$a,x^2-2=0\iff x^2-{\left(\sqrt{2}\right)}^2=0\iff \left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\iff [\begin{array}{c}x=\sqrt{2}\\ x=-\sqrt{2}\end{array}$

Vậy $S=\left\{-\sqrt{2};\sqrt{2}\right\}$

$b,x\left(x-2\right)=0\iff [\begin{array}{c}x=0\\ x=2\end{array}$

Vậy $S=\left\{0;2\right\}$

$c,x^2-2x=0\iff x\left(x-2\right)$ phương trình như câu b, 

$d,x\left(x^2+1\right)\iff [\begin{array}{c}x=0\\ x^2+1=0\end{array}\iff [\begin{array}{c}x=0\\ x^2=-1\left(voli\right)\end{array}$( voli là vô lí )

Vậy $S=\left\{0\right\}$

$A=\genfrac{}{}{0ex}{}{49}{5}-\left(\genfrac{}{}{0ex}{}{1}{2}+\genfrac{}{}{0ex}{}{1}{4}+\genfrac{}{}{0ex}{}{1}{8}+\genfrac{}{}{0ex}{}{1}{16}+\genfrac{}{}{0ex}{}{1}{32}\right)$

$A=\genfrac{}{}{0ex}{}{49}{5}-\left(1-\genfrac{}{}{0ex}{}{1}{2}+\genfrac{}{}{0ex}{}{1}{2}-\genfrac{}{}{0ex}{}{1}{4}+\genfrac{}{}{0ex}{}{1}{4}-\genfrac{}{}{0ex}{}{1}{8}+\genfrac{}{}{0ex}{}{1}{8}-\genfrac{}{}{0ex}{}{1}{16}+\genfrac{}{}{0ex}{}{1}{16}-\genfrac{}{}{0ex}{}{1}{32}\right)$

$A=\genfrac{}{}{0ex}{}{49}{5}-\left(1-\genfrac{}{}{0ex}{}{1}{32}\right)=\genfrac{}{}{0ex}{}{49}{5}-\genfrac{}{}{0ex}{}{31}{32}$

$A=\genfrac{}{}{0ex}{}{1413}{160}$

32,43: 4,6 = 7,05

PHÉP TÍNH ĐÚNG LÀ: 4,6 X70,5 = 324,3

$A=\genfrac{}{}{0ex}{}{49}{5}-\left(\genfrac{}{}{0ex}{}{1}{2}+\genfrac{}{}{0ex}{}{1}{4}+\genfrac{}{}{0ex}{}{1}{8}+\genfrac{}{}{0ex}{}{1}{16}+\genfrac{}{}{0ex}{}{1}{32}\right)$

$A=\genfrac{}{}{0ex}{}{49}{5}-\left(1-\genfrac{}{}{0ex}{}{1}{2}+\genfrac{}{}{0ex}{}{1}{2}-\genfrac{}{}{0ex}{}{1}{4}+\genfrac{}{}{0ex}{}{1}{4}-\genfrac{}{}{0ex}{}{1}{8}+\genfrac{}{}{0ex}{}{1}{8}-\genfrac{}{}{0ex}{}{1}{16}+\genfrac{}{}{0ex}{}{1}{16}-\genfrac{}{}{0ex}{}{1}{32}\right)$

$A=\genfrac{}{}{0ex}{}{49}{5}-\left(1-\genfrac{}{}{0ex}{}{1}{32}\right)=\genfrac{}{}{0ex}{}{49}{5}-\genfrac{}{}{0ex}{}{31}{32}$

$A=\genfrac{}{}{0ex}{}{1413}{160}$

$A=\genfrac{}{}{0ex}{}{49}{5}-\left(\genfrac{}{}{0ex}{}{1}{2}+\genfrac{}{}{0ex}{}{1}{4}+\genfrac{}{}{0ex}{}{1}{8}+\genfrac{}{}{0ex}{}{1}{16}+\genfrac{}{}{0ex}{}{1}{32}\right)$

$A=\genfrac{}{}{0ex}{}{49}{5}-\left(1-\genfrac{}{}{0ex}{}{1}{2}+\genfrac{}{}{0ex}{}{1}{2}-\genfrac{}{}{0ex}{}{1}{4}+\genfrac{}{}{0ex}{}{1}{4}-\genfrac{}{}{0ex}{}{1}{8}+\genfrac{}{}{0ex}{}{1}{8}-\genfrac{}{}{0ex}{}{1}{16}+\genfrac{}{}{0ex}{}{1}{16}-\genfrac{}{}{0ex}{}{1}{32}\right)$

$A=\genfrac{}{}{0ex}{}{49}{5}-\left(1-\genfrac{}{}{0ex}{}{1}{32}\right)=\genfrac{}{}{0ex}{}{49}{5}-\genfrac{}{}{0ex}{}{31}{32}$

$A=\genfrac{}{}{0ex}{}{1413}{160}$

dây điện, lư đồng, nồi, thau, tượng thờ, trống đồng, kèn

đường kính mặt nước hình tròn là : 

31,4 : 3,14 = 10 (m)

bán kính mặt nước hình tròn là : 

10 : 2 = 5 (m)

diện tích mặt hồ nước hình tròn là : 

5 x 5 x 3,14 = 78,5 (m2)

đáp số : 78,5 m2

chứ ai rảnh mà viết dài như z

chép phải để link cho mik tham kaor vs bn :)))))))))))))))))))))))))))))))))))))))))))))))