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Câu 3. Dự đoán dấu "=" khi \(a=b=c=\frac{1}{\sqrt{3}}\)
Dùng phương pháp chọn điểm rơi thôi :)
LG
Áp dụng bđt Cô-si được \(a^2+b^2+c^2\ge3\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow1\ge3\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow\frac{1}{3}\ge\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow\frac{1}{27}\ge a^2b^2c^2\)
\(\Rightarrow\frac{1}{\sqrt{27}}\ge abc\)
Khi đó :\(B=a+b+c+\frac{1}{abc}\)
\(=a+b+c+\frac{1}{9abc}+\frac{8}{9abc}\)
\(\ge4\sqrt[4]{abc.\frac{1}{9abc}}+\frac{8}{9.\frac{1}{\sqrt{27}}}\)
\(=4\sqrt[4]{\frac{1}{9}}+\frac{8\sqrt{27}}{9}=\frac{4}{\sqrt[4]{9}}+\frac{8}{\sqrt{3}}=\frac{4}{\sqrt{3}}+\frac{8}{\sqrt{3}}=\frac{12}{\sqrt{3}}=4\sqrt{3}\)
Dấu "=" \(\Leftrightarrow a=b=c=\frac{1}{\sqrt{3}}\)
Vậy .........
2, \(A=\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\)
\(A=\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\)
\(A=\left[\frac{a^2}{b+c}+\frac{\left(b+c\right)}{4}\right]+\left[\frac{b^2}{a+c}+\frac{\left(a+c\right)}{4}\right]+\left[\frac{c^2}{a+b}+\frac{\left(a+b\right)}{4}\right]-\frac{\left(a+b+c\right)}{2}\)
Áp dụng BĐT AM-GM ta có:
\(A\ge2.\sqrt{\frac{a^2}{4}}+2.\sqrt{\frac{b^2}{4}}+2.\sqrt{\frac{c^2}{4}}-\frac{\left(a+b+c\right)}{2}\)
\(A\ge a+b+c-\frac{6}{2}\)
\(A\ge6-3\)
\(A\ge3\)
Dấu " = " xảy ra \(\Leftrightarrow\)\(\frac{a^2}{b+c}=\frac{b+c}{4}\Leftrightarrow4a^2=\left(b+c\right)^2\Leftrightarrow2a=b+c\)(1)
\(\frac{b^2}{a+c}=\frac{a+c}{4}\Leftrightarrow4b^2=\left(a+c\right)^2\Leftrightarrow2b=a+c\)(2)
\(\frac{c^2}{a+b}=\frac{a+b}{4}\Leftrightarrow4c^2=\left(a+b\right)^2\Leftrightarrow2c=a+b\)(3)
Lấy \(\left(1\right)-\left(3\right)\)ta có:
\(2a-2c=c+b-a-b=c-a\)
\(\Rightarrow2a-2c-c+a=0\)
\(\Leftrightarrow3.\left(a-c\right)=0\)
\(\Leftrightarrow a-c=0\Leftrightarrow a=c\)
Chứng minh tương tự ta có: \(\hept{\begin{cases}b=c\\a=b\end{cases}}\)
\(\Rightarrow a=b=c=2\)
Vậy \(A_{min}=3\Leftrightarrow a=b=c=2\)
giải tạm 1 bài z -,-
2) Cauchy-Schwarz dạng Engel :
\(A=\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}=\dfrac{6}{2}=3\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=2\)
Chúc bạn học tốt ~
4/ Ta có: \(6=a+b+c+ab+bc+ca\ge3\left(\sqrt[3]{\left(abc\right)^2}+\sqrt[3]{abc}\right)\)
Đặt \(\sqrt[3]{abc}=t\Rightarrow t^2+t\le2\Rightarrow t\le1\Rightarrow t^3=C=abc\le1\)
Vậy...
5/ \(D\le\left(\frac{a+b+c}{3}\right)^3.\left[\frac{2\left(a+b+c\right)}{3}\right]^3=\frac{512}{729}\)
Vậy ...
P/s: Em không chắc
Bài 1:
Áp dụng BĐT Bunhiacopxky:
\(M^2=(a\sqrt{9b(a+8b)}+b\sqrt{9a(b+8a)})^2\)
\(\leq (a^2+b^2)(9ab+72b^2+9ab+72a^2)\)
\(\Leftrightarrow M^2\leq (a^2+b^2)(72a^2+72b^2+18ab)\)
Áp dụng BĐT AM-GM: \(a^2+b^2\geq 2ab\Rightarrow 18ab\leq 9(a^2+b^2)\)
Do đó, \(M^2\leq (a^2+b^2)(72a^2+72b^2+9a^2+9b^2)=81(a^2+b^2)^2\)
\(\Leftrightarrow M\leq 9(a^2+b^2)\leq 144\)
Vậy \(M_{\max}=144\Leftrightarrow a=b=\sqrt{8}\)
Bài 6:
\(a+\frac{1}{a-1}=1+(a-1)+\frac{1}{a-1}\)
Vì \(a>1\rightarrow a-1>0\). Do đó áp dụng BĐT Am-Gm cho số dương\(a-1,\frac{1}{a-1}\) ta có:
\((a-1)+\frac{1}{a-1}\geq 2\sqrt{\frac{a-1}{a-1}}=2\)
\(\Rightarrow a+\frac{1}{a-1}=1+(a-1)+\frac{1}{a-1}\geq 3\) (đpcm)
Dấu bằng xảy ra khi \(a-1=1\Leftrightarrow a=2\)
Bài 3:
Xét \(\sqrt{a^2+1}\). Vì \(ab+bc+ac=1\) nên:
\(a^2+1=a^2+ab+bc+ac=(a+b)(a+c)\)
\(\Rightarrow \sqrt{a^2+1}=\sqrt{(a+b)(a+c)}\)
Áp dụng BĐT AM-GM có: \(\sqrt{(a+b)(a+c)}\leq \frac{a+b+a+c}{2}=\frac{2a+b+c}{2}\)
hay \(\sqrt{a^2+1}\leq \frac{2a+b+c}{2}\)
Hoàn toàn tương tự với các biểu thức còn lại và cộng theo vế:
\(\sqrt{a^2+1}+\sqrt{b^2+1}+\sqrt{c^2+1}\leq \frac{2a+b+c}{2}+\frac{2b+a+c}{2}+\frac{2c+a+b}{2}=2(a+b+c)\)
Ta có đpcm. Dấu bằng xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
Bài 4:
Ta có:
\(A=\frac{8a^2+b}{4a}+b^2=2a+\frac{b}{4a}+b^2\)
\(\Leftrightarrow A+\frac{1}{4}=2a+\frac{b+a}{4a}+b^2=2a+b+\frac{b+a}{4a}+b^2-b\)
Vì \(a+b\geq 1, a>0\) nên \(A+\frac{1}{4}\geq a+1+\frac{1}{4a}+b^2-b\)
Áp dụng BĐT AM-GM:
\(a+\frac{1}{4a}\geq 2\sqrt{\frac{1}{4}}=1\)
\(\Rightarrow A+\frac{1}{4}\geq 2+b^2-b=\left(b-\frac{1}{2}\right)^2+\frac{7}{4}\geq \frac{7}{4}\)
\(\Leftrightarrow A\geq \frac{3}{2}\).
Vậy \(A_{\min}=\frac{3}{2}\Leftrightarrow a=b=\frac{1}{2}\)
1/ Ta có: \(x^2-2x-1=\left(\sqrt{2}+1\right)^2-2\left(\sqrt{2}+1\right)-1=0\)
\(\Rightarrow P=\left(x^4-4x^3+4x^2-2\right)^5+\left(x^3-3x^2-x-1\right)^6\)
\(=\left[\left(x^4-2x^3-x^2\right)+\left(-2x^3+4x^2+2x\right)+\left(x^2-2x-1\right)-1\right]^5+\left[\left(x^3-2x^2-x\right)+\left(-x^2+2x+1\right)-2x-2\right]^6\)
\(=\left(-1\right)^5+\left(-2x-2\right)^6\)
Xong
5) Lợi dụng AM-GM :v
\(a^4+a^4+a^4+b^4\ge4a^3b\)
\(b^4+b^4+b^4+a^4\ge4b^3a\)
\(\Rightarrow2a^4+2b^4\ge a^4+a^4+ab^3+a^3b=\left(a^3+b^3\right)\left(a+b\right)\)
\(\Rightarrow P\ge\dfrac{a+b}{2ab}+\dfrac{b+c}{2bc}+\dfrac{c+a}{2ac}=\dfrac{\left(a+b\right)c}{2abc}+\dfrac{\left(b+c\right)a}{2abc}+\dfrac{\left(c+a\right)b}{2abc}=\dfrac{2\left(ab+bc+ca\right)}{2abc}=1\)
Đẳng thức xảy ra \(\Leftrightarrow a=b=c=3\)
Bài 1:
A.\(\left(\sqrt{x}+2\right)\) = -1 (ĐK: \(x\ge0\)
\(\Leftrightarrow\dfrac{1}{x-4}\left(\sqrt{x}+2\right)=-1\)
\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=-1\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}-2}=-1\)
\(\Leftrightarrow\sqrt{x}-2=-1\)
\(\Leftrightarrow\sqrt{x}=1\\ \Leftrightarrow x=1\left(TM\right)\)
Vậy x = 1
Bài 2: ĐK: \(x\ge0\)
Để \(B\in Z\Leftrightarrow\dfrac{3}{\sqrt{x}+2}\in Z\Leftrightarrow\sqrt{x}+2\inƯ\left(3\right)\)\(\Leftrightarrow\sqrt{x}+2\in\left\{\pm1,\pm3\right\}\)\(\Leftrightarrow x\in\left\{1\right\}\)
Bài 3:
a, Ta có: \(x+\sqrt{x}+1=x+2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}+1\\ =\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
Ta có: 2 > 0 và \(x+\sqrt{x}+1>0\Rightarrow C>0\) và \(x\ne1\)
b, ĐK: \(x\ge0,x\ne1\)
\(C=\dfrac{2}{x+\sqrt{x}+1}\)
Ta có: \(x+\sqrt{x}+1=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có: \(\sqrt{x}\ge0\forall x\Rightarrow\sqrt{x}+\dfrac{1}{2}\ge\dfrac{1}{2}\forall x\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{2}\right)^2\ge\dfrac{1}{4}\)
\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge1\Leftrightarrow\dfrac{2}{\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le2\)
Dấu bằng xảy ra \(\Leftrightarrow\sqrt{x}+\dfrac{1}{2}=\dfrac{1}{2}\\ \Leftrightarrow x=0\left(TM\right)\)
Vậy MaxC = 2 khi x = 0
Còn cái GTNN chưa tính ra được, để sau nha
Bài 4: ĐK: \(x\ge0,x\ne1\)
\(D=\left(\dfrac{2x+1}{\sqrt{x^3-1}}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\left(\dfrac{2x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{\left(1+\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\left(\dfrac{2x+1-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(x-\sqrt{x}+1-\sqrt{x}\right)\)
\(=\left(\dfrac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(x-2\sqrt{x}+1\right)\)
\(=\left(\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(\sqrt{x}-1\right)^2\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)}\)
\(=\sqrt{x}-1\)
\(D=3\Leftrightarrow\sqrt{x}-1=3\Leftrightarrow x=2\left(TM\right)\)
\(D=x-3\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}-1=\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(1-\sqrt{x}+2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(3-\sqrt{x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(L\right)\\x=9\left(TM\right)\end{matrix}\right.\)
Bài 5: \(E< -1\Leftrightarrow\dfrac{-3x}{2x+4\sqrt{x}}< -1\)\(\Leftrightarrow\dfrac{-3x}{2x+4\sqrt{x}}+1< 0\Leftrightarrow\dfrac{-3x+2x+4\sqrt{x}}{2x+4\sqrt{x}}< 0\)
\(\Leftrightarrow\dfrac{4\sqrt{x}-x}{2x+4\sqrt{x}}< 0\Leftrightarrow\dfrac{\sqrt{x}\left(4-\sqrt{x}\right)}{2x+4\sqrt{x}}< 0\)
Ta có: \(\sqrt{x}>0\Leftrightarrow x>0\Leftrightarrow2x+4\sqrt{x}>0\) mà \(\dfrac{\sqrt{x}\left(4-\sqrt{x}\right)}{2x+4\sqrt{x}}< 0\)\(\Rightarrow\sqrt{x}\left(4-\sqrt{x}\right)< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}< 0\left(L\right)\\4-\sqrt{x}>0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}>0\\4-\sqrt{x}< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x< 16,x\ne0\\\left\{{}\begin{matrix}x>0\\x< 16\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< 16,x\ne0\\0< x< 16\end{matrix}\right.\)
Câu 4:
Để C chia hết cho D thì \(x^4+a⋮x^2+4\)
\(\Leftrightarrow x^4-16+a+16⋮x^2+4\)
=>a+16=0
hay a=-16