b: \(B=2x\left(x-3\right)-\left(2x-2\right)\left(x-2\right)\)

\(=2x^2-6x-2x^2+4x+2x-4\)

=-4

b: \(\Leftrightarrow\dfrac{-3x^2+36x+12}{3\left(x+4\right)\left(x-1\right)}=\dfrac{36\left(x-1\right)}{3\left(x+4\right)\left(x-1\right)}+\dfrac{12\left(x+4\right)}{3\left(x-1\right)\left(x+4\right)}\)

\(\Leftrightarrow-3x^2+36x+12=36x-36+12x+48\)

\(\Leftrightarrow-3x^2+36x+12-48x-12=0\)

\(\Leftrightarrow3x\left(x+4\right)=0\)

=>x=0(nhận) hoặc x=-4(loại)

b: \(\Leftrightarrow\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}=9+\dfrac{5}{7}-\dfrac{5}{7}=9\)

=>x-1/2=27

hay x=55/2

c: =>1/2x-3/4=42/63=2/3

=>1/2x=17/12

hay x=17/6

\(a,=4x^3\left(x+1\right)-x\left(x+1\right)=x\left(4x^2-1\right)\left(x+1\right)\\ =x\left(2x-1\right)\left(2x+1\right)\left(x+1\right)\\ b,=\left(a-1\right)^2-\left(b-c\right)^2\\ =\left(a-1-b+c\right)\left(a-1+b-c\right)\\ c,=\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72\\ =\left(x^2-9x+17\right)^2-9-72\\ =\left(x^2-9x+17\right)^2-81=\left(x^2-9x+8\right)\left(x^2-9x+26\right)\\ =\left(x-1\right)\left(x-8\right)\left(x^2-9x+26\right)\)

a: \(\left|7-2x\right|+7=2x\)

=>\(\left|2x-7\right|+7=2x\)

=>\(\left|2x-7\right|=2x-7\)

=>2x-7>=0

=>\(x>=\dfrac{7}{2}\)

b: \(\left|1-x\right|=4x+1\)

=>\(\left|x-1\right|=4x+1\)

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

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

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

=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\5x\left(3x+2\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)

c: \(\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\left|3,2+\dfrac{2}{5}\right|\)

=>\(\left|x-\dfrac{1}{3}\right|=\dfrac{16}{5}+\dfrac{2}{5}-\dfrac{4}{5}=\dfrac{14}{5}\)

=>\(\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{14}{5}\\x-\dfrac{1}{3}=-\dfrac{14}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}+\dfrac{1}{3}=\dfrac{42+5}{15}=\dfrac{47}{15}\\x=-\dfrac{14}{5}+\dfrac{1}{3}=\dfrac{-42+5}{15}=-\dfrac{37}{15}\end{matrix}\right.\)

d: \(\left|x-7\right|+2x+5=6\)

=>\(\left|x-7\right|=6-2x-5=-2x+1\)

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

=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(2x-1\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(2x-1+x-7\right)\left(2x-1-x+7\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(3x-8\right)\left(x+6\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left[{}\begin{matrix}x=\dfrac{8}{3}\left(loại\right)\\x=-6\left(nhận\right)\end{matrix}\right.\end{matrix}\right.\)

e: 3x-|2x-1|=2

=>|2x-1|=3x-2

=>\(\left\{{}\begin{matrix}3x-2>=0\\\left(3x-2\right)^2=\left(2x-1\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(3x-2\right)^2-\left(2x-1\right)^2=0\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(x-1\right)\left(5x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left[{}\begin{matrix}x-1=0\\5x-3=0\end{matrix}\right.\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left[{}\begin{matrix}x=1\left(nhận\right)\\x=\dfrac{3}{5}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)

a)ĐKXĐ: \(x\notin\left\{0;-1\right\}\)

Ta có: \(\dfrac{x-1}{x}+\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}+\dfrac{x}{x\left(x+1\right)}=\dfrac{2x-1}{x\left(x+1\right)}\)

Suy ra: \(x^2-1+x-2x+1=0\)

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

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

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

Vậy: S={1}

b) ĐKXĐ: \(x\notin\left\{3;-3\right\}\)

Ta có: \(\dfrac{5}{x-3}-\dfrac{2x-3}{x+3}=\dfrac{2x\left(1-x\right)}{x^2-9}\)

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

Suy ra: \(5x+15-2x^2+6x+3x-9-2x+2x^2=0\)

\(\Leftrightarrow12x+6=0\)

\(\Leftrightarrow12x=-6\)

hay \(x=-\dfrac{1}{2}\)(thỏa ĐK)

Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)

a) 5(2x -1) - 4(8 - 3x) = 7

<=> 10x - 5 - 32 + 12x = 7

<=> 22x = 44 

<=> x =2

Vậy x = 2 là nghiệm phương trình

b) 7(2x - 5) - 5(7x - 2) + 2(5x - 7) = (x - 2) - (x + 4) 

<=> 14x - 35 - 35x + 10 + 10x - 14 = x - 2 - x - 4

<=> -11x - 39 = - 6

<=> -11x = 33

<=> x = -3

Vậy x = -3 là nghiệm phương trình 

\(a,10x-5-32+12x=7\)

\(22x=44\)

\(x=2\)

\(b,14x-35-35x+10+10x-14=x-2-x-4\)

\(-11x-39=-6\)

\(-11x=-33\)

\(x=3\)

sửa lại đoạn cuối

\(-11x=33\)

\(x=-3\)

\(a,=5x-10+2x+6=7x-4\\ b,=x^2+2x+1-x^2+3x+10=5x+11\\ c,=x^2-49-x^2+1=-48\\ d,\text{Đề có sai ko vậy?}\)