B= (x-1).(x-2)....(x-35)

Thay x=34 vào B, ta được:

B=(34-1).(34-2).....(23-34).(34-35)

B= 0

Vậy B=0

Ta có : \(B\text{=}4x^2-12x+9\)

\(B\text{=}\left(2x-3\right)^2\)

Với \(x\text{=}\dfrac{1}{2}\)

\(\Rightarrow B\text{=}\left(2.\dfrac{1}{2}-3\right)^2\)

\(B\text{=}\left(-2\right)^2\text{=}4\)

Ta có : \(A\text{=}5\left(x+3\right)\left(x-3\right)+\left(2x+3\right)^2+\left(x-6\right)^2\)

\(A\text{=}5\left(x^2-9\right)+\left(2x+3\right)^2+\left(x-6\right)^2\)

\(A\text{=}5x^2-45+4x^2+12x+9+x^2-12x+36\)

\(A\text{=}10x^2\)

Với \(x\text{=}-\dfrac{1}{5}\)

\(\Rightarrow A\text{=}10.\left(-\dfrac{1}{5}\right)^2\text{=}\dfrac{2}{5}\)

B = 4x² - 12x + 9

= (2x - 3)²

Tại x = 1/2 ta có:

B = (2.1/2 - 3)²

= (-2)²

= 4

-------------------

A = 5(x + 3)(x - 3) + (2x + 3)² + (x - 6)²

= 5x² - 45 + 4x² + 12x + 9 + x² - 12x + 36

= 10x²

Tại x = 1/5 ta có:

A = 10.(1/5)²

= 2/5

\(B=4x^2-12x+9\)

\(B=\left(2x\right)^2-2\cdot2x\cdot3+3^2\)

\(B=\left(2x-3\right)^2\)

Thay \(x=\dfrac{1}{2}\) vào B ta có: 

\(B=\left(2\cdot\dfrac{1}{2}-3\right)^2=4\)

_______________________

\(A=5\left(x+3\right)\left(x-3\right)+\left(2x+3\right)^2+\left(x-6\right)^2\)

\(A=5\left(x^2-9\right)+4x^2+12x+9+x^2-12x+36\)

\(A=5x^2-45+5x^2+45\)

\(A=10x^2\)

Thay \(x=\dfrac{1}{5}\)vào A ta có:

\(A=10\cdot\left(-\dfrac{1}{5}\right)^2=\dfrac{2}{5}\)

\(\Rightarrow A=2-\left(-3\right)\left(-6\right)+16=2-18+16=0\\ \Rightarrow B=-5+24+\left(-5\right).7=19+\left(-35\right)=-16\\ \Rightarrow C=78-\left(--5\right)+12-\left(-5\right)=78-5+12+5=90\)

a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)

\(=\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\dfrac{x^2-4+10-x^2}{x+2}\)

\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{6}=\dfrac{-1}{x-2}\)

b: Khi x=1/2 thì \(B=\dfrac{-1}{\dfrac{1}{2}-2}=\dfrac{2}{3}\)

Khi x=-1/2 thì B=2/5

c: Để B nguyên thì \(x-2\in\left\{1;-1\right\}\)

hay \(x\in\left\{3;1\right\}\)

a, đk : x khác -2 ; 2 

\(B=\left(\dfrac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{x^2-4+10-x^2}{x+2}\right)\)

\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}:\dfrac{6}{x+2}=\dfrac{1}{2-x}\)

b, Ta có \(\left|x\right|=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{2};x=-\dfrac{1}{2}\)

Với x = 1/2 ta được \(B=\dfrac{1}{2-\dfrac{1}{2}}=\dfrac{2}{3}\)

Với x = -1/2 ta được \(B=\dfrac{1}{2+\dfrac{1}{2}}=\dfrac{2}{5}\)

c, \(\dfrac{1}{2-x}\Rightarrow2-x\inƯ\left(1\right)=\left\{\pm1\right\}\)

\(35\times m+35\times n+35\times p\)

\(=35\times\left(m+n+p\right)\)

Thay \(m=3;n=2;p=5\) vào biểu thức trên ta có:

\(35\times\left(3+2+5\right)=35\times10=350\)

35 x m + 35 x p + 35 x n

Thay số: ⇒ 35 x 3 + 35 x 5 + 35 x 2

= 35 x (3 + 5 + 2)

= 35 x 10

= 350

a: Khi x=2 và y=-3 thì \(x^2+2y=2^2+2\cdot\left(-3\right)=4-6=-2\)

b: \(A=x^2+2xy+y^2=\left(x+y\right)^2\)

Khi x=4 và y=6 thì \(A=\left(4+6\right)^2=10^2=100\)

c: \(P=x^2-4xy+4y^2=\left(x-2y\right)^2\)

Khi x=1 và y=1/2 thì \(P=\left(1-2\cdot\dfrac{1}{2}\right)^2=\left(1-1\right)^2=0\)

Thay x = 1 và y = -2 ta có

1² -2.1.(-2) - (-2)² + 4.1 .(-2)

= 1 - 2.1. (-2) - 4 + 4.1.(-2)

= 1 - (-4) - 4 + (-8)

= -7

`x^2 - 2xy - y^2 + 4xy`

`= x^2 + ( 4xy-2xy)-y^2`

`= x^2 + 2xy -y^2` `(***)`

Thay `x=1;y=2` vào `(***)` được `:`

`1^2 + 2*1*(-2) - (-2)^2`

`= -7` 

Thay x = 1 và y = -2, ta có:

\(1^2-2\times1\times\left(-2\right)-\left(-2\right)^2+4\times1\times\left(-2\right)=-7\)