a.

ĐKXĐ: \(x\ne6\)

\(\dfrac{7}{x-6}=\dfrac{x-6}{7}\)

\(\Leftrightarrow\dfrac{49}{7\left(x-6\right)}=\dfrac{\left(x-6\right)^2}{7\left(x-6\right)}\)

\(\Rightarrow\left(x-6\right)^2=49=7^2\)

\(\Rightarrow\left[{}\begin{matrix}x-6=7\\x-6=-7\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=13\\x=-1\end{matrix}\right.\) (thỏa mãn)

b. ĐKXĐ: \(x\ne\dfrac{1}{2}\)

\(\dfrac{2x-1}{8}=\dfrac{-2}{1-2x}\)

\(\Leftrightarrow\dfrac{\left(2x-1\right)^2}{8\left(2x-1\right)}=\dfrac{16}{8\left(2x-1\right)}\)

\(\Rightarrow\left(2x-1\right)^2=16=4^2\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\) (thỏa mãn)

a. \(\dfrac{8}{7}-\dfrac{1}{7}:\left(\dfrac{x}{3}-2\right)=-1\)

\(-\dfrac{1}{7}:\left(\dfrac{x}{3}-2\right)=\dfrac{15}{7}\)

\(\dfrac{x}{3}-2=\dfrac{-1}{15}\)

\(\dfrac{x}{3}=\dfrac{29}{15}\)

\(x=5,8\)

b. \(\dfrac{5}{8}+\dfrac{1}{4}\left(2x-1\right)=\dfrac{5}{4}\)

\(\dfrac{1}{4}\left(2x-1\right)=\dfrac{5}{8}\)

\(2x-1=\dfrac{5}{2}\)

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

\(x=\dfrac{7}{4}\)

Bài 5: 

a: \(8A=8+8^2+...+8^8\)

\(\Leftrightarrow7A=8^8-1\)

hay \(A=\dfrac{8^8-1}{7}\)

b: \(8B=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)

\(\Leftrightarrow8B=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\)

\(\Leftrightarrow8B=3^{16}-1\)

hay \(B=\dfrac{3^{16}-1}{8}\)

a) x(2x-7)-2x(x+1)=7

<=> 2x² - 7x-2x²-2x=7

<=> -9x=7

<=> x=\(\frac{-7}{9}\)

b) 3x(x+8)-x²-2x(x+1)=2

<=> 3x²+24x-x²-2x²-2x=2

<=> 22x=2

<=> x=\(\frac{1}{11}\)

a/\(\dfrac{7}{2}-2x=5\dfrac{1}{3}:\dfrac{8}{3}\)
\(\dfrac{7}{2}-2x=\dfrac{16}{3}:\dfrac{8}{3}\)
\(\dfrac{7}{2}-2x=2\)
\(2x=\dfrac{7}{2}-2\)
\(2x=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}:2\)
\(x=\dfrac{3}{4}\)
b/Đề là gì ạ?

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

\(\Rightarrow\left(x+1\right)^2-3\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right).\left(x+1-3\right)=0\)

\(\Rightarrow\left(x+1\right).\left(x-2\right)=0\)

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

Vậy \(x\in\left\{-1;2\right\}\)

b, \(\left(2x-7\right)^3=8\left(7-2x\right)^2\)

\(\Rightarrow\left(2x-7\right)^3-8\left(2x-7\right)^2=0\) (do \(\left[A\left(x\right)\right]^2=\left[-A\left(x\right)\right]^2\))

\(\Rightarrow\left(2x-7\right)^2.\left(2x-7-8\right)=0\)

\(\Rightarrow\left(2x-7\right)^3.\left(2x-15\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(2x-7\right)^3=0\\2x-15=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x-7=0\\2x=15\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x=7\\x=7,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3,5\\x=7,5\end{matrix}\right.\)

Vậy \(x\in\left\{3,5;7,5\right\}\)

Chúc bạn học tốt!!!

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

\(-11x=33\)

\(x=-3\)