a) Ta có :  x - 2y = 0

=> x = 2y

Khi đó A = 2.(2y)² - 2y² - 3.2yy - 2.2y.y² + (2y)².y + 5

= 8y² - 2y² - 6y² - 4y³ + 4y³ + 5

= 5

Vậy giá trị của A khi x - 2y = 0 là 5

b)Thay 11 = x - y vào biểu thức B ta có

\(B=\frac{3x-\left(x-y\right)}{2x+y}-\frac{3y+x-y}{2y+x}=\frac{2x+y}{2x+y}-\frac{2y+x}{2y+x}=1-1=0\)

Vậy giá trị của B khi x - y = 11 là 0

Bài 6

\(\left(a-b\right)^2=a^2-2ab+b^2\)

\(=\left(a^2+2ab+b^2\right)-4ab\)

\(=\left(a+b\right)^2-4ab\)

Bài 5 :

\(a,16x^2-\left(4x-5\right)^2=15\)

\(16x^2-16x^2+40x-25-15=0\)

\(40x-40=0\)

\(40x=40\)

\(x=1\)

\(b,\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)

\(4x^2+12x+9-4x^2+4=49\)

\(12x=36\)

\(x=3\)

\(c,\left(2x+1\right)\left(2x-1\right)+\left(1-2x\right)^2=18\)

\(4x^2-1+1-4x+4x^2=18\)

\(8x^2-4x-18=0\)

\(2\left(4x^2-2x-9\right)=0\)

\(x=\frac{1-\sqrt{37}}{4}\)

\(d,2\left(x+1\right)^2-\left(x-3\right)\left(x+3\right)-\left(x-4\right)^2=0\)

\(2x^2+4x+2-x^2+9-x^2+8x-16=0\)

\(12x=4\)

\(x=\frac{1}{3}\)

2) a) Ta có B = \(\frac{x+2}{x-2}-\frac{x-2}{x+2}-\frac{16}{4-x^2}=\frac{\left(x+2\right)^2-\left(x-2\right)^2+16}{\left(x-2\right)\left(x+2\right)}=\frac{8\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{8}{x-2}\)

Khi |x - 1| = 2

=> \(\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

Khi x = 3 (thỏa mãn) => A = \(\frac{3^2-2.3}{3+1}=\frac{3}{4}\)

Khi x = - 1 (không thỏa mãn) => Không tìm được A 

b) Ta có P = \(A.B=\frac{x^2-2x}{x+1}.\frac{8}{x-2}=\frac{8x\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=\frac{8x}{x+1}\)

Đẻ P < 8

=> \(\frac{8x}{x+1}< 8\Leftrightarrow\frac{x}{x+1}< 1\)

=> \(\orbr{\begin{cases}x< x+1\left(x>-1\right)\\x>x+1\left(x< -1\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}0x< 1\left(tm\right)\\0x>1\left(\text{loại}\right)\end{cases}}\)

Vậy x > - 1 thì P < 8 

Thay x = 1/2 vào 

:>>>>>

mệt rời o 

thông cảm 

hihi

Bài 7 

\(a,A=x^2-2x+5\)

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

\(=\left(x-1\right)^2+4\ge4\forall x\)

GTNN \(A=4\) khi \(\left(x-1\right)^2=0\Rightarrow x=1\)

\(b,B=x^2-x+1\)

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

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

\(c,C=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)\)

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

Đặt \(x^2+5x=t\)

\(\Rightarrow C=\left(t-6\right)\left(t+6\right)\)

\(=t^2-36\)

\(\left(x^2+5x\right)^2-36\ge36\forall x\)

\(d,D=x^2+5y^2-2xy+4y-3\)

\(=\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)-4\)

\(=\left(x-y\right)^2+\left(2y+1\right)^2-4\ge-4\)

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

Ta có: \(x^3-y^3-3xy=x^3-y^3-3xy\left(x-y\right)\) (vì x - y = 1)

                                            \(=x^3-3x^2y+3xy^2-y^3\)

                                            \(=\left(x-y\right)^3\)

                                             \(=1^3=1\)  

Bài này dễ mà. Chúc bạn học tốt.

a. \(8x\left(x-2017\right)-2x+4034=0\)

\(8x\left(x-2017\right)-2\left(x-2017\right)=0\)

\(\left(8x-2\right)\left(x-2017\right)=0\)

\(\Rightarrow TH1:8x-2=0\)

\(8x=2\)

\(x=\frac{1}{4}\)

\(TH2:x-2017=0\)

\(x=2017\)

Vậy \(x\in\left\{\frac{1}{4};2017\right\}\)

Bài 1 

a) \(8x\left(x-2017\right)-2x+4034=0\)

\(\Rightarrow8x\left(x-2017\right)-2\left(x-2017\right)=0\)

\(\Rightarrow\left(x-2017\right)\left(4x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=2017\\x=\frac{1}{4}\end{cases}}\)