a) \(\left(x+\frac{7}{4}\right)\times\frac{3}{2}=6\)

\(\Leftrightarrow\left(x+\frac{7}{4}\right)=6\div\frac{3}{2}\)

\(\Leftrightarrow x+\frac{7}{4}=4\)

\(\Leftrightarrow x=4-\frac{7}{4}\)

\(\Leftrightarrow x=\frac{9}{4}\)

b) \(x\div\frac{3}{5}+\frac{2}{5}=\frac{9}{5}\)

\(\Leftrightarrow x\div\frac{3}{5}=\frac{9}{5}-\frac{2}{5}\)

\(\Leftrightarrow x\div\frac{3}{5}=\frac{7}{5}\)

\(\Leftrightarrow x=\frac{7}{5}\times\frac{3}{5}\)

\(\Leftrightarrow x=\frac{21}{25}\)

c) \(\frac{1}{2}\div3+x=\frac{5}{3}\)

\(\Leftrightarrow\frac{1}{6}+x=\frac{5}{3}\)

\(\Leftrightarrow x=\frac{5}{3}-\frac{1}{6}\)

\(\Leftrightarrow x=\frac{3}{2}\)

học đi

\(BĐT\Leftrightarrow\left(\frac{a+1}{a}\right)\left(\frac{b+1}{b}\right)\left(\frac{c+1}{c}\right)\ge64\)(*)

Mà \(\frac{a+1}{a}=\frac{\left(a+a\right)+\left(b+c\right)}{a}\ge\frac{2a+2\sqrt{bc}}{a}\ge\frac{2\sqrt{2a.2\sqrt{bc}}}{a}=\frac{4\sqrt{a\sqrt{bc}}}{a}\) (1)

Tương tự \(\frac{b+1}{b}\ge\frac{4\sqrt{b\sqrt{ac}}}{b}\) (2)  ;           \(\frac{c+1}{c}\ge\frac{4\sqrt{c\sqrt{ab}}}{c}\) (3)

Từ (1), (2) và (3) nhân vế theo vế ta được   (*) \(\ge\frac{4\sqrt{a\sqrt{bc}}.4\sqrt{b\sqrt{ac}}.4\sqrt{c\sqrt{ab}}}{abc}=\frac{64abc}{abc}=64\)

Dấu ''='' xảy ra khi \(\hept{\begin{cases}a+b+c=1\\1+\frac{1}{a}=1+\frac{1}{b}=1+\frac{1}{c}=4\end{cases}\Leftrightarrow a=b=c=\frac{1}{3}}\)

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Cách khác: Áp dụng BĐT AM-GM ta có: 

\(1+\frac{1}{a}=\frac{1}{a}\left(a+b+c+a\right)\ge\frac{1}{4}4\sqrt[4]{a^2bc}\)

\(\Rightarrow1+\frac{1}{a}\ge\frac{4}{a}\sqrt[4]{\frac{a^4bc}{a^2}}=4\sqrt[4]{\frac{bc}{a^2}}\)

Tương tự cũng có: \(1+\frac{1}{b}\ge4\sqrt[4]{\frac{ca}{b^2}};1+\frac{1}{c}\ge4\sqrt[4]{\frac{ab}{c^2}}\)

\(\Rightarrow VT\ge4\sqrt[4]{\frac{bc}{a^2}}4\sqrt[4]{\frac{ca}{b^2}}4\sqrt[4]{\frac{ab}{c^2}}=64\)

Còn tỷ tỷ cách đây cần thì IB nhé !!

Ta cần chứng minh \(\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge\left(1+\sqrt[3]{abc}\right)^3\)

\(\Leftrightarrow1+abc+ab+bc+ca+a+b+c\ge1+3\sqrt[3]{\left(abc\right)^2}+3\sqrt[3]{abc}+abc\)

\(\Leftrightarrow ab+bc+ca+a+b+c\ge3\sqrt[3]{\left(abc\right)^2}+3\sqrt[3]{abc}\)

Đúng theo BĐT AM-GM. Thật vậy ta có:

\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=\frac{\left(1+a\right)\left(1+b\right)\left(1+c\right)}{abc}\)

\(\ge\frac{\left(1+\sqrt[3]{abc}\right)^3}{abc}\ge64\).Từ \(a+b+c=1\Rightarrow abc\le\frac{1}{27}\)

\(\Rightarrow\frac{\left(1+\sqrt[3]{abc}\right)^3}{abc}=\left(\frac{1}{\sqrt[3]{abc}}+1\right)^3\ge64\)

Đẳng thức xảy ra khi a=b=c=1/3

a, \(A=\left(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{a+2}{a-2}\) \(\left(a>0;a\ne2\right)\)

\(=\left[\frac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right]:\frac{a+2}{a-2}\)

\(=\frac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}.\frac{a-2}{a+2}\)

\(=\frac{2\sqrt{a}}{\sqrt{a}}.\frac{a-2}{a+2}\)

\(=\frac{2\left(a-2\right)}{a+2}\)

b, Để: \(A=1\Leftrightarrow\frac{2\left(a-2\right)}{a+2}=1\)

\(\Rightarrow\frac{2a-4-a-2}{a+2}=0\)

\(\Rightarrow\frac{a-6}{a+2}=0\)

\(\Rightarrow a-6=0\)

\(\Rightarrow a=6\left(tm\right)\)

Vậy...........................

Ta cần chứng minh \((1+a)(1+b)(1+c) \geq (1+\sqrt[3]{abc})^3\)

\(\Leftrightarrow 1+abc+ab+bc+ca+a+b+c \geq 1+3\sqrt[3]{(abc)^2}+3\sqrt[3]{abc}+abc\)

\(\Leftrightarrow ab+bc+ca+a+b+c \geq 3\sqrt[3]{(abc)^2}+3\sqrt[3]{abc}\)

Đúng theo BĐT AM-GM. Áp dụng vào ta có:

\(\left(1+\frac{1}{a} \right)\left(1+\frac{1}{b} \right)\left(1+\frac{1}{c} \right)=\dfrac{(1+a)(1+b)(1+c)}{abc} \geq \dfrac{(1+\sqrt[3]{abc})^3}{abc} \geq 64\)
Từ \(a+b+c=1 \Rightarrow abc\le \frac{1}{27}\) \(\Rightarrow \dfrac{(1+\sqrt[3]{abc})^3}{abc}=\bigg(\dfrac{1}{\sqrt[3]{abc}}+1\bigg)^3 \geq 64\)

Đẳng thức xảy ra khi \(a=b=c=\frac{1}{3}\)

có tất cả loại cách từ cấp 2 đến cấp 3 cần thêm cứ bảo

\(VT\ge4\frac{\sqrt[4]{bc}}{\sqrt{a}}.4\frac{\sqrt[4]{ca}}{\sqrt{b}}.4\frac{\sqrt[4]{ab}}{\sqrt{c}}=64\)

Áp dụng bất đẳng thức AM-GM 3 số không âm :

\(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge3\sqrt[3]{\frac{abc}{abc}}=3\sqrt[3]{1}=3\)

Dấu "=" xảy ra \(\Leftrightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\Leftrightarrow a=b=c\)

MInh cam on nhe!

Em có cách này nhưng không chắc. Hình như tên là bán Schur-bán SOS thì phải ạ!

\(f\left(a;b;c\right)=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3=\frac{a}{b}+\frac{b}{a}-2+\frac{b}{c}-\frac{b-c}{a}-1\)

\(=\frac{\left(a-b\right)^2}{ab}+\left(\frac{b}{c}-1\right)-\frac{\left(b-c\right)}{a}\)

\(=\frac{\left(a-b\right)^2}{ab}+\frac{b-c}{c}-\frac{b-c}{a}=\frac{\left(a-b\right)^2}{ab}+\frac{\left(b-c\right)\left(a-c\right)}{ca}\)

Do a,b,c có tính chất hoán vị vòng quanh nên ta giả sử c là số nhỏ nhất (c = min{a;b;c} )suy ra f(a;b;c) \(\ge0\)

Từ đó suy ra \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge3^{\left(đpcm\right)}\)

@Nguyễn Việt Lâm

https://hoc24.vn/id/2782086

Ta có đánh giá \(\frac{b+2}{\left(b+1\right)\left(b+5\right)}\ge\frac{3}{4\left(b+2\right)}\)

Thật vậy, BĐT trên tương đương:

\(4\left(b+2\right)^2\ge3\left(b+1\right)\left(b+5\right)\)

\(\Leftrightarrow b^2-2b+1\ge0\Leftrightarrow\left(b-1\right)^2\ge0\) (luôn đúng)

\(\Rightarrow\frac{\left(a+1\right)\left(b+2\right)}{\left(b+1\right)\left(b+5\right)}\ge\frac{3\left(a+1\right)}{4\left(b+2\right)}\)

Tương tự và cộng lại: \(P\ge\frac{3}{4}\left(\frac{a+1}{b+2}+\frac{b+1}{c+2}+\frac{c+1}{a+2}\right)\)

\(P\ge\frac{3}{4}\left(\frac{\left(a+1\right)^2}{ab+2a+b+2}+\frac{\left(b+1\right)^2}{bc+2b+c+2}+\frac{\left(c+1\right)^2}{ca+2c+a+2}\right)\)

\(P\ge\frac{3}{4}.\frac{\left(a+b+c+3\right)^2}{ab+bc+ca+3a+3b+3c+6}\)

\(P\ge\frac{3}{4}.\frac{a^2+b^2+c^2+2ab+2bc+2ca+6a+6b+6c+9}{ab+bc+ca+3a+3b+3c+6}\)

\(P\ge\frac{3}{4}.\frac{2ab+2bc+2ca+6a+6b+6c+12}{ab+bc+ca+3a+3b+3c+6}=\frac{3}{4}.2=\frac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)