a) \(\left(x-5\right)^{12}=\left(x-5\right)^{10}\)

\(\Rightarrow\left(x-5\right)^{12}-\left(x-5\right)^{10}=0\)

\(\Rightarrow\left(x-5\right)^{10}\left[\left(x-5\right)^2-1\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x-5\right)^{10}=0\\\left(x-5\right)^2-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\left(x-5\right)^{10}=0^{10}\\\left(x-5\right)^2=0+1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\\left(x-5\right)^2=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0+5\\\left(x-5\right)^2=1^2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x-5=\pm1\end{cases}}\)

\(\Rightarrow x=5;\orbr{\begin{cases}x-5=1\\x-5=-1\end{cases}}\)

\(\Rightarrow x=5;\orbr{\begin{cases}x=1+5\\x=-1+5\end{cases}}\)

\(\Rightarrow x=5;\orbr{\begin{cases}x=4\\x=6\end{cases}}\)

Vậy x = 4 hoặc x = 5 hoặc x = 6 

\(a)\left(x-5\right)^{12}=\left(x-5\right)^{10}\)

\(\Leftrightarrow\left(x-5\right)^{12}-\left(x-5\right)^{10}=0\)

\(\Leftrightarrow\left(x-5\right)^{10}\left[\left(x-5\right)^2-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-5\right)^{10}=0\\\left(x-5\right)^2-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\\left(x-4\right)\left(x-6\right)=0\end{cases}}\)

[  ra \(\left(x-4\right)\left(x-6\right)\)do \(\left(x-5\right)^2-1=\left(x-5-1\right)\left(x-5+1\right)=\left(x-6\right)\left(x-4\right)\)]

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=4;x=6\end{cases}}\)

_Minh ngụy_

Câu b thiếu '= ......"

\(\(c)\frac{1}{2.4}+\frac{1}{4.6}+.....+\frac{1}{2x\left(2x+2\right)}=\frac{3}{16}\)\)

\(\(\Rightarrow\frac{1}{4.\left(1.2\right)}+\frac{1}{4.\left(2.3\right)}+......+\frac{1}{4\left(x-1\right).x}=\frac{3}{16}\)\)

\(\(\Leftrightarrow\frac{1}{4}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{2.3}+....+\frac{1}{x\left(x-1\right)}\right)=\frac{3}{16}\)\)

\(\(\Rightarrow\frac{2-1}{1.2}+\frac{3-2}{2.3}+.......+\frac{x-\left(x-1\right)}{x\left(x-1\right)}=\frac{3}{4}\)\)

\(\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-........+\frac{1}{x-1}-\frac{1}{x}=\frac{3}{4}\)\)

\(\(\Leftrightarrow1-\frac{1}{x}=\frac{3}{4}\)\)

\(\(\Leftrightarrow\frac{1}{x}=\frac{1}{4}\)\)

\(\(\Rightarrow x=4\left(t/m:x\ne0\right)\)\)

_Minh ngụy_

a) \(\left(7-14x\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}7-14x=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}14x=7-0\\x=2\end{cases}\Leftrightarrow}}\orbr{\begin{cases}14x=7\\x=2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=2\end{cases}}}\)

Vậy \(x\in\left\{\frac{1}{2};2\right\}\)

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

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=-1\\x=3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}}\)

Vậy \(x\in\left\{-\frac{1}{2};3\right\}\)

^x2-3.(x-1)

(x-1)²

^=>x2-3

x-1

= x³ + 3³ -x(x² -1) -27 =0 ( tổng các lập phuong)

x =0 

CX100%

bạn chỉ cần phá hết hằng đẳng thức ra thôi 

câu 1

(x+3)^2+(x-2)(x+2)-2(x-1)^2=7

<=> x^2+6x+9+x^2-4-2x^2-4x-2-7=0

<=> x^2+x^2-2x^2+6x-4x+9-4-2-7=0

<=> 2x-4=0

<=> x=2

câu 2

(x+3)(x^2-3x+9)-x(x-1)(x+1)-27=0

<=> x^3+27-x(x^2-1)-27=0

<=> x^3-x^3+x+27-27=0

<=>x=0