\(a,\frac{x+2}{6}-\frac{8x+1}{3}=\frac{2-5x}{2}-6\)

\(\Leftrightarrow\frac{x+2}{6}-\frac{\left(8x+1\right)2}{6}=\frac{\left(2-5x\right)3}{6}-\frac{36}{6}\)

=> x + 2 - 16x - 2 = 6 - 15x - 36

<=> x - 16x + 15x = 6 -36 + 2 - 2

<=> 0x = -30

Phương trình vô ngiệm

b, 11 - ( x + 2) = 3(x + 1)

<=> 11 - x - 2= 3x + 3

<=> -x - 3x = 3 - 11 + 2

<=> -4x = -6

<=> x = \(\frac{3}{2}\) 

C,  tương tự a

c) ĐKXĐ: x \(\ne\)0 và x \(\ne\)-1

Ta có: \(\frac{x+3}{x+1}+\frac{x+2}{x}=2\)

=> \(x\left(x+3\right)+\left(x+1\right)\left(x+2\right)=2x\left(x+1\right)\)

<=> x² + 3x + x² + 3x + 2 = 2x² + 2x

<=> 2x² + 6x + 2 - 2x² - 2x = 0

<=> 4x + 2 = 0

<=> 4x = -2

<=> x = -1/2 (tm)

Vậy S = {-1/2}

Đặt  \(ab=x;\)\(bc=y;\)\(ca=z\)

Khi đó:   \(a^3b^3+b^3c^3+c^3a^3=3a^2b^2c^2\)

<=>  \(x^3+y^3+z^3=3xyz\)

<=>  \(x^3+y^3+z^3-3xyz=0\)

<=>  \(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

Nếu:  \(x+y+z=0\)thì:  \(ab+bc+ca=0\)

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

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

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

Nếu:  \(x^2+y^2+z^2-xy-yz-zx=0\)<=>   \(x=y=z\)

<=>  \(ab=bc=ca\)<=>  \(a=b=c\)

\(A=\left(\frac{a}{b}+1\right)\left(\frac{b}{c}+1\right)+\left(\frac{c}{a}+1\right)=2.2+2=6\)

p/s: trg hợp 1 mk lm đc đến có z thôi, bn tham khảo

(2/3×x-1/3)=2/3+1/3

(2/3×x-1/3)=3/3

2/3×x=3/3+1/3

2/3×x=4/3

x=4/3:3/2

x=4/3×2/3

x=8/9

Cảm ơn mn lần nx ạ

Lời giải:

Áp dụng BĐT AM-GM ta có:

\(4x^2+1\geq 4x\)

\(\Rightarrow \left\{\begin{matrix} 5x^2-x+3\geq x^2+3x+2\\ 5x^2+x+\geq x^2+5x+6\\ 5x^2+3x+13\geq x^2+7x+12\\ 5x^2+5x+21\geq x^2+9x+20\end{matrix}\right.\)

\(\text{VT}\leq \frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}\)

\(\Leftrightarrow \text{VT}\leq \frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)}\)

\(\Leftrightarrow \text{VT}\leq \frac{(x+2)-(x+1)}{(x+1)(x+2)}+\frac{(x+3)-(x+2)}{(x+2)(x+3)}+\frac{(x+4)-(x+3)}{(x+3)(x+4)}+\frac{(x+5)-(x+4)}{(x+4)(x+5)}\)

\(\Leftrightarrow \text{VT}\leq \frac{1}{x+1}-\frac{1}{x+5}\)

\(\Leftrightarrow \text{VT}\leq \frac{4}{x^2+6x+5}\)

Dấu "=" xảy ra khi $4x^2=1, x>0$ hay $x=\frac{1}{2}$

Vậy $x=\frac{1}{2}$ là nghiệm của PT.

Nguyễn Việt Lâm anh giúp em pt trên với ạ !!!

Akai Haruma giúp em bài này với ạ ''''

Vì \(a^2+b^2\ge2ab,b^2+1\ge2b\),ta có:

\(\frac{1}{a^2+2b^2+3}=\frac{1}{a^2+b^2+b^2+1+1}\le\frac{1}{2\left(ab+b+1\right)}\)

Tương tự:\(\frac{1}{b^2+2c^2+3}\le\frac{1}{2\left(bc+c+1\right)}\)và \(\frac{1}{b^2+2c^2+3}\le\frac{1}{2\left(bc+c+1\right)}\)

Khi đó\(A\le\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+a}\right)\)

\(\Leftrightarrow A\le\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{ab+b+1}+\frac{b}{ab+b+1}\right)=\frac{1}{2}\)

Dấu"="trg BĐT trên xảy ra khi \(a=b=c=1\)

Vậy \(Max_P=\frac{1}{2}\Leftrightarrow a=b=c=1\)

Chắc không được GP đâu !!

Áp dụng bđt cauchy , ta có :

+) \(a^2+2b^2+3=\left(a^2+b^2\right)+\left(b^2+1\right)+2\ge2ab+2b+2\)

+) \(b^2+2c^2+3\ge2bc+2c+2\)

+) \(c^2+2a^2+3\ge2ac+2a+2\)

Khi đó , ta có :

\(VT\le\frac{1}{2ab+2b+2}+\frac{1}{2bc+2c+2}+\frac{1}{2ac+2a+2}\)

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

\(=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{abc}{bc+c+1}+\frac{abc}{ac+a+1}\right)\)( vì abc= 1 )

\(=\frac{1}{2}=VP\)( đoạn này ban tự phân tích ra nha , mk lmaf hơi tắt )

Vậy .................

Các bn vô link này giúp nốt mk nhé, btvn ý mà:)) https://olm.vn/hoi-dap/detail/261956633251.html?auto=1

\(a,\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{1}-\frac{1}{100}\)

\(=\frac{100}{100}-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(b,\frac{x}{y}=\frac{3}{5}\)

\(\Leftrightarrow\frac{x}{3}=\frac{y}{5}\)

\(\text{Áp dụng tính chất dãy tỉ số bằng nhau ta có :}\)

\(\frac{x}{3}=\frac{y}{5}=\frac{x+y}{3+5}=\frac{18}{8}=\frac{9}{4}\)

\(\Rightarrow\frac{x}{3}=\frac{9}{4}\Rightarrow x=\frac{27}{4}\)

\(\frac{y}{5}=\frac{9}{4}\Rightarrow y=\frac{45}{4}\)

\( \left(5x-2\right)\left(x-3\right)=0\)

\(\left[{}\begin{matrix}5x-2=0\\x-3=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=3\end{matrix}\right.\)