\(c.\left(1-2x\right)^2-\left(3x-2\right)^2=0\)

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

\(\left(-5x+3\right)\left(x-1\right)=0\)

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

\(d.\left(x-2\right)^2-\left(5-2x\right)^2=0\)

\(\left(x-2-5+2x\right)\left(x-2+5-2x\right)=0\)

\(\left(3x-7\right)\left(-x+3\right)=0\)

\(\left[{}\begin{matrix}3x-7=0\\-x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)

\(c,\Leftrightarrow1-4x+4x^2=9x^2-12x+4\\ \Leftrightarrow5x^2-8x+3=0\\ \Leftrightarrow\left(x-1\right)\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{5}\end{matrix}\right.\\ d,\Leftrightarrow\left(x-2-5+2x\right)\left(x-2+5-2x\right)=0\\ \Leftrightarrow\left(3x-7\right)\left(3-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)

c: C={0;3;6;9;12;15;18}

c,Theo đầu bài ta có:

x⋮3 và x<20 => x thuộc tập hợp 3,6,9,12,15,18

d,Theo đầu bài ta có:

16⋮x và 0<x<16 => x thuộc tập hợp 4,8

C, x={0;3;6;9;12;15;18}

D, x={1;2;4;8}

\(c,\Rightarrow\left(x-2\right)-\left(x-2\right)^2=0\\ \Rightarrow\left(x-2\right)\left(1-x+2\right)=0\\ \Rightarrow\left(x-2\right)\left(3-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\\ d,\Rightarrow\left(x^2+3\right)\left(x+1\right)+\left(x+1\right)=0\\ \Rightarrow\left(x^2+3+1\right)\left(x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2+4=0\left(vô.nghiệm\right)\\x+1=0\end{matrix}\right.\Rightarrow x=-1\)

c. \(2\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)-\dfrac{3}{2}=\dfrac{1}{4}\)

\(2\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{1}{4}+\dfrac{3}{2}=\dfrac{7}{4}\)

\(\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{4}:2=\dfrac{7}{8}\)

\(\dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}=\dfrac{29}{24}\)

\(x=\dfrac{29}{24}:\dfrac{1}{2}=\dfrac{29}{12}\)

Vậy : ...

d. \(\dfrac{4}{5}-\dfrac{1}{2}x=\dfrac{1}{10}\)

\(-\dfrac{1}{2}x=\dfrac{1}{10}-\dfrac{4}{5}=-\dfrac{7}{10}\)

\(x=-\dfrac{7}{10}:\left(-\dfrac{1}{2}\right)=\dfrac{7}{5}\)

Vậy : ...

c) \(2\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)-\dfrac{3}{2}=\dfrac{1}{4}\)

\(\Rightarrow2\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{1}{4}+\dfrac{3}{2}\)

\(\Rightarrow2\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{1}{4}+\dfrac{6}{4}\)

\(\Rightarrow2\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{7}{4}\)

\(\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{7}{4}:2\)

\(\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{7}{4}.\dfrac{1}{2}\)

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

\(\Rightarrow\dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\)

\(\Rightarrow\dfrac{1}{2}x=\dfrac{21}{24}+\dfrac{8}{24}\)

\(\Rightarrow\dfrac{1}{2}x=\dfrac{29}{24}\)

\(\Rightarrow x=\dfrac{29}{24}:\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{29}{24}.2\)

\(\Rightarrow x=\dfrac{29}{12}\)

d) \(\dfrac{4}{5}-\dfrac{1}{2}x=\dfrac{1}{10}\)

\(\Rightarrow\dfrac{1}{2}x=\dfrac{4}{5}-\dfrac{1}{10}\)

\(\Rightarrow\dfrac{1}{2}x=\dfrac{8}{10}-\dfrac{1}{10}\)

\(\Rightarrow\dfrac{1}{2}x=\dfrac{7}{10}\)

\(\Rightarrow x=\dfrac{7}{10}:\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{7}{10}.2\)

\(\Rightarrow x=\dfrac{7}{5}\)

c) 140:(x-8)=14

            x-8  =140:14

            x-8  =10

            x     =10+8

            x     =18

vậy...

d) 11(x-9)=88

         x-9  =88:11

         x-9  =8

         x     =8+9

         x     =17

vậy...

18,17 nha bạn

d) 11(𝑥−9)=88

     11𝑥−99=88

    11x-99=8811x−99=88

    11𝑥−99+99=88+99

    11𝑥=187

    11x/11 = 187/11

    𝑥=17

HT~~~~

`c,(x-2)(2x-1)-(2x-3)(x-1)-2`

`=2x^2-x-4x+2-2x^2+2x+3x-3-2`

`=-3`

`->` Biểu thức không phụ thuộc vào biến `x`

\(C=\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\left(x^{32}+1\right)-x^{64}\\ =\left(x^4-1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\left(x^{32}+1\right)-x^{64}\\ =\left(x^8-1\right)\left(x^8+1\right)\left(x^{16}+1\right)\left(x^{32}+1\right)-x^{64}\\ =\left(x^{16}-1\right)\left(x^{16}+1\right)\left(x^{32}+1\right)-x^{64}\\ =\left(x^{32}-1\right)\left(x^{32}+1\right)-x^{64}\\ =\left(x^{64}-1\right)-x^{64}\\ =-1\)

Vậy đa thức ko phụ thuộc vào x

\(C=(x^2-1)(x^2+1)(x^4+1)(x^8+1)(x^{16}+1)(x^{32}+1)-x^{64}\\=(x^4-1)(x^4+1)(x^8+1)(x^{16}+1)(x^{32}+1)-x^{64}\\=(x^8-1)(x^8+1)(x^{16}+1)(x^{32}+1)-x^{64}\\=(x^{16}-1)(x^{16}+1)(x^{32}+1)-x^{64}\\=(x^{32}-1)(x^{32}+1)-x^{64}\\=x^{64}-1-x^{64}\\=-1\)

⇒ Giá trị của C không phụ thuộc vào giá trị của biến

\(C=\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\left(x^{32}+1\right)-x^{64}\)

\(C=\left(x^4-1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\left(x^{32}+1\right)-x^{64}\)

\(C=\left(x^8-1\right)\left(x^8+1\right)\left(x^{16}+1\right)\left(x^{32}+1\right)-x^{64}\)

\(C=\left(x^{16}+1\right)\left(x^{16}-1\right)\left(x^{32}+1\right)-x^{64}\)

\(C=\left(x^{32}-1\right)\left(x^{32}+1\right)-x^{64}\)

\(C=x^{64}-1-x^{64}\)

\(C=-1\)

Vậy: ... 

K CHO MIK TRƯỚC ĐI R MIK TRẢ LỜI CHO NHÁK

hay phết tí nữa minh giải cho nhé

A

Chọn A

A