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

\(x^2=3\Rightarrow\left[{}\begin{matrix}x=-\sqrt{3}\\x=\sqrt{3}\end{matrix}\right.\)

\(x^2=5\Rightarrow\left[{}\begin{matrix}x=-\sqrt{5}\\x=\sqrt{5}\end{matrix}\right.\Rightarrow x=-\sqrt{5}\left(vì.x< 0\right)\)

\(x^2=7\Rightarrow\left[{}\begin{matrix}x=-\sqrt{7}\\x=\sqrt{7}\end{matrix}\right.\Rightarrow x=-\sqrt{7}\left(vì.x< 0\right)\)

\(x^2=9\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)

\(\left(x-2\right)^2=2\Rightarrow\left[{}\begin{matrix}x-2=-\sqrt{2}\\x-2=\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2-\sqrt{2}\\x=2+\sqrt{2}\end{matrix}\right.\)

\(\left(x-4\right)^2=4\Rightarrow\left[{}\begin{matrix}x-2=-2\\x-2=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

\(\left(x-6\right)^2=6\Rightarrow\left[{}\begin{matrix}x-6=-\sqrt{6}\\x-6=\sqrt{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6-\sqrt{6}\\x=6+\sqrt{6}\end{matrix}\right.\)

\(\left(x-8\right)^2=8\Rightarrow\left[{}\begin{matrix}x-8=-2\sqrt{2}\\x-8=2\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8-2\sqrt{2}\\x=2+2\sqrt{2}\end{matrix}\right.\)

\(\left(x-10\right)^2=10\Rightarrow\left[{}\begin{matrix}x-10=-\sqrt{10}\\x-10=\sqrt{10}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10-\sqrt{10}\\x=10+\sqrt{10}\end{matrix}\right.\)

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

\(\left(x-\sqrt{5}\right)^2=5\Rightarrow\left[{}\begin{matrix}x-\sqrt{5}=-\sqrt{5}\\x-\sqrt{5}=\sqrt{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\sqrt{5}\end{matrix}\right.\)

Bằng 0 hết nha bạn

=1 ; 2 ; 3 lần lượt các câu

1+1-1x2:1=2-2=0

2+2-2x2:2=4-2=2

3+3-3x3:3=9-3=6

Bài 2 

a. (x-2y)2 =2x-4y

b. (2x^2 +3)2 =4x^2+6

c. (x-2) (x^2+2x+4) = x^3-8 (hằng đẳng thức)

d. (2x-1)3 = 6x-3

 Xin lỗi mik chỉ lm ổn bài 2 thôi!

Đặt A = 2x2²x2³x....................x2²⁰¹⁹

=> \(2A=2^2.2^3.2^4...2^{2020}\)

\(\Rightarrow A=2A-A=2^{2020}-2\)

\(2.2^2.2^3...2^{2019}=2^{1+2+3+...+2019}=2^{2029095}\)

X²=3                              x²=25     

=> X=\(\pm\sqrt{3}\)             => x=5

X²=36                           

=> x=6

2.(x-1)²+5⁰= 9

2.(x-1)²+1= 9

2.(x-1)²= 8

(x-1)² = 8/2

(x-1)² = 4 

(x-1)² = (2)²

x-1=(\(\pm\)2)

TH1: x-1= 2              TH2: x-1=-2

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

        x= 3                          x = -1

Vậy x\(\in\)\(\left\{3;1\right\}\)

Bạn chú ý đăng lẻ câu hỏi! 1/

a/ \(=x^3-2x^5\)

b/\(=5x^2+5-x^3-x\)

c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)

d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)

e/ \(=x^4-x^2+2x^3-2x\)

f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)

cmm

bài lạ thật

ý bạn là như thế này đúng không ạ:

a/ \(x^2-6x+5=0\)

\(x^2-5x-x+5=0\)

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

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

\(\orbr{\begin{cases}x-5=0\rightarrow x=5\\x-1=0\rightarrow x=1\end{cases}}\)

b/\(2x^2+7x+9=0\)

?!

c/ \(4x^2-7x+3=0\)

\(4x^2-4x-3x+3=0\)

\(4x\left(x-1\right)-3\left(x-1\right)=0\)

\(\left(x-1\right)\left(4x-3\right)=0\)

\(\orbr{\begin{cases}x-1=0\Rightarrow x=1\\4x-3=0\Rightarrow x=\frac{3}{4}\end{cases}}\)

d/ \(2\left(x+5\right)=2x+10\)

-,- mik ko rõ đề ạ, sai thì ibox ạ.Cảm ơn

x2−6x+5=0

2x2+7x+9=0

4x2−7x+3=0

2(x+5)=x2+5x

a) x=0

b) x=0

c) x=0

d)x=x

a b c d 

x=x

\(x^2-6x+5=0\)

\(\Leftrightarrow x^2-x-5x+5=0\)

\(\Leftrightarrow x\left(x-1\right)-5\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=5\end{cases}}\)

\(2x^2+7x+9=0\)

Đề sai??

\(4x^2-7x+3=0\)

\(\Leftrightarrow4x^2-4x-3x+3=0\)

\(\Leftrightarrow4x\left(x-1\right)-3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\4x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{3}{4}\end{cases}}\)

\(2\left(x+5\right)=x^2+5x\)

\(\Leftrightarrow2x+10=x^2+5x\)

\(\Leftrightarrow x^2+5x-2x-10=0\)

\(\Leftrightarrow x^2+3x-10=0\)

\(\Leftrightarrow x^2+5x-2x-10=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x-2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)