\(\frac{1}{x^2-5x+6}+\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}+\frac{1}{x^2-11x+30}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x^2-3x-2x+6}+\frac{1}{x^2-3x-4x+12}+\frac{1}{x^2-4x-5x+20}+\frac{1}{x^2-5x-6x+30}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}+\frac{1}{\left(x-5\right)\left(x-6\right)}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x-6}-\frac{1}{x-5}+\frac{1}{1-5}-\frac{1}{1-4}+\frac{1}{1-4}-\frac{1}{1-3}+\frac{1}{1-3}-\frac{1}{1-2}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x-6}-\frac{1}{x-2}=\frac{1}{8}\)

\(\Leftrightarrow\frac{4}{x^2-8x+12}=\frac{1}{8}\)

\(\Leftrightarrow x^2-8x+12=32\)

\(\Leftrightarrow\left(x-4\right)^2=36\)

\(\Leftrightarrow x=10\) hoặc \(x=-2\)

\(\frac{1}{x^2-5x+6}+\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}+\frac{1}{x^2-11x+30}=\frac{1}{8}\)\(\frac{1}{x^2-2x-3x+6}+\frac{1}{x^2-4x-3x+12}+\frac{1}{x^2-4x-5x+20}+\frac{1}{x^2-6x-5x+30}=\frac{1}{8}\)

\(\frac{1}{x\left(x-2\right)-3\left(x-2\right)}+\frac{1}{x\left(x-4\right)-3\left(x-4\right)}+\frac{1}{x\left(x-4\right)-5\left(x-4\right)}+\frac{1}{x\left(x-6\right)-5\left(x-6\right)}=\frac{1}{8}\)

\(\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}+\frac{1}{\left(x-5\right)\left(x-6\right)}=\frac{1}{8}\)dhjjhhjhhjj

\(\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}+\frac{1}{\left(x-5\right)\left(x-6\right)}=\frac{1}{8}\)

Còn lại tự giải quyết nha

a)\(\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}\)

\(\Leftrightarrow2c+2\sqrt{\left(a+c\right)\left(b+c\right)}=0\)

\(\Leftrightarrow\sqrt{\left(a+c\right)\left(b+c\right)}=-c\)

\(\Leftrightarrow\begin{cases}c< 0\\ab+bc+ca+c^2=c^2\end{cases}\)\(\Leftrightarrow ab+bc+ca=0\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc+ac+ab}{abc}=0\)

Đpcm

phần b chắc quy đồng nó lên quá =))

Xét A= \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=a.\frac{a}{b+c}+b.\frac{b}{c+a}+c.\frac{c}{a+b}\)

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

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

\(=a.\frac{a+b+c}{b+c}-a+b.\frac{a+b+c}{c+a}-b+c.\frac{a+b+c}{a+b}-c\)

\(=\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)-\left(a+b+c\right)\) =0

\(1^2+2^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)(*)

Với n=1, ta có (*) luôn đúng

Giả sử (*) đúng với n=k ta có:

\(1^2+2^2+...+k^2=\frac{k\left(k+1\right)\left(2k+1\right)}{6}\) (1)

Ta sẽ chứng minh (*) đúng với n=k+1, thật vậy từ (1) suy ra:

\(1^2+2^2+...+k^2+\left(k+1\right)^2=\frac{k\left(k+1\right)\left(2k+1\right)}{6}+\left(k+1\right)^2\)

\(=\left(k+1\right)\left[\frac{k\left(2k+1\right)}{6}+\left(k+1\right)\right]\)\(=\frac{\left(k+1\right)\left(2k^2+k+6k+6\right)}{6}\)

\(=\frac{\left(k+1\right)\left(2k^2+7k+6\right)}{6}=\frac{\left(k+1\right)\left(2k^2+4k+3k+6\right)}{6}\)

\(=\frac{\left(k+1\right)\left[2k\left(k+2\right)+3\left(k+2\right)\right]}{6}=\frac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{6}\)

Theo nguyên lý qui nạp (*) đúng với mọi n thuộc N*

Vậy ta có điều phải chứng minh

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\frac{x}{1+y+xz}=\frac{x\left(x^2+y+\frac{z}{x}\right)}{\left(1+y+xz\right)\left(x^2+y+\frac{z}{x}\right)}\le\frac{x^3+xy+z}{\left(x+y+z\right)^2}\)

\(\le\frac{x+y+z}{\left(x+y+z\right)}=\frac{1}{x+y+z}\)

Tương tự ta cũng có: \(\frac{y}{1+z+xy}\le\frac{1}{x+y+z};\frac{z}{1+x+yz}\le\frac{1}{x+y+z}\)

Cộng theo vế ta có: \(\frac{x}{1+y+xz}+\frac{y}{1+z+xy}+\frac{z}{1+x+yz}\le\frac{1+1+1}{x+y+z}=\frac{3}{x+y+z}\)

ff

c: \(=\dfrac{1}{3x-2}-\dfrac{4}{3x+2}+\dfrac{3x-6}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\dfrac{3x+2-12x+8+3x-6}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\dfrac{-6x+4}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{-2}{3x+2}\)

d: \(=\dfrac{x^2-4-x^2+10}{x+2}=\dfrac{6}{x+2}\)

e: \(=\dfrac{1}{2\left(x-y\right)}-\dfrac{1}{2\left(x+y\right)}-\dfrac{y}{\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{x+y-x+y-2y}{2\left(x-y\right)\left(x+y\right)}=\dfrac{0}{2\left(x-y\right)\left(x+y\right)}=0\)

học dạng rút gon này fai linh hoạt, bn để ý cái điều kiện kia kìa

a khác 1 a-1=0 mà mẫu khác 0=>mẫu có 1 nhân tử là a-1, tiếp tục ta có dc

mẫu=(a-1)(a-2)(a-4)

còn tử thì bấm máy ta có Tử=(a-4)(a-1)(a+1)

thay vào rút gọn thôi