a) \(\frac{\left(x+m\right)}{x-5}+\frac{\left(x+5\right)}{x-m}=2\)

<=> \(\frac{\left(x+m\right)\left(x-m\right)}{\left(x-5\right)\left(x-m\right)}+\frac{\left(x+5\right)\left(x-5\right)}{\left(x-5\right)\left(x-m\right)}=2\)

<=>\(\frac{\left(x+m\right)\left(x-m\right)+\left(x+5\right)\left(x-5\right)}{\left(x-5\right)\left(x-m\right)}=2\)

<=>\(\frac{x^2-m^2+x^2-5^2}{\left(x-m\right)\left(x-5\right)}=2\)

<=>2(x-m)(x-5)=2x²-m²-25

Thay m=2, ta có:

2(x-2)(x-5)=2x²-2²-25

2x²-14x+20=2x²-29

20+29=2x²-2x²+14x

49=14x

=>x=3,5

Các câu sau cũng tương tự, dài quá không hi

a) 7(m-11)x-2x+14=5m

<=> 7xm -77x-2x+14=5m

<=> 7xm-79x=5m-14

<=> (7m-79)x=5m-14

* Biện luận pt:

+) Nếu 7m-79=0 <=> m=\(\frac{79}{7}\)<=> 0x=\(\frac{297}{7}\) ( vô lý)

+) Nếu 7m-79\(\ne0\)<=> x=\(\frac{5m-14}{7m-79}\)

Vậy :

Nếu m=\(\frac{79}{7}\) thì pt vô nghiệm.

Nếu m\(\ne\) \(\frac{79}{7}\) thì S = \(\left\{\frac{5m-14}{7m-79}\right\}\)

b) 2xm + 4(2m+1)= m²+ 4 (x-1)

<=> 2xm + 8m + 4= m²+4x-4

<=> 2xm+8m+4-m²-4x+4=0

<=> (2m-4)x -m²+8m+8=0

<=> (2m-4)x=m²-8m-8

*Biện luận:

+) Nếu 2m-4=0 <=> m=2 <=> 0x=-20 (vô lý ) => pt vô nghiệm.

+) Nếu 2m-4 \(\ne0\) <=> x=\(\frac{m^2-8m-8}{2m-4}\)

Vậy :

Nếu m=2 => pt vô nghiệm

Nếu m\(\ne2=>S=\left\{\frac{m^2-8m-8}{2m-4}\right\}\)

a. Với y = 2 ta được:

\(A=\dfrac{x+2}{2-1}\)

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

Ta có pt:

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

\(\Leftrightarrow\dfrac{4\left(x+2\right)}{4}+\dfrac{12}{4}=\dfrac{4x^2+20x}{4}\)

\(\Leftrightarrow4x+8+12=4x^2+20x\)

\(\Leftrightarrow4x+20=4x^2+20x\)

\(\Leftrightarrow-4x^2-16x+20=0\)

\(\Leftrightarrow4x^2+16x-20=0\)

\(\Leftrightarrow\left(4x^2-4x\right)+\left(20x-20\right)=0\)

\(\Leftrightarrow4x\left(x-1\right)+20\left(x-1\right)=0\)

\(\Leftrightarrow\left(4x+20\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)

Vậy..........

a/sửa đề đi

b/\(\Leftrightarrow abx-b^2+2b=2ax+2\)
\(\Leftrightarrow ax\left(b-2\right)-b\left(b-2\right)=2\)

\(\Leftrightarrow\left(ax-b\right)\left(b-2\right)=2\)(*)

PT vô nghiệm khi \(\left[{}\begin{matrix}b=2\\ax=b\end{matrix}\right.\)

Vậy để PT có nghiệm thì \(\left\{{}\begin{matrix}b\ne2\\a\ne0\end{matrix}\right.\)

(*)\(\Leftrightarrow ax-b=\frac{2}{b-2}\)

\(\Leftrightarrow ax=\frac{b^2-2b+2}{b-2}\)

\(\Leftrightarrow x=\frac{b^2-2b+2}{ab-2a}\)

a/ \(\left(m+1\right)^2x=\left(3m+7\right)x+2+m\)

\(\Leftrightarrow\left[\left(m+1\right)^2-\left(3m+7\right)\right]x=m+2\Leftrightarrow\left(m^2-m-6\right)x=m+2\)

* Với \(m=3\Rightarrow x\in\varnothing\)

* Với \(m=-2\Rightarrow x\in R\)

* Với \(m\ne3;m\ne-2\)\(\Rightarrow x=\frac{m+2}{m^2-m-6}=\frac{m+2}{\left(m+2\right)\left(m-3\right)}=\frac{1}{m-3}\)

KL: ...............................

b/ \(b\left(ax-b+2\right)=2\left(ax+1\right)\)

\(\Leftrightarrow\left(ab-2a\right)x=b^2-2b+2\)

Với \(ab-2a=0\Rightarrow b^2-2b+2=0.x\Leftrightarrow x\in\varnothing\)

Với \(ab-2a\ne0\Rightarrow x=\frac{b^2-2b+2}{ab-2a}\)

KL: ..........................

d)

\(x\ne a,x\ne b\)

đặt \(\frac{x-a}{x-b}=t\Leftrightarrow t+\frac{1}{t}=2\Leftrightarrow\frac{t^2-2t+1}{t}=0\Rightarrow t=1\)

\(\frac{x-a}{x-b}=1\Leftrightarrow\frac{\left(x-a\right)-\left(x-b\right)}{x-b}=\frac{b-a}{x-b}=0\)

Vậy: \(a\ne b\) Pt vô nghiệm

a=b phương trinhg nghiệm với mọi x khác a, b

cảm ơn bạn nha

\(a)\) ĐKXĐ: \(a\ne-b;a\ne-c;b\ne-c\)

\(\dfrac{x-ab}{a+b}+\dfrac{x-ac}{a+c}+\dfrac{x-bc}{b+c}=a+b+c\)

\(\Leftrightarrow\left(\dfrac{x-ab}{a+b}-c\right)+\left(\dfrac{x-ac}{a+c}-b\right)+\left(\dfrac{x-bc}{b+c}-a\right)=0\)

\(\Leftrightarrow\dfrac{x-ab-ac-bc}{a+b}+\dfrac{x-ac-ab-bc}{a+c}+\dfrac{x-bc-ab-ac}{b+c}=0\)

\(\Leftrightarrow\left(x-ab-ac-bc\right)\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)=0\)

Vì \(a,b,c>0\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{a+c}+\dfrac{1}{b+c}>0\)

\(\Leftrightarrow x-ab-ac-bc=0\)

\(\Leftrightarrow x=ab+ac+bc\)