\(a.3x-5y+1=3.\dfrac{1}{3}-5.\left(-\dfrac{1}{5}\right)+1=1+1+1=3\)

b.x=1

\(\Rightarrow3.1^2-2.1-5=-4\)

x=-1

\(\Rightarrow3.\left(-1\right)^2-2.\left(-1\right)-5=3+2-5=0\)

a.\(x=0;y=-1\)

\(\Rightarrow2.0-\dfrac{-1\left(0^2-2\right)}{0.-1-1}=0-\dfrac{2}{-1}=2\)

b.\(x=2\)

\(\Rightarrow4.2^2-3\left|2\right|-2=16-6-2=8\)

\(x=-3\)

\(\Rightarrow4.\left(-3\right)^2-3\left|-3\right|-2=36-9-2=25\)

c.\(x=-\dfrac{1}{5};y=-\dfrac{3}{7}\)

\(\Rightarrow5.\left(-\dfrac{1}{5}\right)^2-7.\left(-\dfrac{3}{7}\right)+6=5.\dfrac{1}{25}+3+6=\dfrac{1}{5}+3+6=\dfrac{46}{5}\)

thay x=2 và biểu thức A ta đc

\(A=4.2^2-3.\left|2\right|-2=4.4-6-2=16-6-2=8\)

thay x=-3  biểu thức A ta đc

\(A=4.\left(-3\right)^2-3.\left|-3\right|-2=4.9-9-2=36-9-2=25\)

thay x=-1/5 ; y=-3/7  biểu thức B ta đc

\(B=5.\left(-\dfrac{1}{5}\right)^2-7.\left(-\dfrac{3}{7}\right)+6\)

\(B=5\cdot\dfrac{1}{25}+3+6\)

\(B=\dfrac{1}{5}+3+6=\dfrac{46}{5}\)

 thay x =0 , y= -1 và biểu thức ta  đc

\(0-\dfrac{\left(-1\right)\left(0^2-2\right)}{0.\left(-1\right)+\left(-1\right)}=0-\dfrac{2}{-1}=0+2=2\)

a) Thay `x=1/2` vào A được:

`A=(5. 1/2 -7)(2. 1/2 +3)-(7 . 1/2 +2)(1/2 -4)=5/4`

b) Thay `x=2;y=-2` vào B được:

`B=(2+2.2)(-2-2.2)+(2-2.2)(-2+2.2)=-40`.

a) Với \(x=\dfrac{1}{2}\) ta được:

\(\Leftrightarrow A=\left(\dfrac{5.1}{2}-7\right)\left(\dfrac{2.1}{2}+3\right)-\left(\dfrac{7.1}{2}+2\right)\left(\dfrac{1}{2}-4\right)\)

\(\Leftrightarrow A=-\dfrac{9}{2}.4-\dfrac{11}{2}.\left(-\dfrac{7}{2}\right)\)

\(\Rightarrow A=\dfrac{5}{4}\)

b) Với \(x = 2; y = - 2 \) ta được :

\(\Leftrightarrow B=\left(2-2\left(-2\right)\right)\left(\left(-2\right)-2.2\right)+\left(2+2\left(-2\right)\right)\left(\left(-2\right)+2.2\right)\)

\(\Leftrightarrow B=-40\)

\(a,N=\dfrac{x^2+xy+y^2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\left(x-y\right)\left(x^4-y^4\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\\ N=\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x+y\right)}=x^2+y^2\\ b,N=\left(x+y\right)^2-2xy=0-2\cdot1=-2\)

ĐKXĐ: \(x\ne y\)

a) \(N=\dfrac{x^2+y\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}:\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{x^4\left(x-y\right)-y^4\left(x-y\right)}=\dfrac{x^2+xy+y^2}{\left(x-y\right)\left(x+y\right)}.\dfrac{\left(x-y\right)^2\left(x+y\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}=x^2+y^2\)

b) \(x+y=0\Leftrightarrow\left(x+y\right)^2=0\Leftrightarrow x^2+y^2-2xy=0\)

\(\Leftrightarrow N=x^2+y^2=0+2xy=2.1=2\)

Bài 1: 

b) ĐKXĐ: \(x\ne3\)

Ta có: \(\dfrac{3-x}{20}=\dfrac{-5}{x-3}\)

\(\Leftrightarrow\dfrac{x-3}{-20}=\dfrac{-5}{x-3}\)

\(\Leftrightarrow\left(x-3\right)^2=100\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=10\\x-3=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\left(nhận\right)\\x=-7\left(nhận\right)\end{matrix}\right.\)

Vậy: \(x\in\left\{13;-7\right\}\)

Bài 1:

|\(x\)| = 1 ⇒ \(x\) \(\in\) {-\(\dfrac{1}{3}\); \(\dfrac{1}{3}\)}

A(-1) = 2(-\(\dfrac{1}{3}\))² - 3.(-\(\dfrac{1}{3}\)) + 5

A(-1) = \(\dfrac{2}{9}\) + 1 + 5

A (-1) = \(\dfrac{56}{9}\)

A(1) = 2.(\(\dfrac{1}{3}\) )²- \(\dfrac{1}{3}\).3 + 5

A(1) = \(\dfrac{2}{9}\) - 1 + 5

A(1) = \(\dfrac{38}{9}\)

|y| = 1 ⇒ y \(\in\) {-1; 1} 

⇒ (\(x;y\)) = (-\(\dfrac{1}{3}\); -1); (-\(\dfrac{1}{3}\); 1); (\(\dfrac{1}{3};-1\)); (\(\dfrac{1}{3};1\))

B(-\(\dfrac{1}{3}\);-1) = 2.(-\(\dfrac{1}{3}\))² - 3.(-\(\dfrac{1}{3}\)).(-1) + (-1)²

B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) - 1 + 1

B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\)

B(-\(\dfrac{1}{3}\); 1) = 2.(-\(\dfrac{1}{3}\))² - 3.(-\(\dfrac{1}{3}\)).1 + 1²

B(-\(\dfrac{1}{3};1\)) = \(\dfrac{2}{9}\) + 1 + 1

B(-\(\dfrac{1}{3}\); 1) = \(\dfrac{20}{9}\) 

B(\(\dfrac{1}{3};-1\)) = 2.(\(\dfrac{1}{3}\))² - 3.(\(\dfrac{1}{3}\)).(-1) + (-1)²

B(\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) + 1 + 1

B(\(\dfrac{1}{3}\); -1) = \(\dfrac{20}{9}\)

B(\(\dfrac{1}{3}\); 1) = 2.(\(\dfrac{1}{3}\))² - 3.(\(\dfrac{1}{3}\)).1 + (1)²

B(\(\dfrac{1}{3}\); 1) = \(\dfrac{2}{9}\) - 1 + 1

B(\(\dfrac{1}{3}\);1) = \(\dfrac{2}{9}\)

Bài 2:

\(x+y+1=0\Rightarrow x+y=-1\)

A = \(x\)(\(x+y\)) - y².(\(x+y\)) + \(x^2\) - y² + 2(\(x+y\)) + 3

Thay \(x\) + y  = -1 vào biểu thức A ta có:

A = \(x\).( -1) - y² .(-1) + \(x^2\)  - y² + 2(-1) + 3

A = -\(x\) + y² + \(x^2\) - y² - 2 + 3

A = \(x^2\) - \(x\) + 1

`a)100x^2-20x+1`

`=(10x-1)^2`

Thay `x=1/10`

`=>100x^2-20x+1=(1-1)^2=0`

`b)49x^2-42x+10`

`=49*4/49-42*2/7+10`

`=4-12+10=2`

`c)25x^2+40x+16y^2`

`=(5x+4y)^2=(2+3)^2=25`