Lời giải;
Từ PT(2) suy ra $5x^2=\frac{16y^2}{5}$

$\Leftrightarrow x^2=(\frac{4}{5}y)^2$

$\Leftrightarrow x=\frac{4}{5}y$ hoặc $x=\frac{-4}{5}y$

Nếu $x=\frac{4}{5}y$ thì: thay vào PT(1):

$4y+\frac{16}{y}=360$

$\Leftrightarrow y+\frac{4}{y}=90$

$\Leftrightarrow y^2-90y+4=0$

$\Leftrightarrow (y-45)^2=2021$

$\Leftrightarrow y-45=\pm \sqrt{2021}$

$\Leftrightarrow y=45\pm \sqrt{2021}$

$\Rightarrow  x=36\pm \frac{4}{5}\sqrt{2021}$ (tương ứng)

Trường hợp $x=\frac{-4}{5}y$ giải tương tự.

1.

PT $\Leftrightarrow 2^{x^2-5x+6}+2^{1-x^2}-2^{7-5x}-1=0$

$\Leftrightarrow (2^{x^2-5x+6}-2^{7-5x})-(1-2^{1-x^2})=0$

$\Leftrightarrow 2^{7-5x}(2^{x^2-1}-1)-(2^{x^2-1}-1)2^{1-x^2}=0$

$\Leftrightarrow (2^{x^2-1}-1)(2^{7-5x}-2^{1-x^2})=0$

$\Rightarrow 2^{x^2-1}-1=0$ hoặc $2^{7-5x}-2^{1-x^2}=0$

Nếu $2^{x^2-1}=1\Leftrightarrow x^2-1=0$

$\Leftrightarrow x^2=1\Leftrightarrow x=\pm 1$

$2^{7-5x}-2^{1-x^2}=0$

$\Leftrightarrow 7-5x=1-x^2\Leftrightarrow x^2-5x+6=0$

$\Leftrightarrow (x-2)(x-3)=0\Leftrightarrow x=2; x=3$

2. Đặt $\sin ^2x=a$ thì $\cos ^2x=1-a$. PT trở thành:

$16^a+16^{1-a}=10$

$\Leftrightarrow 16^a+\frac{16}{16^a}=10$

$\Leftrightarrow (16^a)^2-10.16^a+16=0$

Đặt $16^a=x$ thì:

$x^2-10x+16=0$

$\Leftrightarrow (x-2)(x-8)=0$

$\Leftrightarrow x=2$ hoặc $x=8$

$\Leftrightarrow 16^a=2$ hoặc $16^a=8$

$\Leftrightarrow 2^{4a}=2$ hoặc $2^{4a}=2^3$

$\Leftrightarroww 4a=1$ hoặc $4a=3$

$\Leftrightarrow a=\frac{1}{4}$ hoặc $a=\frac{3}{4}$

Nếu $a=\frac{1}{4}\Leftrightarrow \sin ^2x=\frac{1}{4}$

$\Leftrightarrow \sin x=\pm \frac{1}{2}$

Nếu $a=\sin ^2x=\frac{3}{4}\Rightarrow \sin x=\pm \frac{\sqrt{3}}{2}$

Đến đây thì đơn giản rồi.

Thay \(5=x+1\) vào biểu thức B ,có :

\(x^{19}-\left(x+1\right)x^{18}+\left(x+1\right)x^{17}-\left(x+1\right)x^{16}+...-\left(x+1\right)x^2+\left(x+1\right)x+1886\)

\(=x^{19}-x^{19}-x^{18}+x^{18}+x^{17}-x^{17}+x^{16}+...-x^3-x^2+x^2+x+1886\)

\(=x+1886\)

\(=4+1886\)

\(=1890\)

Vậy tại x = 4 giá trị của biểu thức B là 1890

\(\text{Ta có : }x=4\Rightarrow\left\{{}\begin{matrix}x+1\\-x+1890\end{matrix}\right.\\ \Rightarrow B=x^{19}-5x^{18}+5x^{17}-...-5x^2+5x+1886\\ =x^{19}-\left(x+1\right)x^{18}+\left(x+1\right)x^{17}-...-\left(x+1\right)x^2+\left(x+1\right)x-x+1890\\ =x^{19}-x^{19}-x^{18}+x^{18}+x^{17}-...-x^3-x^2+x^2+x-x+1890\\ \\=1890\)

5x+x+x=16-x

=> 7x=16-x

=> 8x=16

=> x=2

\(5x+x+x=16-x\)

\(7x=16-x\)

\(7x+x=16\)

\(8x=16\)

\(x=\frac{16}{8}\)

\(x=2\)

5x + x+ x = 16 - x 

5x + x + x + x = 16 

5x + x * 1 + x *1 + x*1 = 16

x * ( 5 + 1 + 1 + 1 ) = 16

x * 8 =16 

x = 16 : 8 

x = 2 

1) \(2x^4+3x^3-x^2+3x+2=0\)

\(\Rightarrow2x^4+x^3+2x^3+x^2-2x^2-x+4x+2=0\)

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

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

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

\(\Rightarrow\left(2x+1\right)\left[x^2\left(x+2\right)-x\left(x+2\right)+\left(x+2\right)\right]=0\)

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

\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\x+2=0\\x^2-x+1=0\end{matrix}\right.\)

Ta có:

\(x^2-x+1\)

\(=x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+1\)

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

Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\) với mọi x

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\) với mọi x

\(\Rightarrow x^2-x+1\) vô nghiệm

\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\x+2=0\end{matrix}\right.\)

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

3) \(\left(x+2\right)^4+\left(x+4\right)^4=16\)

Đặt x + 3 = a, ta được

\(\left(a-1\right)^4+\left(a+1\right)^4=16\)

\(\Rightarrow\left[\left(a-1\right)^2\right]^2+\left[\left(a+1\right)^2\right]^2=16\)

\(\Rightarrow\left(a^2-2a+1\right)^2+\left(a^2+2a+1\right)^2=16\)

\(\Rightarrow a^4+4a^2+1+2a^2-4a^3-4a+a^4+4a^2+1+2a^2+4a^3+4a=16\)

\(\Rightarrow2a^4+2.4a^2+2+2.2a^2=16\)

\(\Rightarrow2a^4+8a^2+4a^2+2=16\)

\(\Rightarrow2a^4+12a^2+2-16=0\)

\(\Rightarrow2a^4+12a^2-14=0\)

\(\Rightarrow2a^4-2a^2+14a^2-14=0\)

\(\Rightarrow2a^2\left(a^2-1\right)+14\left(a^2-1\right)=0\)

\(\Rightarrow\left(a^2-1\right)\left(2a^2+14\right)=0\)

\(\Rightarrow\left(a-1\right)\left(a+1\right).2\left(a^2+7\right)=0\)

\(\Rightarrow\left(a-1\right)\left(a+1\right)\left(a^2+7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a-1=0\\a+1=0\\a^2+7=0\end{matrix}\right.\)

Vì \(a^2\ge0\) với mọi a

\(\Rightarrow a^2+7\ge7\) với mọi a

\(\Rightarrow a^2+7\) vô nghiệm

\(\Rightarrow\left[{}\begin{matrix}a-1=0\\a+1=0\end{matrix}\right.\)

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

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

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

\(\Rightarrow x^2-5x-36=0\Rightarrow x\left(x-9\right)+4\left(x-9\right)=0\Rightarrow\left(x-9\right)\left(x+4\right)=0\Rightarrow\left[{}\begin{matrix}x=9\\x=-4\end{matrix}\right.\)

quy đồng mẫu số hộ mình

\(\dfrac{x}{4x-16}=\dfrac{x}{4\left(x-4\right)}=\dfrac{-5x}{20\left(4-x\right)}\)

\(\dfrac{1+x}{20-5x}=\dfrac{1+x}{5\left(4-x\right)}=\dfrac{4+4x}{20\left(4-x\right)}\)

\(\dfrac{x}{4x-16}=\dfrac{x}{4\left(x-4\right)}=\dfrac{5x}{20\left(x-4\right)}\\ \dfrac{1+x}{20-5x}=\dfrac{-\left(x+1\right)}{5\left(x-4\right)}=\dfrac{-4\left(x+1\right)}{20\left(x-4\right)}\)

\(\left(\dfrac{4}{3}\right)^{5x}.\left(\dfrac{3}{16}\right)^{5x}=0\)

\(\left(\dfrac{4}{3}.\dfrac{3}{16}\right)^{5x}=0\)

\(\left(\dfrac{1}{4}\right)^{5x}=0\)

⇔\(\dfrac{1}{4}=0\) (vô lí)

⇒không tồn tại x