Bài 5: Phương trình chứa ẩn ở mẫu

Thay x=-3 vào pt, ta được:

\(\dfrac{-m-3}{-3+2}=2\)

=>\(\dfrac{m+3}{1}=2\)

=>m+3=2

=>m=-1

\(\dfrac{x-m}{x+2}=2\left(x\ne-2\right)\) đề ntn phải ko ạ

để pt có nghiệm `x=-3`

thì `(x-m)/(x+2)` có nghiệm `x=-3`

 \(=>\dfrac{-3-m}{-3+2}=2\\ < =>\dfrac{-3-m}{-1}=2\\ < =>-3-m=-2\\ < =>-m=-2+3\\ < =>-m=1\\ < =>m=-1\)

vậy `m=-1` thì pt có nghiệm `x=-3`

=>x^2-4x+2y^2-4y+6=0

=>x^2-4x+4+2y^2-4y+2=0

=>(x-2)^2+2(y-1)^2=0

=>x=2 và y=1

Số số hạng là (1-x+5):1+1=6-x+1=7-x(số)

Tổng là (7-x)*(x-5+1)/2(x-5)

=(7-x)(x-4)/2(x-5)

Theo đề, ta có: (7-x)(x-4)/2(x-5)=4

=>(7x-28-x^2+4x)=8(x-5)

=>-x^2+11x-28-8x+40=0

=>-x^2+3x+12=0

mà x là số tự nhiên

nên \(x\in\varnothing\)

\(\dfrac{5x+3}{x-4}=2\)   (1)

ĐKXĐ: \(x\ne4\)

(1) \(\Leftrightarrow5x+3=2\left(x-4\right)\)

\(\Leftrightarrow5x+3=2x-8\)

\(\Leftrightarrow5x-2x=-8-3\)

\(\Leftrightarrow3x=-11\)

\(\Leftrightarrow x=\dfrac{-11}{3}\) (nhận)

Vậy \(S=\left\{\dfrac{-11}{3}\right\}\)

\(\dfrac{5x+3}{x-4}=2\text{ĐKXĐ:}x\ne4\)

\(\Leftrightarrow\dfrac{5x+3}{x-4}=\dfrac{2\left(x-4\right)}{x-4}MTC:x-4\)

\(\Rightarrow5x+3=2x-8\)

\(\Leftrightarrow5x+3-2x+8=0\)

\(\Leftrightarrow3x+11=0\)

\(\Leftrightarrow3x=-11\)

\(\Leftrightarrow x=\dfrac{-11}{3}\left(\text{nhận}\right)\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{\dfrac{-11}{3}\right\}\)

\(\dfrac{9}{x+4}=\dfrac{3}{x-2}\) (ĐK: \(x\ne2;-4\))

\(\Rightarrow9\left(x-2\right)=3\left(x+4\right)\)

\(\Leftrightarrow9x-18=3x+12\)

\(\Leftrightarrow9x-3x=12+18\)

\(\Leftrightarrow6x=30\)

\(\Leftrightarrow x=5\) (TMĐK)

Vậy \(S=\left\{5\right\}\).

\(\dfrac{9}{x+4}=\dfrac{3}{x-2}\text{ĐKXĐ:}x\ne-4;2\)

\(\Leftrightarrow\dfrac{9\left(x-2\right)}{\left(x+4\right)\left(x-2\right)}=\dfrac{3\left(x+4\right)}{\left(x+4\right)\left(x-2\right)}MTC:\left(x+4\right)\left(x-2\right)\)

\(\Rightarrow9x-18=3x+12\)

\(\Leftrightarrow9x-18-3x-12=0\)

\(\Leftrightarrow6x-30=0\)

\(\Leftrightarrow6x=30\)

\(\Leftrightarrow x=5\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{5\right\}\)

\(\dfrac{2}{x-5}=\dfrac{3}{x+1}\) (ĐK: \(x\ne-1;5\))

\(\Rightarrow2\left(x+1\right)=3\left(x-5\right)\)

\(\Leftrightarrow2x+2=3x-15\)

\(\Leftrightarrow2x-3x=-15-2\)

\(\Leftrightarrow-x=-17\)

\(\Leftrightarrow x=17\) (TMĐK)

\(\dfrac{2}{x-5}=\dfrac{3}{x+1}\text{ĐKXĐ}:x\ne5;-1\)

\(\Leftrightarrow\dfrac{2\left(x+1\right)}{\left(x-5\right)\left(x+1\right)}=\dfrac{3\left(x-5\right)}{\left(x-5\right)\left(x+1\right)}MTC:\left(x-5\right)\left(x+1\right)\)

\(\Rightarrow2x+2=3x-15\)

\(\Leftrightarrow2x+2-3x+15=0\)

\(\Leftrightarrow17-x=0\) 

\(\Leftrightarrow x=17\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{17\right\}\)

=>5(x-3)=6(x+2)

=>6x+12=5x-15

=>x=-27

\(\dfrac{5}{x+2}=\dfrac{6}{x-3}\) (ĐK: \(x\ne-2;x\ne3\))

\(\Rightarrow5\left(x-3\right)=6\left(x+2\right)\)

\(\Leftrightarrow5x-15=6x+12\)

\(\Leftrightarrow-15-12=6x-5x\)

\(\Leftrightarrow-27=x\)

\(\Leftrightarrow x=-27\) (TMĐK)

Vậy \(S=\left\{-27\right\}\).

\(\dfrac{5}{x+2}=\dfrac{6}{x-3}\text{ĐKXĐ:}x\ne-2;3\)

\(\Leftrightarrow\dfrac{5\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}=\dfrac{6\left(x+2\right)}{\left(x+2\right)\left(x-3\right)}MTC:\left(x+2\right)\left(x-3\right)\)

\(\Rightarrow5x-15=6x+12\)

\(\Leftrightarrow5x-15-6x-12=0\)

\(\Leftrightarrow-27-x=0\)

\(\Leftrightarrow x=-27\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{-27\right\}\)

\(\dfrac{6x-2}{x+4}=5\Leftrightarrow6x-2=5x+20\)

\(\Leftrightarrow x=22\)

\(\dfrac{6x-2}{x+4}=5\)

ĐKXĐ:  \(x\ne-4\)

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

\(6x-2=5x+20\)

\(6x-5x=20+2\)

\(x=22\)

Vậy \(S=\left\{22\right\}\)

\(\dfrac{6x-2}{x+4}=5\left(dkxd:x\ne-4\right)\)

Suy ra :

\(6x-2=5\left(x+4\right)\)

\(\Rightarrow6x-2=5x+20\)

\(\Rightarrow x=22\left(tmdk\right)\)

Vậy \(S=\left\{22\right\}\)

\(\dfrac{x+3}{x-2}=4\left(dkxd:x\ne2\right)\)

Suy ra :

\(x+3=4\left(x-2\right)\)

\(\Rightarrow x+3=4x-8\)

\(\Rightarrow-3x=-11\)

\(\Rightarrow x=\dfrac{11}{3}\left(tmdk\right)\)

Vậy \(S=\left\{\dfrac{11}{3}\right\}\)

\(\dfrac{x+3}{x-2}=4\text{ĐKXĐ:}x\ne2\)

\(\Leftrightarrow\dfrac{x+3}{x-2}=\dfrac{4\left(x-2\right)}{x-2}MTC:x-2\)

\(\Rightarrow x+3=4x-8\)

\(\Leftrightarrow x+3-4x+8=0\)

\(\Leftrightarrow11-3x=0\)

\(\Leftrightarrow3x=11\)

\(\Leftrightarrow x=\dfrac{11}{3}\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{\dfrac{11}{3}\right\}\)

\(\dfrac{7}{x-6}=\dfrac{2}{x+2}\left(x\ne6;x\ne-2\right)\)

suy ra

`7(x+2)=2(x-6)`

`<=>7x+14=2x-12`

`<=> 7x-2x=-12-14`

`<=> 5x=-26`

`<=> x=-26/5 (tm)`

\(\dfrac{7}{x-6}=\dfrac{2}{x+2}\left(dkxd:x\ne6;-2\right)\)

\(\Leftrightarrow7\left(x+2\right)=2\left(x-6\right)\)

\(\Leftrightarrow7x+14=2x-12\)

\(\Leftrightarrow5x=-26\)

\(\Leftrightarrow x=-\dfrac{26}{5}\left(tmdk\right)\)