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1.
\(f\left(x\right)=\dfrac{4}{x}+\dfrac{x-1+1}{1-x}=\dfrac{2^2}{x}+\dfrac{1}{1-x}-1\ge\dfrac{\left(2+1\right)^2}{x+1-x}-1=8\)
\(f\left(x\right)_{min}=8\) khi \(x=\dfrac{2}{3}\)
2.
\(f\left(x\right)=\dfrac{1}{x}+\dfrac{1}{1-x}\ge\dfrac{4}{x+1-x}=4\)
\(f\left(x\right)_{min}=4\) khi \(x=\dfrac{1}{2}\)
f(x)=4x+x−1+11−x=22x+11−x−1≥(2+1)2x+1−x−1=8f(x)=4x+x−1+11−x=22x+11−x−1≥(2+1)2x+1−x−1=8
f(x)min=8f(x)min=8 khi x=23x=23
2.
f(x)=1x+11−x≥4x+1−x=4f(x)=1x+11−x≥4x+1−x=4
f(x)min=4f(x)min=4 khi x=12
\(P=\dfrac{x}{\sqrt{y}}+\dfrac{y}{\sqrt{x}}\Rightarrow P^2=\dfrac{x^2}{y}+\dfrac{y^2}{x}+2\sqrt{xy}\)
\(P^2=\left(\dfrac{x^2}{y}+\sqrt{xy}+\sqrt{xy}\right)+\left(\dfrac{y^2}{x}+\sqrt{xy}+\sqrt{xy}\right)-2\sqrt{xy}\)
\(P^2\ge3x+3y-2\sqrt{xy}\ge3\left(x+y\right)-\left(x+y\right)=2\left(x+y\right)=4038\)
\(\Rightarrow P\ge\sqrt{4038}\)
Dấu "=" xảy ra khi \(x=y=\dfrac{2019}{2}\)
Ta có:
\(P=\dfrac{x}{\sqrt{2019-x}}+\dfrac{y}{\sqrt{y-2019}}=\dfrac{x}{\sqrt{y}}+\dfrac{y}{\sqrt{x}}\ge\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2}{\sqrt{x}+\sqrt{y}}=\sqrt{x}+\sqrt{y}\)
Lại có:
\(P=\dfrac{x}{\sqrt{2019-x}}+\dfrac{y}{\sqrt{2019-y}}=\dfrac{2019-y}{\sqrt{y}}+\dfrac{2019-x}{\sqrt{x}}\\ =\dfrac{2019}{\sqrt{x}}+\dfrac{2019}{\sqrt{y}}-\sqrt{x}-\sqrt{y}\)
\(\Rightarrow2P=\dfrac{2019}{\sqrt{x}}+\dfrac{2019}{\sqrt{y}}=2019\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)\ge2019\cdot\dfrac{2}{\sqrt[4]{xy}}\\ \ge2019\dfrac{2}{\sqrt[2]{\dfrac{x+y}{2}}}=2019\cdot\dfrac{2}{\sqrt{\dfrac{2019}{2}}}=2\sqrt{2}\sqrt{2019}\)
\(\Rightarrow P\ge\sqrt{2}\sqrt{2019}\)
Dấu = khi \(x=y=\dfrac{2019}{2}\)
1) Áp dụng bđt Cauchy cho 3 số dương ta có
\(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}+x^3\ge4\sqrt[4]{\dfrac{1}{x}.\dfrac{1}{x}.\dfrac{1}{x}.x^3}=4\) (1)
\(\dfrac{3}{y^2}+y^2\ge2\sqrt{\dfrac{3}{y^2}.y^2}=2\sqrt{3}\) (2)
\(\dfrac{3}{z^3}+z=\dfrac{3}{z^3}+\dfrac{z}{3}+\dfrac{z}{3}+\dfrac{z}{3}\ge4\sqrt[4]{\dfrac{3}{z^3}.\dfrac{z}{3}.\dfrac{z}{3}.\dfrac{z}{3}}=4\sqrt{3}\) (3)
Cộng (1);(2);(3) theo vế ta được
\(\left(\dfrac{3}{x}+\dfrac{3}{y^2}+\dfrac{3}{z^3}\right)+\left(x^3+y^2+z\right)\ge4+2\sqrt{3}+4\sqrt{3}\)
\(\Leftrightarrow3\left(\dfrac{1}{x}+\dfrac{1}{y^2}+\dfrac{1}{z^3}\right)\ge3+4\sqrt{3}\)
\(\Leftrightarrow P\ge\dfrac{3+4\sqrt{3}}{3}\)
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{x}=x^3\\\dfrac{3}{y^2}=y^2\\\dfrac{3}{z^3}=\dfrac{z}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\sqrt[4]{3}\\z=\sqrt{3}\end{matrix}\right.\) (thỏa mãn giả thiết ban đầu)
2) Ta có \(4\sqrt{ab}=2.\sqrt{a}.2\sqrt{b}\le a+4b\)
Dấu"=" khi a = 4b
nên \(\dfrac{8}{7a+4b+4\sqrt{ab}}\ge\dfrac{8}{7a+4b+a+4b}=\dfrac{1}{a+b}\)
Khi đó \(P\ge\dfrac{1}{a+b}-\dfrac{1}{\sqrt{a+b}}+\sqrt{a+b}\)
Đặt \(\sqrt{a+b}=t>0\) ta được
\(P\ge\dfrac{1}{t^2}-\dfrac{1}{t}+t=\left(\dfrac{1}{t^2}-\dfrac{2}{t}+1\right)+\dfrac{1}{t}+t-1\)
\(=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\)
Có \(\dfrac{1}{t}+t\ge2\sqrt{\dfrac{1}{t}.t}=2\) (BĐT Cauchy cho 2 số dương)
nên \(P=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\ge\left(\dfrac{1}{t}-1\right)^2+1\ge1\)
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{t}-1=0\\t=\dfrac{1}{t}\end{matrix}\right.\Leftrightarrow t=1\)(tm)
khi đó a + b = 1
mà a = 4b nên \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)
Vậy MinP = 1 khi \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)
\(Q=x^2+\dfrac{4}{x}+\dfrac{4}{x}\ge3\sqrt[3]{\dfrac{16x^2}{x^2}}=3\sqrt[3]{16}\)
Dấu "=" xảy ra khi \(x=\sqrt[3]{4}\)
1.
Đặt \(x^2-2x+m=t\), phương trình trở thành \(t^2-2t+m=x\)
Ta có hệ \(\left\{{}\begin{matrix}x^2-2x+m=t\\t^2-2t+m=x\end{matrix}\right.\)
\(\Rightarrow\left(x-t\right)\left(x+t-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=t\\x=1-t\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=x^2-2x+m\\x=1-x^2+2x-m\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}m=-x^2+3x\\m=-x^2+x+1\end{matrix}\right.\)
Phương trình hoành độ giao điểm của \(y=-x^2+x+1\) và \(y=-x^2+3x\):
\(-x^2+x+1=-x^2+3x\)
\(\Leftrightarrow x=\dfrac{1}{2}\Rightarrow y=\dfrac{5}{4}\)
Đồ thị hàm số \(y=-x^2+3x\) và \(y=-x^2+x+1\):
Dựa vào đồ thị, yêu cầu bài toán thỏa mãn khi \(m< \dfrac{5}{4}\)
Mà \(m\in\left[-10;10\right]\Rightarrow m\in[-10;\dfrac{5}{4})\)
Có cách nào lm bài này bằng cách lập bảng biến thiên k ạ
\(\Rightarrow f\left(x\right)=\dfrac{7}{4}x+\dfrac{1}{8}x+\dfrac{1}{8}x+\dfrac{8}{x^2}\)
Áp dụng bđt Cô-si :
\(\dfrac{1}{8}x+\dfrac{1}{8}x+\dfrac{8}{x^2}\ge3\sqrt[3]{\dfrac{1}{8}x\cdot\dfrac{1}{8}x\cdot\dfrac{8}{x^2}}=\dfrac{3}{2}\)
\(\Rightarrow f\left(x\right)=\dfrac{7}{4}x+\dfrac{1}{8}x+\dfrac{1}{8}x+\dfrac{8}{x^2}\ge7+\dfrac{3}{2}=\dfrac{17}{2}\)
Dấu bằng xảy ra \(\Leftrightarrow x=4\)
\(f\left(x\right)=\dfrac{x}{8}+\dfrac{x}{8}+\dfrac{8}{x^2}+\dfrac{7}{4}x\ge3\sqrt[3]{\dfrac{8x^2}{64x^2}}+\dfrac{7}{4}.4=\dfrac{17}{2}\)
Dấu "=" xảy ra khi \(x=4\)
c) \(h\left(x\right)=\left(x+1\right)^2+\left(\dfrac{x^2+2x+2}{x+1}\right)^2=\left(x+1\right)^2+\left(x+1+\dfrac{1}{x+1}\right)^2=2\left(x+1\right)^2+\dfrac{1}{\left(x+1\right)^2}+2\ge_{AM-GM}2\sqrt{2}+2\).
Đẳng thức xảy ra khi \(2\left(x+1\right)^2=\dfrac{1}{\left(x+1\right)^2}\Leftrightarrow x=\pm\sqrt{\dfrac{1}{2}}-1\).
b) \(g\left(x\right)=\dfrac{\left(x+2\right)\left(x+3\right)}{x}=\dfrac{x^2+5x+6}{x}=\left(x+\dfrac{6}{x}\right)+5\ge_{AM-GM}2\sqrt{6}+5\).
Đẳng thức xảy ra khi x = \(\sqrt{6}\).
2.
b, \(-4< \dfrac{2x^2+mx-4}{-x^2+x-1}< 6\)
\(\Leftrightarrow\left\{{}\begin{matrix}-4< \dfrac{2x^2+mx-4}{-x^2+x-1}\left(1\right)\\\dfrac{2x^2+mx-4}{-x^2+x-1}< 6\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow4\left(x^2-x+1\right)>2x^2+mx-4\)
\(\Leftrightarrow2x^2-\left(m+4\right)x+8>0\)
Yêu cầu bài toán thỏa mãn khi \(\Delta=m^2+8m-48< 0\Leftrightarrow-12< m< 4\)
\(\left(2\right)\Leftrightarrow-6\left(x^2-x+1\right)< 2x^2+mx-4\)
\(\Leftrightarrow8x^2+\left(m-6\right)x+2>0\)
Yêu cầu bài toán thỏa mãn khi \(\Delta=m^2-12m-28< 0\Leftrightarrow-2< x< 14\)
Vậy \(m\in\left(-2;4\right)\)
2.
a, Yêu cầu bài toán thỏa mãn khi phương trình \(\left(m-4\right)x^2+\left(1+m\right)x+2m-1>0\) có nghiệm đúng với mọi x
\(\Leftrightarrow\left\{{}\begin{matrix}m-4>0\\\Delta=m^2+2m+1-4\left(m-4\right)\left(2m-1\right)< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m>4\\\left[{}\begin{matrix}m< \dfrac{3}{7}\\m>5\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow m>5\)
Có điều kiện không bạn.