em cần câu C thôi ạ

\(\left(3x-5\right)^2-\left(x-1\right)^2=0\)

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

\(\Rightarrow\left(2x-4\right).\left(4x-6\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-4=0\\4x-6=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=4\\4x=6\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{3}{2}\end{cases}}\)

\(16.\left(2-3x\right)+x^2.\left(3x-2\right)=0\)

\(\Rightarrow16.\left(2-3x\right)-x^2.\left(2-3x\right)=0\)

\(\Rightarrow\left(16-x^2\right).\left(2-3x\right)=0\)

\(\Rightarrow\left(4-x\right).\left(4+x\right).\left(2-3x\right)=0\)

Trường hợp 1: \(4-x=0\Rightarrow x=4\)

Trường hợp 2: \(4+x=0\Rightarrow x=-4\)

Trường hợp 3: \(2-3x=0\Rightarrow3x=2\Rightarrow x=\frac{2}{3}\)

\(x^2.\left(x^2+4\right)-x^2-4=0\)

\(\Rightarrow x^2.\left(x^2+4\right)-\left(x^2+4\right)=0\)

\(\Rightarrow\left(x^2+4\right).\left(x^2-1\right)=0\)

\(\Rightarrow\left(x^2+4\right).\left(x-1\right).\left(x+1\right)=0\)

Với mọi \(x\) ta có: \(x^2+4\ge4\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

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

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

\(\Rightarrow\left(x+2\right).\left(3x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+2=0\\3x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{4}{3}\end{cases}}\)

Đặt \(t=sinx\). 

Do \(x\in\left(0,\frac{\pi}{2}\right)\)nên \(t\in\left(0,1\right)\).

\(P=\frac{2}{1-t}+\frac{1}{t}\ge\frac{\left(\sqrt{2}+1\right)^2}{1-t+t}=3+2\sqrt{2}\)

Dấu \(=\)khi \(\frac{\sqrt{2}}{1-t}=\frac{1}{t}\Leftrightarrow t=\sqrt{2}-1\)

kết quả là 1/3628800

1:2:3:4:5:6:7:8:9:10 = 1/3621600

TL;

\(\left(2x+3y^2\right)\)

=  2 + ( 3 + xy² )

= 2 + 3⁴

= 5⁵