^{2023}\)
=1-1
=0
c: \(\left|x-3\right|>=0\forall x\)
=>\(\left|x-3\right|+2>=2\forall x\)
=>\(\left(\left|x-3\right|+2\right)^2>=4\forall x\)
mà \(\left|y+3\right|>=0\forall y\)
nên \(\left(\left|x-3\right|+2\right)^2+\left|y+3\right|>=4\forall x,y\)
=>\(P=\left(\left|x-3\right|+2\right)^2+\left|y-3\right|+2019>=4+2019=2023\forall x,y\)
Dấu '=' xảy ra khi x-3=0 và y-3=0
=>x=3 và y=3
Sửa đề: \(B=\left(\dfrac{2008}{2023}-\dfrac{2023}{2008}\right)-\left(\dfrac{-15}{2003}-\dfrac{15}{2008}\right)\)
\(=\dfrac{2008}{2023}-\dfrac{2023}{2008}+\dfrac{15}{2003}+\dfrac{15}{2008}\)
=1-1
=0
a: \(\left|a-2b+3\right|^{2023}>=0\forall a,b\)
\(\left(b-1\right)^{2024}>=0\forall b\)
Do đó: \(\left|a-2b+3\right|^{2023}+\left(b-1\right)^{2024}>=0\forall a,b\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}a-2b+3=0\\b-1=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}b=1\\a=2b-3=2\cdot1-3=-1\end{matrix}\right.\)
Thay a=-1 và b=1 vào P, ta được:
\(P=\left(-1\right)^{2023}\cdot1^{2024}+2024=2024-1=2023\)
\(\left(x-1\right)^3-\left(\dfrac{2}{2023}-\dfrac{7}{247}+\dfrac{1}{8}\right)=\dfrac{7}{247}-\dfrac{2}{2023}\)
\(\Rightarrow\left(x-1\right)^3-\dfrac{2}{2023}+\dfrac{7}{247}-\dfrac{1}{8}=\dfrac{7}{247}-\dfrac{2}{2023}\)
\(\Rightarrow\left(x-1\right)^3=\dfrac{7}{247}-\dfrac{7}{247}-\dfrac{2}{2023}+\dfrac{2}{2023}+\dfrac{1}{8}\)
\(\Rightarrow\left(x-1\right)^3=\dfrac{1}{8}\)
\(\Rightarrow\left(x-1\right)^3=\left(\dfrac{1}{2}\right)^3\)
\(\Rightarrow x-1=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{2}+1\)
\(\Rightarrow x=\dfrac{3}{2}\)
Lời gải:
$(x-1)^3=\frac{7}{247}-\frac{2}{2023}+\frac{2}{2023}-\frac{7}{247}+\frac{1}{8}=\frac{1}{8}$
$x-1=\frac{1}{2}$
$x=\frac{1}{2}+1=\frac{3}{2}$
tìm giá trị lớn nhất của P = \(\dfrac{|x-2022|-|x-2023|+|x-2024|+2022}{|x-2022|+|x-2023|+|x-2024|}\)
Mình đùa chút nhé:
Cần j chứng minh, thấy nó đúng là đc mà!
mình nghĩ c/m là cái điều đấy nó đã đúng sẵn rồi
nên chắc chẳng cần c/m đâu nhỉ =)
\(\dfrac{2023^{2024x}+2023^{2025x}}{2023^{2023x}+2023^{2024x}}=2023\)
=>\(\dfrac{2023^{2024x}\left(1+2023^x\right)}{2023^{2023x}\left(1+2023^x\right)}=2023\)
=>\(2023^x=2023\)
=>x=1