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1, a,\(\left(-7x^2\right)\left(3x^2-x-2\right)\)
\(=-21x^4+7x^3+14x^2\)
\(b,\left(2x^3-3x^2-10x+3\right):\left(x-3\right)\)
2x^3-3x^2-10x+3 x-3 2x^2+3x-1 2x^3-6x^2 - 3x^2-10x+3 3x^2-9x - -x+3 -x+3 - 0
2,\(a,\left(x-3\right)\left(x^2+1\right)-\left(x-3\right)\left(x^2+3x+9\right)\)
\(=x^3+x-3x^2-3-x^3+27\)
\(=-3x^2+x+24\)
\(b,\left(2x+1\right)^2+\left(2x-1\right)^2+2\left(4x^2-1\right)\)
\(=4x^2+4x+1+4x^2-4x+1+8x^2-2\)
\(=24x^2\)
\(3,a,x^3-x^2-x+1\)
\(=x^2\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)^2\left(x+1\right)\)
\(b,3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(=3x\left(x+1\right)-10\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-10\right)\)
4, a. Bn kiểm tra lại đề bài nhé
b,\(4x^2-12xy+10y^2\)
\(=\left(4x^2-12xy+9y^2\right)+y^2\)
\(=\left(2x-3y\right)^2+y^2\ge0\forall x,y\)
1.
\(y\left(0\right)=-4\) ; \(y\left(5\right)=-4\) ; \(y\left(\frac{5}{3}\right)=\frac{392}{27}\)
\(\Rightarrow y_{max}=\frac{392}{27}\) khi \(x=\frac{5}{3}\)
2.
\(2x-1\ge0\Rightarrow x\ge\frac{1}{2}\)
\(3x+m\le0\Rightarrow x\le-\frac{m}{3}\)
Hệ có nghiệm khi \(-\frac{m}{3}\ge\frac{1}{2}\Rightarrow m\le-\frac{3}{2}\)
3.
\(P=a+b+\frac{1}{a}+\frac{1}{b}\ge a+b+\frac{4}{a+b}=a+b+\frac{1}{a+b}+\frac{3}{a+b}\)
\(P\ge2\sqrt{\frac{a+b}{a+b}}+\frac{3}{1}=5\)
\(P_{min}=5\) khi \(a=b=\frac{1}{2}\)
4.
\(y=2x+\frac{3}{x}\ge2\sqrt{\frac{6x}{x}}=2\sqrt{6}\)
Dấu "=" xảy ra khi \(2x=\frac{3}{x}\Leftrightarrow x=\sqrt{\frac{3}{2}}=\frac{\sqrt{6}}{2}\)
1.
\(6=\frac{\sqrt{2}^2}{x}+\frac{\sqrt{3}^2}{y}\ge\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{x+y}=\frac{5+2\sqrt{6}}{x+y}\)
\(\Rightarrow x+y\ge\frac{5+2\sqrt{6}}{6}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\frac{x}{\sqrt{2}}=\frac{y}{\sqrt{3}}\\x+y=\frac{5+2\sqrt{6}}{6}\end{matrix}\right.\)
Bạn tự giải hệ tìm điểm rơi nếu thích, số xấu quá
2.
\(VT\ge\sqrt{\left(x+y+z\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\ge\sqrt{\left(x+y+z\right)^2+\frac{81}{\left(x+y+z\right)^2}}\)
Đặt \(x+y+z=t\Rightarrow0< t\le1\)
\(VT\ge\sqrt{t^2+\frac{81}{t^2}}=\sqrt{t^2+\frac{1}{t^2}+\frac{80}{t^2}}\ge\sqrt{2\sqrt{\frac{t^2}{t^2}}+\frac{80}{1^2}}=\sqrt{82}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)
3.
\(\frac{a^2}{b^5}+\frac{a^2}{b^5}+\frac{a^2}{b^5}+\frac{1}{a^3}+\frac{1}{a^3}\ge5\sqrt[5]{\frac{a^6}{b^{15}.a^6}}=\frac{5}{b^3}\)
Tương tự: \(\frac{3b^2}{c^5}+\frac{2}{b^3}\ge\frac{5}{a^3}\) ; \(\frac{3c^2}{d^5}+\frac{2}{c^3}\ge\frac{5}{d^3}\) ; \(\frac{3d^2}{a^5}+\frac{2}{d^2}\ge\frac{5}{a^3}\)
Cộng vế với vế và rút gọn ta được: \(3VT\ge3VP\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c=d=1\)
4.
ĐKXĐ: \(-2\le x\le2\)
\(y^2=\left(x+\sqrt{4-x^2}\right)^2\le2\left(x^2+4-x^2\right)=8\)
\(\Rightarrow y\le2\sqrt{2}\Rightarrow y_{max}=2\sqrt{2}\) khi \(x=\sqrt{2}\)
Mặt khác do \(\left\{{}\begin{matrix}x\ge-2\\\sqrt{4-x^2}\ge0\end{matrix}\right.\) \(\Rightarrow x+\sqrt{4-x^2}\ge-2\)
\(y_{min}=-2\) khi \(x=-2\)
Bài 1:
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{a^2}{a+2b}+\frac{b^2}{2a+b}\geq \frac{(a+b)^2}{a+2b+2a+b}=\frac{(a+b)^2}{3(a+b)}=\frac{a+b}{3}=\frac{1}{3}\) (đpcm)
Dấu "=" xảy ra khi \(\left\{\begin{matrix} \frac{a}{a+2b}=\frac{b}{2a+b}\\ a+b=1\end{matrix}\right.\Leftrightarrow a=b=\frac{1}{2}\)
Bài 2:
Vì $x+y=2019$ nên $2019-x=y; 2019-y=x$
Áp dụng BĐT Cauchy-Schwarz ta có:
\(P=\frac{x}{\sqrt{2019-x}}+\frac{y}{\sqrt{2019-y}}=\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}=\frac{x^2}{x\sqrt{y}}+\frac{y^2}{y\sqrt{x}}\geq \frac{(x+y)^2}{x\sqrt{y}+y\sqrt{x}}\)
Mà theo BĐT AM-GM và Bunhiacopxky:
\((x\sqrt{y}+y\sqrt{x})^2\leq (xy+yx)(x+y)=2xy(x+y)\leq \frac{(x+y)^2}{2}.(x+y)=\frac{(x+y)^3}{2}\)
\(\Rightarrow P\geq \frac{(x+y)^2}{\sqrt{\frac{(x+y)^3}{2}}}=\sqrt{2(x+y)}=\sqrt{2.2019}=\sqrt{4038}\)
Vậy \(P_{\min}=\sqrt{4038}\Leftrightarrow x=y=\frac{2019}{2}\)