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a: \(\left|3x-1\right|\ge0\forall x\)
=>\(\left|3x-1\right|+2025\ge2025\forall x\)
Dấu '=' xảy ra khi 3x-1=0
=>3x=1
=>\(x=\frac13\)
b: Sửa đề: \(\left|2x+1\right|+\left|2y-1\right|+2\)
Ta có: \(\left|2x+1\right|\ge0\forall x\)
\(\left|2y-1\right|\ge0\forall y\)
Do đó: \(\left|2x+1\right|+\left|2y-1\right|\ge0\forall x,y\)
=>\(\left|2x+1\right|+\left|2y-1\right|+2\ge2\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}2x+1=0\\ 2y-1=0\end{cases}\Rightarrow\begin{cases}x=-\frac12\\ y=\frac12\end{cases}\)
Câu a:
A = |3\(x\) - 1| + 2025
A = |3\(x\) - 1| ≥ 0 ∀ \(x\)
A = |3\(x\) - 1| + 2025 ≥ 2025; Dấu = xảy ra khi:
3\(x\) - 1 = 0 ⇒ 3\(x\) = 1 ⇒ \(x=\frac13\)
Vậy Amin = 2025 khi \(x\) = \(\frac13\)
Câu b:
B = |2\(x\) + 1| - |2y - 1| + 2
|2\(x\) + 1| ≥ 0 ∀ \(x\) ; |2y - 1| ≥ 0 ∀ y
⇒ |2\(x\) + 1| - |2y - 1| + 2 ≥ 2 Dấu bằng xảy ra khi:
\(\begin{cases}2x+1=0\\ 2y-1=0\end{cases}\)
⇒ \(\begin{cases}2x=-1\\ 2y=1\end{cases}\)
⇒ \(\begin{cases}x=-\frac12\\ y=\frac12\end{cases}\)
Vậy Bmin = 2 khi (\(x;y\)) = (- \(\frac12\); \(\frac12\))
a. Tại x=\(\frac{-1}{2}\), ta có:
\(\left(\frac{-1}{2}\right)^2+4.\left(\frac{-1}{2}\right)+3=\frac{1}{4}+\left(-2\right)+3=\frac{5}{4}\)
b. Ta có:
\(x^2+4x+3=0\)
\(\Rightarrow x^2+x+3x+3=0\)
\(\Rightarrow\left(x^2+x\right)+\left(3x+3\right)=0\)
\(\Rightarrow x\left(x+1\right)+3\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x+3\right)=0\)
\(\Rightarrow\hept{\begin{cases}x+1=0\\x+3=0\end{cases}\Rightarrow\hept{\begin{cases}x=-1\\x=-3\end{cases}}}\)
Vậy \(x=-1;x=-3\)
a/ \(M=x^4-xy^3+x^3y-y^4-1\)
\(\Leftrightarrow M=x^3\left(x+y\right)-y^3\left(x+y\right)-1\)
Mà \(x+y=0\)
\(\Leftrightarrow M=x^3.0-y^3.0-1\)
\(\Leftrightarrow M=-1\)
Vậy ...
\(A=\frac{\left(1+2+...+100\right)\left(\frac{1}{2}^2-...-\frac{1}{5}\right)\left(2,4.42-21.4,8\right)}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)
=> \(A=\frac{\left(1+2+...+100\right)\left(\frac{1}{2}-...-\frac{1}{5}\right).0}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)= 0
a: \(A=\left|x+1\right|+5\ge5\forall x\)
Dấu '=' xảy ra khi x=-1
b: \(B=\dfrac{x^2+3+12}{x^2+3}=1+\dfrac{12}{x^2+3}\le\dfrac{12}{3}+1=4+1=5\)
Dấu '=' xảy ra khi x=0
\(D=\frac{x^2+8}{x^2+3}=\frac{x^2+3+5}{x^2+3}=1+\frac{5}{x^2+3}\)
ta có x^2+3>=3 => 5/(x^2+3)<=5/3
=> D = 8/3 tại x=0
câu b)
2(x-1)2 +3 >=3
=> C <= 1/3 tại x=1
\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{2025^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{2025}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{2025}+1\right)\)
\(=-\dfrac{1}{2}\cdot-\dfrac{2}{3}\cdot...\cdot\dfrac{-2024}{2025}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2026}{2025}\)
\(=\dfrac{1}{2025}\cdot\dfrac{2026}{2}=\dfrac{1013}{2025}\)
=-\(\left(\frac12\cdot\frac32\right)\cdot\left(\frac23\cdot\frac43\right)\ldots\left(\frac{2024}{2025}\cdot\frac{2026}{2025}\right)\)
=-\(\left(\frac{1}{2025}\cdot\frac{2026}{2}\right)\)
=-\(\frac{1013}{2025}\)