Lời giải:

a. Gọi 2 số cần tìm là $a$ và $b$

Theo bài ra: 

$a+b=2\Rightarrow b=2-a$

$ab=\frac{3}{4}$

Thay $b=2-a$ thì:

$a(2-a)=\frac{3}{4}$

$\Leftrightarrow a^2-2a+\frac{3}{4}=0$

$\Leftrightarrow (a-\frac{3}{2})(a-\frac{1}{2})=0$

$\Leftrightarrow a=\frac{3}{2}$ hoặc $a=\frac{1}{2}$

Nếu $a=\frac{3}{2}$ thì $b=2-a=\frac{1}{2}$

Nếu $a=\frac{1}{2}$ thì $b=2-a=\frac{3}{2}$

b,c: Tương tự

d. 

Gọi hai số cần tìm là $a$ và $b$

Theo bài ra ta có:

$ab=12$

$a^2+b^2=25$

$\Leftrightarrow (a+b)^2-2ab=25$

$\Leftrightarrow (a+b)^2=25+2ab=25+2.12=49$

$\Leftrightarrow a+b=\pm 7$

Đến đây lại đưa về dạng tìm 2 số biết tổng và tích giống như phần a.

1997²-1996.1998-1

= 1997.1997-(1997-1).(1997+1)-1

= 1997.1997-(1997.1997+1997-1997-1)-1

= 1997.1997-1997.1997-1997+1997+1-1

= 0

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

\(5x^3-15x^2+5x+x-5x^2-x+2=0\)

\(5x^3-20x^2+5x+2=0\)

\(5x\left(x^2-4x+1\right)+2=0\)

Chịu .... 

\(3x\left(\dfrac{4}{3x}+1\right)-4x\left(x-2\right)=10\\ \Rightarrow4+3x-4x^2+8x-10=0\\\Rightarrow -4x^2+11x-6=0\\ \Rightarrow-4x^2+8x+3x-6=0\\\Rightarrow \left(-4x+3\right)\left(x-2\right)=0\\\Rightarrow \left[{}\begin{matrix}x=2\\x=\dfrac{3}{4}\end{matrix}\right.\)

Ta có :

x²+2x-y²+2y = x² - y² + 2x+2y

= (x-y)(x+y) + 2(x+y) 

= (x-y+2)(x+y)

a, \(\dfrac{2-5x}{3}+\dfrac{2x^2-x}{2}>\dfrac{x\left(3x-1\right)}{3}-\dfrac{5x}{4}\)

\(\Rightarrow8-20x+12x^2-6x>4x\left(3x-1\right)-15x\)

\(\Leftrightarrow8-26x+12x^2>12x^2-19x\Leftrightarrow8>7x\Leftrightarrow x< \dfrac{8}{7}\)

b, \(\Rightarrow20x+10x+5>30x-2\Leftrightarrow5>-2\)(luôn đúng) 

Vậy bft luôn có nghiệm 

Ta có \(\left(x-5\right)^4+\left(x-2\right)^4=1^4+2^4=2^4+1^4\)

TH1 \(\left\{{}\begin{matrix}\left(x-5\right)^4=1^4\\\left(x-2\right)^4=2^4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-5=1\\x-5=-1\end{matrix}\right.\\\left[{}\begin{matrix}x-2=2\\x-2=-2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=6\\x=4\end{matrix}\right.\\\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\end{matrix}\right.\)

TH2 \(\left\{{}\begin{matrix}\left(x-5\right)^4=2^4\\\left(x-2\right)^4=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\\\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\\\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\end{matrix}\right.\)

Đặt $x-\frac{7}{2}=a$. Khi đó PT trở thành:

$(a-\frac{3}{2}{)}^4+(a+\frac{3}{2}{)}^4=17$

$\iff 2a^4+27a^2+\frac{81}{8}=17$

$\iff 2a^4+27a^2=\frac{55}{8}$

$\iff a^4+\frac{27}{2}{a}^2=\frac{55}{16}$

$\iff (a^2+\frac{27}{4}{)}^2=49$

$\Rightarrow [\begin{array}{c}{a}^2+\frac{27}{4}=7\\ a^2+\frac{27}{4}=-7<0\left(\text{vô lý}\right)\end{array}$

$\Rightarrow a^2=\frac{1}{4}\Rightarrow a=\pm \frac{1}{2}$

$\Rightarrow x=a+\frac{7}{2}=[\begin{array}{c}4\\ 3\end{array}$