rất tiếc em mới học lớp 6

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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: 3x + y = 1 => y = 1 - 3x

a, Thay y = 1 - 3x vào M, ta có:

\(\Rightarrow M=3x^2+\left(1-3x\right)^2=3x^2+1-6x+9x^2=12x^2-6x+1=3\left(4x^2-2x+\frac{1}{3}\right)\)

\(=3\left(4x^2-2x+\frac{1}{4}+\frac{1}{12}\right)=3\left(2x-\frac{1}{2}\right)^2+\frac{3}{12}=3\left(2x-\frac{1}{2}\right)^2+\frac{1}{4}\)

Vì \(\left(2x-\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow3\left(2x-\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow3\left(2x-\frac{1}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\forall x\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}2x-\frac{1}{2}=0\\3x+y=1\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\y=1-3x=1-3.\frac{1}{4}=\frac{1}{4}\end{cases}}\)\(\Leftrightarrow x=y=\frac{1}{4}\)

Vậy GTNN M = 1/4 khi x = y = 1/4

b, Thay y = 1 - 3x vào N

\(\Rightarrow N=x\left(1-3x\right)=x-3x^2=-3\left(x^2-\frac{x}{3}+\frac{1}{36}-\frac{1}{36}\right)\)

\(=-3\left(x-\frac{1}{6}\right)^2-3.\left(-\frac{1}{36}\right)=-3\left(x-\frac{1}{6}\right)^2+\frac{1}{12}\)

Vì \(\left(x-\frac{1}{6}\right)^2\ge0\forall x\)

\(\Rightarrow-3\left(x-\frac{1}{6}\right)^2\le0\forall x\)

\(\Rightarrow-3\left(x-\frac{1}{6}\right)^2+\frac{1}{12}\le\frac{1}{12}\forall x\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-\frac{1}{6}=0\\3x+y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{6}\\y=1-3x=1-3.\frac{1}{6}=\frac{1}{2}\end{cases}}\)

Vậy GTLN N = 1/12 khi x = 1/6 và y = 1/2

We have : \(A=x+y+\dfrac{1}{2x}+\dfrac{2}{y}=\dfrac{x+y}{2}+\left(\dfrac{y}{2}+\dfrac{2}{y}\right)+\left(\dfrac{1}{2x}+\dfrac{x}{2}\right)\)

\(Applying\) C-S we have : \(\dfrac{y}{2}+\dfrac{2}{y}\ge2;\dfrac{1}{2x}+\dfrac{x}{2}\ge1\)

x + y \(\ge3\)  \(\Rightarrow\dfrac{x+y}{2}\ge\dfrac{3}{2}\)

So : \(A\ge\dfrac{3}{2}+2+1=\dfrac{9}{2}\)

" = " \(\Leftrightarrow x=1;y=2\)

bạn họ cái này chưa?

/a+b/=</a/+/b/

Vì /y-2/=/2-y/( cái này thực ra là không cần đâu nhưng mk diễn giải ra cho bn dễ hiểu)

A=/x+1/+/y-2/=/x+1/+/2-y/>=/x-y+1+2/=/3+3/=6

Dấu '=' xảy ra khi (x+1)(2-y)>=0 (1)

x-y=3 => x=3+y

=> (1) <=> (y+4)(2-y)>=0

=> -4=<y=<2

y=x-3 =>(1)<=>(x+1)(5-x)>=0

                     =>-1=<x=<5