cho x>0 tim Max \(A=\dfrac{1200x\left(12+x\right)}{\left(48+16x\right)^2}\)
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Đề này còn có lý, lần sau chú ý đọc kĩ đề trước khi đăng lên, tránh làm mất thời gian vô ích:
\(\left|x-2y\right|\le\dfrac{1}{\sqrt{x}}\Rightarrow1\ge\sqrt{x}\left|x-2y\right|\Rightarrow1\ge x\left(x-2y\right)^2\)
\(\Rightarrow1\ge x^3-4x^2y+4xy^2\)
Tương tự: \(\dfrac{1}{\sqrt{y}}\ge\left|y-2x\right|\Rightarrow1\ge y^3-4xy^2+4xy^2\)
Cộng vế:
\(\Rightarrow2\ge x^3+y^3=\dfrac{1}{2}\left(x^3+x^3+1\right)+\left(y^3+1+1\right)-\dfrac{5}{2}\ge\dfrac{1}{2}.3x^2+3y-\dfrac{3}{2}=\dfrac{3}{2}\left(x^2+2y\right)-\dfrac{5}{2}\)
\(\Rightarrow\dfrac{3}{2}\left(x^2+2y\right)\le\dfrac{9}{2}\Rightarrow x^2+2y\le3\)
\(E=\dfrac{4\left|x\right|+9}{\left|x\right|+1}\)
\(\left\{{}\begin{matrix} \left|x\right|\ge0\Rightarrow4\left|x\right|\ge0\Rightarrow4\left|x\right|+9\ge9\\\left|x\right|\ge0\Rightarrow x+1\ge1\end{matrix}\right.\)
\(MAX_E\Rightarrow MIN_{\left|x\right|+1}\)
\(MIN_{\left|x\right|+1}=1\)
\(\Rightarrow\left|x\right|=0\Rightarrow x=0\)
\(\Rightarrow MAX_E=\dfrac{4.\left|0\right|+9}{\left|0\right|+1}=\dfrac{9}{1}=9\)
\(F=\dfrac{2\left|x\right|+8}{3\left|x\right|+1}\)
\(\left\{{}\begin{matrix}\left|x\right|\ge0\Rightarrow2\left|x\right|\ge0\Rightarrow2\left|x\right|+8\ge8\\\left|x\right|\ge0\Rightarrow3\left|x\right|\ge0\Rightarrow3\left|x\right|+1\ge1\end{matrix}\right.\)
\(MAX_F\Rightarrow MIN_{3\left|x\right|+1}\)
\(MIN_{3\left|x\right|+1}=1\)
\(\Rightarrow\left|x\right|=0\Rightarrow x=0\)
\(\Rightarrow MAX_F=\dfrac{2.\left|0\right|+8}{3.\left|0\right|+1}=\dfrac{8}{1}=8\)
\(\)
(1) \(\Leftrightarrow\left(x+1\right)\left(\sqrt{16x+17}-x+\dfrac{23}{8}\right)=0\)
cái này đâu ra z ???
nguyen van tuan: hì, xin lỗi, làm hơi tắt ^^!
\(\left(1\right)\Leftrightarrow\left(x+1\right)\sqrt{16x+17}=\left(x+1\right)\left(x-\dfrac{23}{8}\right)\Leftrightarrow\left(x+1\right)\sqrt{16x+17}-\left(x+1\right)\left(x-\dfrac{23}{8}\right)=0\Leftrightarrow\left(x+1\right)\left(\sqrt{16x+17}-x+\dfrac{23}{8}\right)=0\)
Vì 3 ≤ x ≤ 7 => x - 3 ≥ 0; 7 - x ≥ 0
=> C ≥ 0
Dấu = xảy ra khi và chỉ khi x = 3 hoặc x = 7
C = (x - 3)(7 - x) ≤ \(\dfrac{1}{4}\)(x - 3 + 7 - x)2 = \(\dfrac{1}{4}\).42 = 4
Dấu "=" xảy ra <=> x - 3 = 7 - x <=> x = 5
\(G=\left(x^2+\sqrt[3]{3}\right)+\left(\dfrac{2}{x^3}+\dfrac{2}{\sqrt{3}}+\dfrac{2}{\sqrt{3}}\right)-\sqrt[3]{3}-\dfrac{4}{\sqrt{3}}\ge2\sqrt{x^2.\sqrt[3]{3}}+3\sqrt[3]{\dfrac{2}{x^3}.\dfrac{2}{\sqrt{3}}.\dfrac{2}{\sqrt{3}}}-\sqrt[3]{3}-\dfrac{4}{\sqrt{3}}=2\sqrt[6]{3}.x+\dfrac{6}{\sqrt[3]{3}x}-\sqrt[3]{3}-\dfrac{4}{\sqrt{3}}\ge2\sqrt{2\sqrt[6]{3}.x.\dfrac{6}{\sqrt[3]{3}x}}-\sqrt[3]{3}-\dfrac{4}{\sqrt{3}}=2\sqrt{\dfrac{12\sqrt[6]{3}}{\sqrt[3]{3}}}-\sqrt[3]{3}-\dfrac{4}{\sqrt{3}}\)
Dấu "=" xảy ra khi và chỉ khi \(x=\sqrt[6]{3}\)
a) ĐK: \(x\ge0\)
PT \(\Leftrightarrow\sqrt{4x}\left(\dfrac{3}{4}-1-\dfrac{1}{4}\right)+5=0\)
\(\Leftrightarrow2\sqrt{x}.\left(-\dfrac{1}{2}\right)+5=0\)
\(\Leftrightarrow x=25\) (thỏa)
Vậy \(x=25\)
b) Đk: \(x\le3\)
PT \(\Leftrightarrow\sqrt{3-x}-\sqrt{9\left(3-x\right)}+\dfrac{5}{4}\sqrt{16\left(3-x\right)}=6\)
\(\Leftrightarrow\sqrt{3-x}\left(1-\sqrt{9}+\dfrac{5}{4}.\sqrt{16}\right)=6\)
\(\Leftrightarrow\sqrt{3-x}=2\Leftrightarrow x=-1\) (thỏa)
Vậy \(x=-1\)
2:
a:
Sửa đề: \(P=\left(\dfrac{2}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\dfrac{2}{\sqrt{1-a^2}}+1\right)\)
\(P=\dfrac{2+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}:\dfrac{2+\sqrt{1-a^2}}{\sqrt{1-a^2}}\)
\(=\dfrac{2+\sqrt{1-a^2}}{\sqrt{1+a}}\cdot\dfrac{\sqrt{1-a^2}}{2+\sqrt{1-a^2}}=\sqrt{\dfrac{1-a^2}{1+a}}\)
\(=\sqrt{1-a}\)
b: Khi a=24/49 thì \(P=\sqrt{1-\dfrac{24}{49}}=\sqrt{\dfrac{25}{49}}=\dfrac{5}{7}\)
c: P=2
=>1-a=4
=>a=-3
\(A=\left(\dfrac{x^2}{\left(x+1\right)^3}\right)^2=\left(\dfrac{x^2}{x^3+3x^2+3x+1}\right)^2=\left(\dfrac{1}{x+\dfrac{3}{x}+\dfrac{1}{x^2}+3}\right)^2\)
\(A=\left(\dfrac{1}{x+\dfrac{4}{x}+\left(\dfrac{1}{x}-\dfrac{1}{2}\right)^2+\dfrac{11}{4}}\right)^2\le\left(\dfrac{1}{x+\dfrac{4}{x}+\dfrac{11}{4}}\right)^2\)
\(A\le\left(\dfrac{1}{2\sqrt{\dfrac{4x}{x}}+\dfrac{11}{4}}\right)^2=\dfrac{16}{729}\)
Dấu "=" xảy ra khi \(x=2\)
\(A=\dfrac{75x\left(12+x\right)}{\left(12+4x\right)^2}\);\(A>0\forall x>0\)
Gọi \(A_0\in MGT\) của A
\(\Rightarrow A_0=\dfrac{75x\left(12+x\right)}{\left(12+4x\right)^2}\) có nghiệm
\(\Rightarrow A_0\left(12+4x\right)^2=75x\left(12+x\right)\)
\(\Leftrightarrow x^2\left(16A_0-75\right)+x\left(96A_0-900\right)+144A_0=0\) có nghiệm
\(\Leftrightarrow\Delta\ge0\Leftrightarrow-4A_0+25\ge0\)\(\Leftrightarrow A_0\le\dfrac{25}{4}\)
\(\Rightarrow maxA=\dfrac{25}{4}\)