bai 1

1 thay k=0 vao pt ta co 4x^2-25+0^2+4*0*x=0

<=>(2x)^2-5^2=0

<=>(2x+5)*(2x-5)=0

<=>2x+5=0 hoăc 2x-5 =0 tiếp tục giải ý 2 tương tự

a)   ta có :(x-1)(x-2)(x+3)(x+4)=24

           <=>[(x-1)(x+3)].[(x-2)(x+4)] =24

          <=>(x^2 +2x -3)(x^2+2x -8)=24

         đặt x^2  +2x -3  =a =>  (x^2 +2x -3)(x^2 +2x-8)=a(a-5) =24

                                                                                 <=>a^2 -5a-24=0

                                                                                <=>(a-8)(a+3)=0  <=> a-8=0 hoặc a+3=0 <=>a=8 hoặc a=-3

+) với a=8 => x^2 +2x-3=8 <=>x^2 +2x-11=0<=>(x+1)^2 -10=0   (vô nghiệm)  vì (x+1)^2  >=0

+) với a=-3=>x^2 +2x-3=-3<=>x^2 +2x=0<=>x.(x+2)=0  <=> x=0 hoặc x+2=0 <=>x=0 hoặc x=-2

Vậy tập nghiệm của pt là S={0;-2}

\(\Leftrightarrow\frac{1}{x}+\frac{1}{x+2}+\frac{1}{x+5}+\frac{1}{x+7}-\frac{1}{x+1}-\frac{1}{x+3}-\frac{1}{x+4}-\frac{1}{x+6}=0\)

\(\Leftrightarrow\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{\left(x+4\right)\left(x+5\right)}-\frac{1}{\left(x+6\right)\left(x+7\right)}=0\)

\(\Leftrightarrow\frac{8x+20}{x\left(x+1\right)\left(x+4\right)\left(x+5\right)}+\frac{8x+36}{\left(x+2\right)\left(x+3\right)\left(x+6\right)\left(x+7\right)}=0\).Đến đây mk chịu

Câu 1 :

a, Ta có : \(3\left(x-1\right)-2\left(x+3\right)=-15\)

=> \(3x-3-2x-6=-15\)

=> \(3x-3-2x-6+15=0\)

=> \(x=-6\)

Vậy phương trình có nghiệm là x = -6 .

b, Ta có : \(3\left(x-1\right)+2=3x-1\)

=> \(3x-3+2=3x-1\)

=> \(3x-3+2-3x+1=0\)

=> \(0=0\)

Vậy phương trình có vô số nghiệm .

c, Ta có : \(7\left(2-5x\right)-5=4\left(4-6x\right)\)

=> \(14-35x-5=16-24x\)

=> \(14-35x-5-16+24x=0\)

=> \(-35x+24x=7\)

=> \(x=\frac{-7}{11}\)

Vậy phương trình có nghiệm là \(x=\frac{-7}{11}\) .

Bài 2 :

a, Ta có : \(\frac{x}{30}+\frac{5x-1}{10}=\frac{x-8}{15}-\frac{2x+3}{6}\)

=> \(\frac{x}{30}+\frac{3\left(5x-1\right)}{30}=\frac{2\left(x-8\right)}{30}-\frac{5\left(2x+3\right)}{30}\)

=> \(x+3\left(5x-1\right)=2\left(x-8\right)-5\left(2x+3\right)\)

=> \(x+15x-3=2x-16-10x-15\)

=> \(x+15x-3-2x+16+10x+15=0\)

=> \(24x+28=0\)

=> \(x=\frac{-28}{24}=\frac{-7}{6}\)

Vậy phương trình có nghiệm là \(x=\frac{-7}{6}\) .

b, Ta có : \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)

=> \(\frac{6\left(x+4\right)}{30}-\frac{30x}{30}+\frac{120}{30}=\frac{10x}{30}-\frac{15\left(x-2\right)}{30}\)

=> \(6\left(x+4\right)-30x+120=10x-15\left(x-2\right)\)

=> \(6x+24-30x+120=10x-15x+30\)

=> \(6x+24-30x+120-10x+15x-30=0\)

=> \(-19x+114=0\)

=> \(x=\frac{-114}{-19}=6\)

Vậy phương trình có nghiệm là x = 6 .

\(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-2}+\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-....+\dfrac{1}{x-5}-\dfrac{1}{x-6}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-6}=\dfrac{1}{10}\Leftrightarrow\dfrac{x-6-x+1}{\left(x-1\right)\left(x-6\right)}=\dfrac{1}{10}\)

\(\Leftrightarrow x^2-7x+56=0\Leftrightarrow x^2-2.\dfrac{7}{2}x+\dfrac{49}{4}+\dfrac{175}{4}=\left(x-\dfrac{7}{2}\right)^2+\dfrac{175}{4}>0\)

Vậy phương trình vô nghiệm 

ĐKXĐ: \(x\notin\left\{1;2;3;4;5;6\right\}\)

Ta có: \(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-2}+\dfrac{1}{x-4}+\dfrac{1}{x-3}+\dfrac{1}{x-5}-\dfrac{1}{x-4}+\dfrac{1}{x-6}-\dfrac{1}{x-5}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{1}{x-6}-\dfrac{1}{x-1}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{10\left(x-1\right)}{10\left(x-6\right)\left(x-1\right)}-\dfrac{10\left(x-6\right)}{10\left(x-1\right)\left(x-6\right)}=\dfrac{\left(x-1\right)\left(x-6\right)}{10\left(x-1\right)\left(x-6\right)}\)

Suy ra: \(x^2-7x+6=10x-10-10x+60\)

\(\Leftrightarrow x^2-7x+6=50\)

\(\Leftrightarrow x^2-7x-44=0\)

\(\Leftrightarrow x^2-11x+4x-44=0\)

\(\Leftrightarrow x\left(x-11\right)+4\left(x-11\right)=0\)

\(\Leftrightarrow\left(x-11\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-11=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

Vậy: S={11;-4}

ĐKXĐ : \(x\notin\left\{1;2;...;6\right\}\)

\(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+...+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{10}\\ \Leftrightarrow\dfrac{\left(x-1\right)-\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}+\dfrac{\left(x-2\right)-\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+...+\dfrac{\left(x-5\right)-\left(x-6\right)}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{10}\\ \Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-2}+...+\dfrac{1}{x-6}-\dfrac{1}{x-5}=\dfrac{1}{10}\\ \Leftrightarrow\dfrac{1}{x-6}-\dfrac{1}{x-1}=\dfrac{1}{10}\\ \Leftrightarrow\dfrac{5}{\left(x-1\right)\left(x-6\right)}=\dfrac{5}{50}\\ \Rightarrow\left(x-1\right)\left(x-6\right)=50\\ \Leftrightarrow x^2-7x-44=0\\ \Leftrightarrow\left(x-11\right)\left(x+4\right)=0\\ \Leftrightarrow\begin{matrix}x=-4\\x=11\end{matrix}\left(t.m\right)\)

Bài 2 :

a, Ta có : \(\left(x+4\right)\left(x-1\right)=0\)

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

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

b, Ta có : \(\left(3x-2\right)\left(4x-7\right)=0\)

=> \(\left[{}\begin{matrix}3x-2=0\\4x-7=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}3x=2\\4x=7\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{2}{3}\\x=\frac{7}{4}\end{matrix}\right.\)

c, Ta có : \(\left(x+5\right)\left(x^2+1\right)=0\)

=> \(\left[{}\begin{matrix}x+5=0\\x^2+1=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-5\\x^2+1=0\left(VL\right)\end{matrix}\right.\)

d, Ta có : \(x\left(x-1\right)\left(x^2+4\right)=0\)

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

=> \(\left[{}\begin{matrix}x=0\\x=1\\x^2+4=0\left(VL\right)\end{matrix}\right.\)

e, Ta có : \(\left(3x+2\right)\left(x+\frac{1}{2}\right)=0\)

=> \(\left[{}\begin{matrix}3x+2=0\\x+\frac{1}{2}=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-\frac{2}{3}\\x=-\frac{1}{2}\end{matrix}\right.\)

f, Ta có : \(\left(x+2\right)\left(x+3\right)\left(x^2+7\right)=0\)

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

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

Bài 1 :

a, Ta có : \(1-\frac{x+3}{4}-\frac{x-2}{6}=0\)

=> \(\frac{12}{12}-\frac{3\left(x+3\right)}{12}-\frac{2\left(x-2\right)}{12}=0\)

=> \(12-3\left(x+3\right)-2\left(x-2\right)=0\)

=> \(12-3x-9-2x+4=0\)

=> \(-5x=-7\)

=> \(x=\frac{7}{5}\)