a) Tìm GTNN Của:
A=\(\left(2x+\dfrac{1}{3}\right)^4-1\)
a) Tìm GTLN Của:
B=\(-\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6+3\)
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\(AB=5\left(cm\right)< CD=k^2\)
\(MN=11\left(cm\right)>CD=k^2\)
\(\Rightarrow CD=k^2=9\left(cm\right)\)
\(A=\dfrac{3x-1}{x+2}\inℕ\left(x\inℕ;x\ne-2\right)\)
\(\Rightarrow3x-1⋮x+2\)
\(\Rightarrow3x-1-3\left(x+2\right)⋮x+2\)
\(\Rightarrow3x-1-3x-6⋮x+2\)
\(\Rightarrow-7⋮x+2\)
\(\Rightarrow x+2\in U\left(7\right)=\left\{1;7\right\}\)
\(\Rightarrow x\in\left\{-1;5\right\}\)
\(\Rightarrow x\in\left\{5\right\}\left(x\inℕ\right)\)
\(\dfrac{x-6}{1998}\) + \(\dfrac{x-4}{2000}\) = \(\dfrac{x-2000}{4}\) + \(\dfrac{x-1998}{6}\)
\(\dfrac{x-6}{1998}\) - 1 + \(\dfrac{x-4}{2000}\) - 1 = \(\dfrac{x-2000}{4}\) - 1 + \(\dfrac{x-1998}{6}\) - 1
\(\dfrac{x-6-1998}{1998}\) + \(\dfrac{x-4-2000}{2000}\) = \(\dfrac{x-2000-4}{4}\) + \(\dfrac{x-1998-6}{6}\)
\(\dfrac{x-2004}{1998}\) + \(\dfrac{x-2004}{2000}\) = \(\dfrac{x-2004}{4}\) + \(\dfrac{x-2004}{6}\)
(\(x-2004\)).[\(\dfrac{1}{1998}\) + \(\dfrac{1}{2000}\) - \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\)] = 0
\(x\) - 2004 = 0
\(x\) = 2004
\(\dfrac{x+1}{65}+\dfrac{x+3}{63}+\dfrac{x+5}{61}+\dfrac{x+7}{59}\)
\(\Leftrightarrow\dfrac{x+1}{65}+\dfrac{x+3}{63}-\dfrac{x+5}{61}-\dfrac{x+7}{59}=0\)
\(\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+3}{63}+1\right)-\left(\dfrac{x+5}{61}+1\right)-\left(\dfrac{x+7}{59}+1\right)\)
\(\Leftrightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}+\dfrac{x+66}{61}+\dfrac{x+66}{59}=0\)
\(\Leftrightarrow\left(x+66\right).\left[\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)\right]\)\(=0\)
Do \(\dfrac{1}{65}< \dfrac{1}{63}< \dfrac{1}{61}< \dfrac{1}{59}\)
\(\Rightarrow\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)< 0\)
Vậy để \(\left(x+66\right).\left[\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)\right]=0\)
\(\Leftrightarrow x+66=0\)
\(\Leftrightarrow x=-66\)
Vậy \(x\in\left\{-66\right\}\)
\(\left(\dfrac{1}{10}\right)^{15}=\left[\left(\dfrac{1}{10}\right)^3\right]^5=\left(\dfrac{1}{1000}\right)^5=\left(\dfrac{10}{10000}\right)^5\)
\(\left(\dfrac{3}{10}\right)^{20}=\left[\left(\dfrac{3}{10}\right)^4\right]^5=\left(\dfrac{81}{10000}\right)^5\)
\(\dfrac{10}{10000}< \dfrac{81}{10000}\)
\(\Rightarrow\left(\dfrac{10}{10000}\right)^5< \left(\dfrac{81}{10000}\right)^5\)
\(\Rightarrow\left(\dfrac{1}{10}\right)^{15}< \left(\dfrac{3}{10}\right)^{20}\)
Ta có:
\(\left(\dfrac{1}{10}\right)^{15}=\left[\left(\dfrac{1}{10}\right)^3\right]^5=\left(\dfrac{1}{1000}\right)^5\)
\(\left(\dfrac{3}{10}\right)^{20}=\left[\left(\dfrac{3}{10}\right)^4\right]^5=\left(\dfrac{81}{10000}\right)^5\)
Ta thấy: \(\dfrac{1}{1000}< \dfrac{81}{10000}\)
\(\Rightarrow\left(\dfrac{1}{1000}\right)^5< \left(\dfrac{81}{10000}\right)^5\)
\(\Rightarrow\left(\dfrac{1}{10}\right)^{15}< \left(\dfrac{3}{10}\right)^{20}\)
a) Vẽ tia By' là tia đối của tia By
Ta có:
∠ABy' + ∠ABy = 180⁰ (kề bù)
⇒ ∠ABy' = 180⁰ - ∠ABy
= 180⁰ - 135⁰
= 45⁰
⇒ ∠ABy' = ∠BAx = 45⁰
Mà ∠ABy' và ∠BAx là hai góc so le trong
⇒ By // Ax
b) Ta có:
∠CBy' = ∠ABC - ∠ABy'
= 75⁰ - 45⁰
= 30⁰
⇒ ∠CBy' = ∠BCz = 30⁰
Mà ∠CBy' và ∠BCz là hai góc so le trong
⇒ By // Cz
\(a^m=a^n\)
\(\Rightarrow m=n\)
Với \(a^m=a^n\) mọi \(m=n\)
Vậy: \(m=n\in\left\{1;2;3;4;...\right\}\)
Tìm \(x\) biết: |\(x\) + 1| + |\(x\) + 4| = 3\(x\) ( đk \(x\) ≥ 0)
|\(x\) + 1| + | \(x\) + 4| = 3\(x\)
Với \(x\) ≥ 0 ta có: \(x\) + 1 + \(x\) + 4 = 3\(x\)
2\(x\) + 5 = 3\(x\)
3\(x\) - 2\(x\) = 5
\(x\) = 5 (thỏa mãn)
Vậy \(x\) = 5
\(\left|x+1\right|+\left|x+4\right|=3x\left(1\right)\)
Ta có :
\(\left|x+1\right|+\left|x+4\right|\ge\left|x+1+x+4\right|=\left|2x+5\right|\)
\(pt\left(1\right)\Leftrightarrow\left|2x+5\right|=3x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+5=3x\\2x+5=-3x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\5x=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
\(B=-\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6+3\)
vì \(B=-\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6\le0,\forall x\inℝ\)
\(\Rightarrow B=-\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6+3\le3\)
Dấu "=" xảy ra khi và chỉ khi
\(\dfrac{4}{9}x-\dfrac{2}{15}=0\Rightarrow\dfrac{4}{9}x=\dfrac{2}{15}\Rightarrow x=\dfrac{9}{15}\)
Vậy \(GTLN\left(B\right)=3\left(tạix=\dfrac{9}{15}\right)\)
\(A=\left(2x+\dfrac{1}{3}\right)^4-1\)
vì \(\left(2x+\dfrac{1}{3}\right)^4\ge0,\forall x\inℝ\)
\(\Rightarrow A=\left(2x+\dfrac{1}{3}\right)^4-1\ge-1\)
Dấu "=" xảy ra khi và chỉ khi
\(2x+\dfrac{1}{3}=0\Rightarrow2x=-\dfrac{1}{3}\Rightarrow x=-\dfrac{1}{6}\)
\(\Rightarrow GTNN\left(A\right)=-1\left(tạix=-\dfrac{1}{6}\right)\)