Ta có: x \(\le\) \(\dfrac{1}{4}\)

\(\Rightarrow\) \(\dfrac{1}{x}\ge4\)

Lại có: B = \(\dfrac{x+1}{x}=1+\dfrac{1}{x}\)

\(\Rightarrow\) 1 + \(\dfrac{1}{x}\) \(\ge\) 1 + 4 = 5

hay B \(\ge\) 5

Dấu "=" xảy ra khi và chỉ khi x = \(\dfrac{1}{4}\)

Chúc bn học tốt!

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 \(x=\dfrac{1}{2}\)

x=23x=23

2.

x=12

Ta có: \(\dfrac{a+b}{2}\ge\sqrt{ab}\)

\(\Leftrightarrow a+b\ge2\sqrt{ab}\)

\(\Leftrightarrow a-2\sqrt{ab}+b\ge0\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\)(luôn đúng)

*Chứng minh bất đẳng thức

Ta có: \(\forall a,b\ge0\) thì \(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\)

\(\Leftrightarrow a+b-2\sqrt{ab}\ge0\) \(\Leftrightarrow a+b\ge2\sqrt{ab}\) \(\Leftrightarrow\dfrac{a+b}{2}\ge\sqrt{ab}\)  (đpcm)

Ta có: \(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\forall a,b>0\)

\(\Leftrightarrow a-2\sqrt{ab}+b\ge0\forall a,b>0\)

\(\Leftrightarrow a+b\ge2\sqrt{ab}\forall a,b>0\)

\(\Leftrightarrow\dfrac{a+b}{2}\ge\sqrt{ab}\forall a,b>0\)(đpcm)

\(1,A=5^{n+2}+26\cdot5^n+8^{2n+1}\\ A=5^n\cdot25+26\cdot5^n+8\cdot8^{2n+1}\\ A=51\cdot5^n+8\cdot64^n\)

Ta có \(64:59R5\Rightarrow64^n:59R5\)

Vì vậy \(51\cdot5^n+8\cdot64^n:59R=5^n\cdot51+8\cdot5^n=5^n\left(51+8\right)=5^n\cdot59⋮59\)

Vậy \(A⋮59\)

(\(R\) là dư)

\(2,\\ a,2x\ge0;\left(x+2\right)^2\ge0,\forall x\\ \Leftrightarrow P=\dfrac{\left(x+2\right)^2}{2x}\ge0\\ P_{min}=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)

\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)

Lời giải:

$A=(x-y)+\frac{4}{x-y}+y+\frac{1}{y}$

Áp dụng BĐT Cô-si:

$(x-y)+\frac{4}{x-y}\geq 2\sqrt{(x-y).\frac{4}{x-y}}=4$
$y+\frac{1}{y}\geq 2$

$\Rightarrow A\geq 4+2=6$

Vậy $A_{\min}=6$ khi $(x,y)=(3,1)$

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 \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)

Ta có : \(\dfrac{a+b}{2}\ge\sqrt{ab}\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) ( luôn đúng )

Dấu "=" xảy ra khi \(a=b\)

Bài tập :

Có : \(A=\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{x}+\dfrac{x+y}{y}=2+\dfrac{x}{y}+\dfrac{y}{x}\) ( do \(x+y=1\) )

Theo BĐT trên có : \(\dfrac{x}{y}+\dfrac{y}{x}\ge2.\sqrt{\dfrac{x}{y}\cdot\dfrac{y}{x}}=2\)

Nên \(A\ge2+2=4\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)