Lời giải:

Với điều kiện đã cho thì hiển nhiên mẫu dương.

Áp dụng BĐT Cauchy-Schwarz ta có:

\(M=\frac{a^2}{2a\sqrt{b}-3a}+\frac{b^2}{2b\sqrt{c}-3b}+\frac{c^2}{2c\sqrt{a}-3c}\)\(\geq \frac{(a+b+c)^2}{2(a\sqrt{b}+b\sqrt{c}+c\sqrt{a})-3(a+b+c)}\)

Áp dụng BĐT Bunhiacopxky kết hợp BĐT AM-GM:

\((a\sqrt{b}+b\sqrt{c}+c\sqrt{a})^2\leq (a+b+c)(ab+bc+ac)\)

\(\leq (a+b+c).\frac{(a+b+c)^2}{3}=\frac{(a+b+c)^3}{3}\)

\(\Rightarrow a\sqrt{b}+b\sqrt{c}+c\sqrt{a}\leq \sqrt{\frac{(a+b+c)^3}{3}}\)

\(\Rightarrow M\geq \frac{(a+b+c)^2}{2\sqrt{\frac{(a+b+c)^3}{3}}-3(a+b+c)}\)

Đặt \(\sqrt{\frac{a+b+c}{3}}=t(t>\frac{3}{2})\)\(\Rightarrow a+b+c=3t^2\)

Ta có:

\(P\geq\frac{9t^4}{6t^3-9t^2}=\frac{3t^2}{2t-3}\)

\(\Leftrightarrow P\geq \frac{\frac{3}{4}(2t-3)(2t+3)}{2t-3}+\frac{27}{4(2t-3)}\)

\(\Leftrightarrow P\geq \frac{3}{4}(2t+3)+\frac{27}{4(2t-3)}=\frac{3}{4}(2t-3)+\frac{27}{4(2t-3)}+\frac{9}{2}\)

Áp dụng BĐT AM-GM:

\(\frac{3}{4}(2t-3)+\frac{27}{4(2t-3)}\geq 2\sqrt{\frac{3}{4}.\frac{27}{4}}=\frac{9}{2}\)

\(\Rightarrow P\geq \frac{9}{2}+\frac{9}{2}=9\)

Vậy \(P_{\min}=9\)

Đặt \(\left\{{}\begin{matrix}\sqrt{a}=x\\\sqrt{b}=y\\\sqrt{c}=z\end{matrix}\right.\)

\(\Rightarrow P=\dfrac{x^2}{2y-3}+\dfrac{y^2}{2z-3}+\dfrac{z^2}{2x-3}\)

\(\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)-9}\ge9\)

Vì \(\dfrac{t^2}{2t-9}-9=\dfrac{\left(t-9\right)^2}{2t-9}\ge0\) (với \(t=x+y+z\))

# cách khác:

Áp dụng AM-GM: \(\dfrac{a}{2\sqrt{b}-3}+\left(2\sqrt{b}-3\right)\ge2\sqrt{a}\)

Thiết lập tương tự rồi cộng lại ta được

\(VT+2\sqrt{a}+2\sqrt{b}+2\sqrt{c}-9\ge2\sqrt{a}+2\sqrt{b}+2\sqrt{c}\)

\(\Rightarrow VT\ge9\)

Dấu = xảy ra tại a=b=c=9

A=-3; B=

1) Thay x=1x=1 vào biểu thức: A=√1+2√1−2A=1+21−2
A=−3A=−3
2) Chứng minh B=√x√x+2B=xx+2 với x≥0,x≠4x≥0,x≠4.
B=√x+2(√x−2)(√x+2)+(√x+1)(√x−2)(√x+2)(√x−2)−2√x(√x+2)(√x−2)B=x+2(x−2)(x+2)+(x+1)(x−2)(x+2)(x−2)−2x(x+2)(x−2)
=√x+2+x−√x−2−2√x(√x+2)(√x−2)=x−2√x(√x+2)(√x−2)=x+2+x−x−2−2x(x+2)(x−2)=x−2x(x+2)(x−2)
=√x(√x−2)(√x+2)(√x−2)=√x√x+2=x(x−2)(x+2)(x−2)=xx+2
3) Tìm xx để A⋅B≥0A⋅B≥0
A⋅B=√x+2√x−2⋅√x√x+2=√x√x−2A⋅B=x+2x−2⋅xx+2=xx−2.
TH1: x=0⇒√x=0⇒A⋅B=0x=0⇒x=0⇒A⋅B=0 (TM)
TH2: x>0⇒√x>0⇒√x−2>0⇒x>4x>0⇒x>0⇒x−2>0⇒x>4
Kết hợp điêu kiện: x=0x=0 hoặc x>4x>4 thỏa mãn yêu cầu.

x

các bạn ơi giúp mình với

a: \(=ab+2\cdot\sqrt{\dfrac{b}{a}\cdot ab}-\sqrt{ab\cdot\left(\dfrac{a}{b}+\dfrac{1}{\sqrt{ab}}\right)}\)

\(=ab+2b-\sqrt{ab\cdot\dfrac{a\sqrt{a}+\sqrt{b}}{b\sqrt{a}}}\)

\(=ab+2b-\sqrt{\sqrt{a}\cdot\left(a\sqrt{a}+\sqrt{b}\right)}\)

b: \(=\left(\sqrt{\dfrac{a^2m^2\cdot n}{b^2\cdot m}}-\sqrt{mn\cdot\dfrac{a^2b^2}{n^2}}+\sqrt{\dfrac{a^4}{b^4}\cdot\dfrac{m}{n}}\right)\cdot a^2b^2\cdot\sqrt{\dfrac{n}{m}}\)

\(=\left(\dfrac{a\sqrt{mn}}{b}-\sqrt{a^2b^2\cdot\dfrac{m}{n}}+\dfrac{a^2}{b^2}\cdot\sqrt{\dfrac{m}{n}}\right)\cdot\sqrt{\dfrac{n}{m}}\cdot a^2b^2\)

\(=\left(\dfrac{an}{b}-ab+\dfrac{a^2}{b^2}\right)\cdot a^2b^2\)

\(=a^3nb-a^3b^3+a^4\)

Lời giải:

Hiển nhiên $a-b>0$.

Ta có:

\(P=\sqrt{ab}.\sqrt{ab}+\frac{a-b}{\sqrt{ab}}=\sqrt{ab}.\frac{a+b}{a-b}+\frac{a-b}{\sqrt{ab}}\geq 2\sqrt{a+b}\) theo BĐT AM-GM.

Mặt khác:

Từ ĐKĐB suy ra \(ab(a-b)^2=(a+b)^2\)

\(\Leftrightarrow ab[(a+b)^2-4ab]=(a+b)^2\)

Đặt $a+b=x; ab=y$ với $x,y>0; x^2\geq 4y$ thì:

\(y(x^2-4y)=x^2\Leftrightarrow x^2(y-1)=4y^2\)

Hiển nhiên $y>1$

$\Rightarrow x^2=\frac{4y^2}{y-1}=\frac{4(y^2-1)}{y-1}+\frac{4}{y-1}$

$=4(y+1)+\frac{4}{y-1}=4(y-1)+\frac{4}{y-1}+8$

$\geq 2\sqrt{4(y-1).\frac{4}{y-1}}+8=16$ (AM-GM)

$\Rightarrow x\geq 4$ hay $a+b\geq 4$

Do đó: $P\geq 2\sqrt{a+b}\geq 2\sqrt{4}=4$

Vậy $P_{\min}=4$
Giá trị này đạt tại $(a,b)=(2+\sqrt{2}, 2-\sqrt{2})$

ta có \(\sqrt{ab}=\dfrac{a+b}{a-b}=>ab=\dfrac{\left(a+b\right)^2}{\left(a-b\right)^2}\)

=>P=\(ab+\dfrac{a-b}{\sqrt{ab}}=\dfrac{\left(a+b\right)^2}{\left(a-b\right)^2}+\dfrac{a-b}{\dfrac{a+b}{a-b}}=\dfrac{\left(a+b\right)^2}{\left(a-b\right)^2}+\dfrac{\left(a-b\right)^2}{a+b}\)

áp dụng BDT AM-GM ta có \(\dfrac{\left(a+b\right)^2}{\left(a-b\right)^2}+\dfrac{\left(a-b\right)^2}{a+b}\ge\sqrt{\dfrac{\left(a+b\right)^2}{\left(a-b\right)^2}.\dfrac{\left(a-b\right)^2}{a+b}}=2\sqrt{a+b}\left(1\right)\)

lại có \(\sqrt{ab}=\dfrac{a+b}{a-b}=>a+b=\sqrt{ab}.\left(a-b\right)=>2.\left(a+b\right)=2.\sqrt{ab}.\left(a-b\right)\)

áp dụng BDT AM-GM ta được \(2\left(a+b\right)=2.\sqrt{ab}.\left(a-b\right)\le\dfrac{\left(2\sqrt{ab}\right)^2+\left(a-b\right)^2}{2}=\dfrac{4ab+a^2-2ab+b^2}{2}\)

=\(\dfrac{\left(a+b\right)^2}{2}\)

=>\(2\left(a+b\right)\le\dfrac{\left(a+b\right)^2}{2}=>a+b\ge4\left(2\right)\)

từ (1)(2)=>\(\dfrac{\left(a+b\right)^2}{\left(a-b\right)^2}+\dfrac{\left(a-b\right)^2}{a+b}\ge2\sqrt{a+b}\ge4\)

dấu '=' xảy ra \(\Leftrightarrow\)a=2\(+\sqrt{2}\), b=\(2-\sqrt{2}\)

vậy MIn P=4 khi (a,b)=(2+\(\sqrt{2};2-\sqrt{2}\))

Bài 1: 

a: \(=\dfrac{1}{mn^2}\cdot\dfrac{n^2\cdot\left(-m\right)}{\sqrt{5}}=\dfrac{-\sqrt{5}}{5}\)

b: \(=\dfrac{m^2}{\left|2m-3\right|}=\dfrac{m^2}{3-2m}\)

c: \(=\left(\sqrt{a}+1\right):\dfrac{\left(a-1\right)^2}{\left(1-\sqrt{a}\right)}=\dfrac{-\left(a-1\right)}{\left(a-1\right)^2}=\dfrac{-1}{a-1}\)

Áp dụng  \(x^2+y^2+z^2\ge xy+yz+zx\) và \(x^2+y^2+z^2\ge\dfrac{1}{3}\left(x+y+z\right)^2\)

\(N\ge\dfrac{a^2b}{c}+\dfrac{b^2c}{a}+\dfrac{c^2a}{b}\ge\dfrac{1}{3}\left(a\sqrt{\dfrac{b}{c}}+b\sqrt{\dfrac{c}{a}}+c\sqrt{\dfrac{a}{b}}\right)^2=3\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

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 \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)