a, (3x-1)(x²+2)=(3x-1)(7x-10)

<=>(3x-1)(x²+2)-(3x-1)(7x-10)=0

<=>(3x-1)(x²+2-7x+10)=0

<=>(3x-1)(x²-7x+12)=0

<=>(3x-1)(x²-3x-4x+12)=0

<=>(3x-1)(x-3)(x-4)=0

<=>\(\left[{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\)<=>\(\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)

Vậy ft có tập nghiệm S=\(\left\{\dfrac{1}{3},3,4\right\}\)

b,\(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\) (ĐKXĐ:t\(\ne2;t\ne-3\))

<=>\(\dfrac{\left(t+3\right)^2+\left(t-2\right)^2}{\left(t-2\right)\left(t+3\right)}\)=\(\dfrac{5t+15}{t^2-2t+3t-6}\)

<=>\(\dfrac{t^2+6t+9+t^2-4t+4}{\left(t-2\right)\left(t+3\right)}\)=\(\dfrac{5t+15}{\left(t-2\right)\left(t+3\right)}\)

=>2t²+2t+13=5t+15

<=>2t²+2t-5t+13-15=0

<=>2t²-3t-2=0

<=>2t²-4t+t-2=0

<=>(t-2)(2t+1)=0

<=>\(\left[{}\begin{matrix}t-2=0\\2t+1=0\end{matrix}\right.< =>\left[{}\begin{matrix}t=2\left(loại\right)\\t=\dfrac{-1}{2}\left(tmđkxđ\right)\end{matrix}\right.\)

Vậy ft có nghiệm duy nhất x=\(\dfrac{-1}{2}\)

Giải:

a) \(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)

Chia cả hai vế cho 3x-1, ta được:

\(x^2+2=7x-10\)

\(\Leftrightarrow x^2-7x+10+2=0\)

\(\Leftrightarrow x^2-7x+12=0\)

\(\Leftrightarrow x^2-4x-3x+12=0\)

\(\Leftrightarrow x\left(x-4\right)-3\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

Vậy ...

b) \(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\) (1)

ĐKXĐ: \(t\ne2;t\ne-3\)

\(\left(1\right)\Leftrightarrow\dfrac{\left(t+3\right)\left(t+3\right)}{\left(t-2\right)\left(t+3\right)}+\dfrac{\left(t-2\right)\left(t-2\right)}{\left(t-2\right)\left(t+3\right)}=\dfrac{5t+15}{\left(t-2\right)\left(t+3\right)}\)

\(\Rightarrow\left(t+3\right)^2+\left(t-2\right)^2=5t+15\)

\(\Leftrightarrow t^2+6t+9+t^2-4t+4=5t+15\)

\(\Leftrightarrow2t^2+2t+13=5t+15\)

\(\Leftrightarrow2t^2+2t+13-5t-15=0\)

\(\Leftrightarrow2t^2-3t-2=0\)

\(\Leftrightarrow2t^2-4t+t-2=0\)

\(\Leftrightarrow2t\left(t-2\right)+\left(t-2\right)=0\)

\(\Leftrightarrow\left(2t+1\right)\left(t-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2t+1=0\\t-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=-\dfrac{1}{2}\left(tm\right)\\t=2\left(ktm\right)\end{matrix}\right.\)

Vậy ...

a,\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x^2-7x+12=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=1\\\left(x-4\right)\left(x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x-4=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\\x=3\end{matrix}\right.\)

Vậy...

b,\(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\)

\(\Leftrightarrow\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{\left(t+3\right)\left(t-2\right)}\)

\(\Leftrightarrow\dfrac{\left(t+3\right)^2}{\left(t-2\right)\left(t+3\right)}+\dfrac{\left(t-2\right)^2}{\left(t-2\right)\left(t+3\right)}=\dfrac{5t+15}{\left(t+3\right)\left(t-2\right)}\)

\(\Leftrightarrow t^2+6t+9+t^2-4t+4=5t+15\)

\(\Leftrightarrow t^2+t^2+6t-4t-5t=15-9-4\)

\(\Leftrightarrow2t^2-3t=2\)

\(\Leftrightarrow\left(2t+1\right)\left(t+2\right)=0\)

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

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

Vậy...

a,

\(\dfrac{1+x+3-3x-3+x}{1-x}=0\\ \dfrac{1-x}{1-x}=0\\ =>1-x=0\\ =>x=1\\ \)

b =>x-3 =10x -15

=>x-10x=-15+3

=>-9x=-12

=>x=4/3

a, \(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\) ĐKXĐ: t\(\ne\)2,t\(\ne\)-3

\(\Leftrightarrow\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{\left(t+3\right)\left(t-2\right)}\)

\(\Rightarrow\left(t+3\right)\left(t+3\right)+\left(t-2\right)\left(t-2\right)=5t+15\)

\(\Leftrightarrow t^2+6t+9+t^2-4t+4-5t-15=0\)

\(\Leftrightarrow-3t-2=0\)

\(\Leftrightarrow-3t=2\)

\(\Leftrightarrow t=\dfrac{-2}{3}\) (tđk)

\(\Rightarrow S=\left\{\dfrac{-2}{3}\right\}\)

b, \(\left(2x+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)ĐKXĐ: x\(\ne\)\(\dfrac{2}{7}\)

\(\Leftrightarrow\) \(\left(2x+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)-\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)=0\)

\(\Rightarrow\left(\dfrac{3x+8}{2-7x}+1\right)\left(2x+3-x+5\right)=0\)

\(\Leftrightarrow\) \(\Rightarrow\left(\dfrac{3x+8}{2-7x}+1\right)\left(x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3x+8}{2-7x}+1=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x+8+2-7x=0\\x=-8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-4x+10=0\\x=-8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-8\end{matrix}\right.\)

\(\Rightarrow S=\left\{\dfrac{5}{2};-8\right\}\)

ĐKXĐ: x khác 2 và x khác -3

\(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\)

\(\Leftrightarrow\dfrac{\left(t+3\right)\left(t+3\right)}{\left(t+3\right)\left(t-2\right)}+\dfrac{\left(t-2\right)\left(t-2\right)}{\left(t+3\right)\left(t-2\right)}=\dfrac{5t+15}{t^2+t-6}\)

\(\Rightarrow t^2+6t+9+t^2-4=5t+15\)

\(\Leftrightarrow2t^2+t-10=0\)

\(\Leftrightarrow2t^2-4t+5t-10=0\)

\(\Leftrightarrow2t\left(t-2\right)+5\left(t-2\right)=0\)

\(\Leftrightarrow\left(2t+5\right)\left(t-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=2\left(loại\right)\\t=\dfrac{-5}{2}\end{matrix}\right.\)

Vậy..................

ĐKXĐ: x khác \(\dfrac{2}{7}\)\(\left(2x+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)

\(\Leftrightarrow\left(2x+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)-\left(x+5\right)\left(\dfrac{3x+8}{2-7x}+1\right)=0\)

\(\Leftrightarrow\left(\dfrac{3x+8}{2-7x}+1\right)\left(2x+3-x-5\right)=0\)

\(\Leftrightarrow\left(\dfrac{-4x+10}{2-7x}\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{-4x+10}{2-7x}=0\\x-2=0\end{matrix}\right.\)

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

Vậy...............

1) \(\dfrac{2}{15}\cdot6\dfrac{5}{11}+\dfrac{5}{11}\cdot\dfrac{-2}{15}-\dfrac{2}{15}\cdot2015^0\)

\(=\dfrac{2}{15}\cdot\dfrac{71}{11}-\dfrac{1}{11}\cdot\dfrac{2}{3}-\dfrac{2}{15}\cdot1\)

\(=\dfrac{142}{165}-\dfrac{2}{33}-\dfrac{2}{15}\)

\(=\dfrac{2}{3}\)

2) \(\dfrac{5}{2\cdot7}+\dfrac{3}{14\cdot11}+\dfrac{4}{11\cdot7}+\dfrac{1}{14\cdot15}+\dfrac{13}{15\cdot16}\)

\(=\dfrac{5}{14}+\dfrac{3}{154}+\dfrac{4}{77}+\dfrac{1}{210}+\dfrac{13}{240}\)

\(=\dfrac{39}{80}\)

\(\dfrac{2}{15}.6\dfrac{5}{11}+\dfrac{5}{11}.\dfrac{-2}{15}-\dfrac{2}{15}.2015^0\)

\(=\dfrac{2}{15}.\dfrac{71}{11}+\dfrac{-5}{11}.\dfrac{2}{5}-\dfrac{2}{15}.1\)

\(=\dfrac{2}{15}\left(\dfrac{71}{11}+\dfrac{-5}{11}-1\right)=\dfrac{2}{15}.5=\dfrac{2}{3}\)

\(\dfrac{5}{2.7}+\dfrac{3}{14.11}+\dfrac{4}{11.7}+\dfrac{1}{14.15}+\dfrac{13}{15.16}\)

\(=\dfrac{5}{14}+\dfrac{3}{154}+\dfrac{4}{77}+\dfrac{1}{210}+\dfrac{13}{240}\)

\(=\dfrac{39}{80}\)

Nhận thấy \(t=0\) ko phải nghiệm

Với \(t\ne0\) pt tương đương:

\(\dfrac{3}{t+3+\dfrac{2}{t}}+\dfrac{2}{t+1+\dfrac{2}{t}}=1\)

Đặt \(t+\dfrac{1}{t}+1=x\Rightarrow t+\dfrac{2}{t}+3=x+2\)

Pt trở thành:

\(\dfrac{3}{x+2}+\dfrac{2}{x}=1\)

\(\Rightarrow3x+2\left(x+2\right)=x\left(x+2\right)\)

\(\Leftrightarrow x^2-3x-4=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}t+\dfrac{2}{t}+1=-1\\t+\dfrac{2}{t}+1=4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}t^2+2t+2=0\left(vn\right)\\t^2-3t+2=0\end{matrix}\right.\)

\(\Rightarrow t=\left\{1;2\right\}\)

a) \(\left(\dfrac{3}{4}-5x\right)-\left(2x-\dfrac{17}{4}\right)\)=\(\dfrac{44}{7}\)

\(\left(\dfrac{3}{4}-\dfrac{17}{4}\right)-\left(5x-2x\right)\) = \(\dfrac{44}{7}\)

\(\dfrac{-14}{4}-3x\) = \(\dfrac{44}{7}\)

3x= \(\dfrac{-14}{4}-\dfrac{44}{7}\)=\(\dfrac{-98}{28}-\dfrac{176}{28}\)= \(\dfrac{-274}{28}=\dfrac{-137}{14}\)

x= \(\dfrac{-137}{14}:3\)

x= \(\dfrac{-137}{42}\)

3.

\(A=1-3+3^2-3^3+...-3^{2009}-3^{2010}\)

\(\Rightarrow3A=3-3^2+3^3-3^4+...-3^{2010}+3^{2011}\)

\(\Rightarrow4A=3-3^2+3^3-3^4+...-3^{2010}+3^{2011}+\left(1-3+3^2-3^3+...-3^{2009}+3^{2010}\right)\)\(\Rightarrow4A=3^{2011}-1\)

\(\Rightarrow4A=3^{2011}\)

\(\Rightarrow\)ĐPCM

Bài 1:

a)=2.( \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+......+\dfrac{1}{97}-\dfrac{1}{99}\)

=2. (1/3-1/99)

=2. (33/99-1/99)

=2. 32/99

=64/99

b) tương tự như trên.

Bài 1 :

a) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)

\(=2\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\right)\)

\(=2\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)

\(=2\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)

\(=2\left(\dfrac{33}{99}-\dfrac{1}{99}\right)\)

\(=2.\dfrac{32}{99}\)

\(=\dfrac{2.32}{99}\)

\(=\dfrac{64}{99}\)

b) \(\dfrac{3}{1.3}+\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{49.51}\)

\(=2\left(\dfrac{3}{1.3}+\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{49.51}\right)\)

\(=3\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\right)\)

\(=3\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)

\(=3\left(1-\dfrac{1}{51}\right)\)

\(=3.\dfrac{50}{51}\)

\(=\dfrac{3.50}{51}\)

\(=\dfrac{1.50}{17}\)

\(=\dfrac{50}{17}\)

Bài 2:

a) \(\dfrac{2}{3}.x+\dfrac{1}{4}=\dfrac{7}{12}\)

\(\dfrac{2}{3}.x=\dfrac{7}{12}-\dfrac{1}{4}\)

\(\dfrac{2}{3}.x=\dfrac{1}{3}\)

\(x=\dfrac{1}{3}:\dfrac{2}{3}\)

\(x=\dfrac{1}{2}\)

Vậy \(x=\dfrac{1}{2}\)

b) \(\dfrac{1}{2}+\dfrac{1}{3}:2.x=-1\)

\(\dfrac{1}{3}:2.x=-1-\dfrac{1}{2}\)

\(\dfrac{1}{3}:2.x=\dfrac{-3}{2}\)

\(2.x=\dfrac{1}{3}:\dfrac{-3}{2}\)

\(2.x=\dfrac{-2}{9}\)

\(x=\dfrac{-2}{9}:2\)

\(x=\dfrac{-1}{9}\)

Vậy \(x=\dfrac{-1}{9}\)