\(\dfrac{2bz-3cy}{a}=\dfrac{3cx-az}{2b}=\dfrac{ay-2bx}{3c}\)

\(\Leftrightarrow\dfrac{2abz-3acy}{a^2}=\dfrac{6bcx-2abz}{4b^2}=\dfrac{3acy-6bcx}{9c^2}\)

\(=\dfrac{2abz-3acy+6bcx-2abz+3acy-6bcx}{a^2+4b^2+9c^2}=0\)

\(\Rightarrow2bz-3cy=0\Rightarrow\dfrac{z}{3c}=\dfrac{y}{2b}\left(1\right)\)

\(\Rightarrow3cx-az=0\Rightarrow\dfrac{x}{a}=\dfrac{z}{3c}\left(2\right)\)

Từ (1) và (2) suy ra: \(\dfrac{x}{a}=\dfrac{y}{2b}=\dfrac{z}{3c}\)

\(y=\dfrac{3\left(\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{101}\right)}{5\left(\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{101}\right)}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{2}.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}\right)}=\dfrac{3}{5}+\dfrac{2}{5}=1\)

\(|2x-3|=\left(\dfrac{1}{2}\right)^1=\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}2x-3=\dfrac{1}{2}\Rightarrow x=\dfrac{7}{4}\\2x-3=-\dfrac{1}{2}\Rightarrow x=\dfrac{5}{4}\end{matrix}\right.\)

Gọi hai số cần tìm là \(a,b\left(a>b>0\right)\) theo đầu bài ta có:

\(15\left(a+b\right)=60\left(a-b\right)=8ab\) hay \(\dfrac{a+b}{8}=\dfrac{a-b}{2}=\dfrac{ab}{15}=k\Rightarrow k=1\Rightarrow\left\{{}\begin{matrix}a=5\\b=3\end{matrix}\right.\)

\(a,\) Xét \(-\dfrac{34}{35}+\dfrac{2}{5}-\dfrac{3}{7}+1=\dfrac{-34+14-15+35}{35}=0\)

Thay vào ta có: \(\left(2x-3\right)\left(3x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}2x-3=0\Rightarrow x=\dfrac{3}{2}\\3x+1=0\Rightarrow x=-\dfrac{1}{3}\end{matrix}\right.\)

\(b,\) Ta có: \(x^2-5x+6=0\Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\)

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

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

\(S\) có \(\left(n-1\right)\) số hạng:

\(S=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{n^2-1}{n^2}=\left(1-\dfrac{1}{2^2}\right)+\left(1-\dfrac{1}{3^2}\right)+\left(1-\dfrac{1}{4^2}\right)+...+\left(1-\dfrac{1}{n^2}\right)\)

\(S=n-1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}\right)< n-1\left(1\right)\)

Mặt khác \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right).n}=1-\dfrac{1}{n}\)

\(S>n-1-1+\dfrac{1}{n}=n-2+\dfrac{1}{n}>n-2\left(2\right)\)

Từ (1) và (2) ta có: \(n-2< S< n-1\)

Vậy S không có giá trị nguyên với mọi số tự nhiên

\(B>1\)

\(\Leftrightarrow\dfrac{6x}{x-9}>1\)

\(\Leftrightarrow\dfrac{6x}{x-9}-1>0\)

\(\Leftrightarrow\dfrac{6x-\left(x-9\right)}{x-9}>0\)

\(\Leftrightarrow\dfrac{5x+9}{x-9}>0\)

Vì \(x\ge0,x\ne9\Rightarrow5x+9>0\)

Để  \(\dfrac{5x+9}{x-9}>0\) thì \(x-9>0\Leftrightarrow x>9\)

Kết hợp điều kiện \(x\ge0,x\ne9\)

Vậy \(x>9\) thì \(B>1\)

\(a,A=\sqrt{27}-\sqrt{12}\)

\(A=\sqrt{3^3}-\sqrt{2^2.3}=3\sqrt{3}-2\sqrt{3}=\sqrt{3}\)

\(b,B=\sqrt{a}-\dfrac{a-4}{\sqrt{a}+2}\) với \(a\ge0\)

\(B=\sqrt{a}-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}+2}\)

\(B=\sqrt{a}-\left(\sqrt{a}-2\right)\)

\(B=\sqrt{a}-\sqrt{a}+2\)

\(B=2\)

Ta có: \(\Delta=7^2-4.12=1>0\) nên pt có 2 nghiệm phân biệt \(\left[{}\begin{matrix}x_1=\dfrac{-7+1}{2}=-3\\x_2=\dfrac{-7-1}{2}=-4\end{matrix}\right.\)

Vậy tập nghiệm của pt là \(S=\left\{-3;-4\right\}\)

\(\left(x-1\right)^2-x+1=0\)

\(\Leftrightarrow\left(x-1\right)^2-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-1-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(5\sqrt{9}-\sqrt{36}\)

\(=5\sqrt{3^2}-\sqrt{6^2}\)

\(=5.3-6\)

\(=9\)