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 \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)

Có điều kiện không bạn.

\(Q=\dfrac{x+4\sqrt{x}+20}{2\left(\sqrt{x}+2\right)}=\dfrac{x+4\sqrt{x}+4+16}{2\left(\sqrt{x}+2\right)}=\dfrac{\left(\sqrt{x}+2\right)^2+16}{2\left(\sqrt{x}+2\right)}\)

\(=\dfrac{1}{2}\left(\sqrt{x}+2\right)+\dfrac{16}{2\left(\sqrt{x}+2\right)}\ge2\sqrt{\dfrac{1}{2}\left(\sqrt{x}+2\right).\dfrac{16}{2\left(\sqrt{x}+2\right)}}\)

\(=2\sqrt{4}=4\)

\(\Rightarrow Q_{min}=4\) khi \(\dfrac{1}{2}\left(\sqrt{x}+2\right)=\dfrac{16}{2\left(\sqrt{x}+2\right)}\Rightarrow\left(\sqrt{x}+2\right)^2=16\)

mà \(\sqrt{x}+2>0\Rightarrow\sqrt{x}+2=4\Rightarrow x=4\)

\(y=x+\dfrac{1}{x}-5\ge2\sqrt{\dfrac{x}{x}}-5=-3\)

\(y_{min}=-3\) khi \(x=1\)

\(y=4x^2+\dfrac{1}{2x}+\dfrac{1}{2x}-4\ge3\sqrt[3]{\dfrac{4x^2}{2x.2x}}-4=-1\)

\(y_{min}=-1\) khi \(x=\dfrac{1}{2}\)

\(y=x+\dfrac{4}{x}\Rightarrow y'=1-\dfrac{4}{x^2}=0\Rightarrow x=-2\)

\(y\left(-2\right)=-4\Rightarrow\max\limits_{x>0}y=-4\) khi \(x=-2\)

\(\dfrac{x^3}{4\left(y+2\right)}+\dfrac{x\left(y+2\right)}{16}\ge\dfrac{x^2}{4}\) ; \(\dfrac{y^3}{4\left(x+2\right)}+\dfrac{y\left(x+2\right)}{16}\ge\dfrac{y^2}{4}\)

\(\Rightarrow Q+\dfrac{2xy+2x+2y}{16}\ge\dfrac{x^2+y^2}{4}\ge\dfrac{\left(x+y\right)^2}{8}\)

\(\Rightarrow Q\ge\dfrac{\left(x+y\right)^2-\left(x+y\right)}{8}-\dfrac{1}{2}=\dfrac{\left(x+y-4\right)^2+7\left(x+y\right)-16}{8}-\dfrac{1}{2}\)

\(\Rightarrow Q\ge\dfrac{7\left(x+y\right)-16}{8}-\dfrac{1}{2}\ge\dfrac{14\sqrt{xy}-16}{8}-\dfrac{1}{2}=1\)

\(Q_{min}=1\) khi \(x=y=2\)

a) x² - 2 = 0

x² = 2

x = -√2 (loại) hoặc x = √2 (loại)

Vậy không tìm được x Q thỏa mãn đề bài

b) x² + 7/4 = 23/4

x² = 23/4 - 7/4

x² = 4

x = 2 (nhận) hoặc x = -2 (nhận)

Vậy x = -2; x = 2

c) (x - 1)² = 0

x - 1 = 0

x = 1 (nhận)

Vậy x = 1

bạn linh này giỏi dữ ta

a) x² - 2 = 0

x² = 2

x = √2 hoặc -√2 (loại) (x ϵ Q)

Vậy x ϵ rỗng

b) x² + 7/4 =23/4

x² = 23/4 - 7/4

x² = 16/4 = 4

x² = 4 = (-2)² = 2²

x² = (-2)²

x = -2 (Nhận)

x² = 2

x = 2 (Nhận)

Vậy x ϵ ( 2 , -2 )

c) (x-1)² = 0

x-1 = 0

x = 1

Vậy x = 1