5.

\(\sqrt{\left(a+b\right)\left(c+a\right)}\ge\sqrt{\left(\sqrt{ac}+\sqrt{ab}\right)^2}=\sqrt{ac}+\sqrt{ab}\)

\(\Rightarrow\dfrac{a}{a+\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{a}{a+\sqrt{ac}+\sqrt{ab}}=\dfrac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

Tương tự:

\(\dfrac{b}{b+\sqrt{\left(a+b\right)\left(b+c\right)}}\le\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

\(\dfrac{c}{c+\sqrt{\left(a+c\right)\left(b+c\right)}}\le\dfrac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

Cộng vế với vế:

\(P\le\dfrac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1\)

Dấu "=" xảy ra khi \(a=b=c\)

6.

\(P=\dfrac{a}{1+b-a}+\dfrac{b}{1+c-b}+\dfrac{c}{1+a-c}\)

Thay \(1=a+b+c\)

\(\Rightarrow P=\dfrac{a}{2b+c}+\dfrac{b}{2c+a}+\dfrac{c}{2a+b}\)

\(P=\dfrac{a^2}{2ab+ac}+\dfrac{b^2}{2bc+ab}+\dfrac{c^2}{2ac+bc}\)

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

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)

Thầy Nguyễn Việt Lâm ơi giúp em mấy bài này với.Em sắp phải nộp rồi ạ  - Hoc24

4.

\(ab+bc+ca=3abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)

Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=3\)

\(S=\sum\dfrac{\dfrac{1}{y^2}}{\dfrac{1}{x}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)}=\sum\dfrac{x^3}{x^2+y^2}=\sum\left(x-\dfrac{xy^2}{x^2+y^2}\right)\)

\(S\ge\sum\left(x-\dfrac{xy^2}{2xy}\right)=\sum\left(x-\dfrac{y}{2}\right)=\dfrac{x+y+z}{2}=\dfrac{3}{2}\)

\(S_{min}=\dfrac{3}{2}\) khi \(x=y=z=1\) hay \(a=b=c=1\)

5.

Đặt \(\left(\dfrac{1}{a};\dfrac{2}{b};\dfrac{3}{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=3\)

Đặt vế trái là P

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

Quay lại dòng 3 của bài số 4

1.

\(\left(x+y\right)^2=\left(\dfrac{1}{2}.2x+\dfrac{1}{3}.3y\right)^2\le\left(\dfrac{1}{4}+\dfrac{1}{9}\right)\left(4x^2+9y^2\right)=\dfrac{169}{36}\)

\(\Rightarrow-\dfrac{13}{6}\le x+y\le\dfrac{13}{6}\)

Dấu "=" lần lượt xảy ra tại \(\left(-\dfrac{3}{2};-\dfrac{2}{3}\right)\) và \(\left(\dfrac{3}{2};\dfrac{2}{3}\right)\)

2.

\(\left(y-2x\right)^2=\left(\dfrac{1}{4}.4y+\left(-\dfrac{1}{3}\right).6x\right)^2\le\left(\dfrac{1}{16}+\dfrac{1}{9}\right)\left(16y^2+36x^2\right)=\dfrac{25}{16}\)

\(\Rightarrow\left|y-2x\right|\le\dfrac{5}{4}\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(\mp\dfrac{2}{5};\pm\dfrac{9}{20}\right)\)

3.

\(B^2=\left(6.\sqrt{x-1}+8\sqrt{3-x}\right)^2\le\left(6^2+8^2\right)\left(x-1+3-x\right)=200\)

\(\Rightarrow B\le2\sqrt{10}\)

Dấu "=" xảy ra khi \(\dfrac{\sqrt{x-1}}{6}=\dfrac{\sqrt{3-x}}{8}\Leftrightarrow x=\dfrac{43}{25}\)

\(B=6\sqrt{x-1}+6\sqrt{3-x}+2\sqrt{3-x}\ge6\sqrt{x-1}+6\sqrt{3-x}\)

\(B\ge6\left(\sqrt{x-1}+\sqrt{3-x}\right)\ge6\sqrt{x-1+3-x}=6\sqrt{2}\)

\(B_{min}=6\sqrt{2}\) khi \(\sqrt{3-x}=0\Rightarrow x=3\)

4.

\(49=\left(3a+4b\right)^2=\left(\sqrt{3}.\sqrt{3}a+2.2b\right)^2\le\left(3+4\right)\left(3a^2+4b^2\right)\)

\(\Rightarrow3a^2+4b^2\ge\dfrac{49}{7}=7\)

Dấu "=" xảy ra khi \(a=b=1\)

Đoàn Gia Khánh câu hỏi đâu?

Chỉ muốn biết bạn đăng câu hỏi này để làm gì?

a) \(3x^2\left(2x^3+7xy-5y^3\right)=6x^5+21x^3y-15x^2y^3\)

b) \(\frac{4x}{7\left(x-y\right)}.\frac{x-y}{x^2}=\frac{4x\left(x-y\right)}{7x^2\left(x-y\right)}=\frac{4}{7x}\)

hoài có cái so tài không à mà cứ kéo dài mãi

a) \(3x^2\left(2x^3+7xy-5y^3\right)\)

\(=6x^5+21x^3y-15x^2y\)

b) \(\frac{4x}{7\left(x-y\right)}.\frac{x-y}{x^2}\)

\(=\frac{4x.\left(x-y\right)}{7\left(x-y\right).x^2}\)\(=\frac{4x}{7x^2}\)

Bài 1 : (4a - b).(4a + b)    = 16a² + (-b²)

       (\(x^2y\) + 2y)(\(x^2\)y - 2y    = \(x^4\).y² + (- 4y²)

   (\(\dfrac{3}{4}\)\(x\) + \(\dfrac{3}{5}\)y)(\(\dfrac{3}{5}\)y - \(\dfrac{3}{4}\)\(x\))     = \(\dfrac{9}{25}\)y² + (- \(\dfrac{9}{16}\)\(x^2\))

    2; (\(x+2\))(\(x^2\) - 2\(x\) + 4)  =       \(x^3\) + 8

         (3\(x\) + 2y)(9\(x^2\) - 6\(xy\) + 4y²)   = 27\(x^3\) + 8y³

       3,  (5- 3\(x\))(25 + 15\(x\) + 9\(x^2\))      = 125 + ( -27\(x^3\))

        (\(\dfrac{1}{2}\)\(x\) - \(\dfrac{1}{5}\)y).(\(\dfrac{1}{4}\)\(x^2\) + \(\dfrac{1}{10}\)\(xy\) + \(\dfrac{1}{25}\)y² =  \(\dfrac{1}{8}\)\(x^3\) + (-\(\dfrac{1}{125}\)y³)

em cảm ơn cô ạ

cậu đọc full đề đi

bỏ dâu "-" đi là oke ! tag thầy cô bừa bải vải chưởng