Bài 1:

a) \(\Leftrightarrow x^2-8x+16-x^2+4=6\\ \Leftrightarrow-8x=-14\\ \Leftrightarrow x=\dfrac{7}{4}\)

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

c) \(\Leftrightarrow\left[{}\begin{matrix}3x-1=x+2\\3x-1=-x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=3\\4x=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

Bài 2:

a) \(=5\left(x^2-2xy+y^2-4z^2\right)=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\)

b) \(=\left(x+1\right)^3-\left(3z\right)^3=\left(x-3z+1\right)\left[\left(x+1\right)^2+3z\left(x+1\right)+\left(3z\right)^2\right]=\left(x-3z+1\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)

c) \(=x\left(x^2-16\right)-15x\left(x-4\right)=x\left(x+4\right)\left(x-4\right)-15x\left(x-4\right)=\left(x-4\right)\left(x^2+4x-15x\right)=\left(x-4\right)\left(x^2-11x\right)=x\left(x-4\right)\left(x-11\right)\)

để đây làm cho

3

1)

a) 4y²-4xy+x²= x²-4xy+4y²= (x-2y)²

b) 9x²-12xy+4y²= (3x)²-2.3x.2y+(2y)²= (3x-2y)²

c) 16x²-25=(4x)²-5²= (4x-5)(4x+5)

d) 1-9y²= 1²-(3y)²=(1-3y)(1+3y)

g) x³-27y³= (x-3y)(x²+3xy+9y²)

h) 64 + 8x³=(4+2x)(16+8x+4x²)

i) x⁴+4y⁴-4x²y²= (x²)²-2x²2y²+(y²)²=(x²-y²)²

Bài 1: 

a: =8xy/2x=4y

b: \(=\dfrac{4x-1-7x+1}{3x^2y}=\dfrac{-3x}{3x^2y}=\dfrac{-1}{xy}\)

c: \(=\dfrac{3x-x+6}{2x\left(x+3\right)}=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}=\dfrac{1}{x}\)

e: \(=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-10}{4}=-\dfrac{5}{2}\)

câu 3 bài 1: = (x-y)(y-1)

b: \(=8+2\cdot3-7\cdot1.3+3\cdot\dfrac{5}{4}=8.65\)

Bài 1 :

\(a,P=\dfrac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}\left(dkxd:x\ne-1;3\right)\)

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

\(=\dfrac{3x}{2x-6}\)

Để P = 1 thì :

\(\dfrac{3x}{2x-6}=1\)

\(\Leftrightarrow\dfrac{3x-2x+6}{2x-6}=0\)

\(\Leftrightarrow x=-6\)

Bài 1 : 

a) ĐKXĐ : `x \ne -1 ; x \ne 3`

b)

`P = ( 3x^2 + 3x )/((x+1)(2x-6))`

`P=(3x( x + 1 ))/((x+1)(2x-6))`

`P=(3x)/(2x-6)`
Để `P = 1 <=> (3x)/(2x-6)=1`

`<=> 3x=2x-6`

`<=> 2x-3x=6`

`<=> -x=6`

`<=> x=-6` ( tm )

Vậy `x=-6` 

a, ĐKXĐ: 

x+1≠0 => x≠-1

2x-6≠0 => x≠-3

b, 

Ta có: P=1

=>\(\dfrac{3x}{2x-6}\)

=>3x=2x-6

=>3x-2x=-6

=>x=-6

Vậy x=-6 để P=1

đề là tìm n nguyên để biểu thức nguyên hả bạn ? 

d, \(\frac{3n+1}{n-2}=\frac{3\left(n-2\right)+7}{n-2}=3+\frac{7}{n-2}\)ĐK : \(n\ne2\)

\(\Rightarrow n-2\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

tương tự 

ừ làm hộ tôi nốt hai câu đi

okk, mình nghĩ bạn tự làm tốt hơn, nhưng dạng mới tiếp cận chưa quen nên thôi ;-; 

e, \(\frac{5n}{n+1}=\frac{5\left(n+1\right)-5}{n+1}=5-\frac{5}{n+1}\)ĐK : \(n\ne-1\)

\(\Rightarrow n+1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

f, \(\frac{n+8}{n+1}=\frac{n+1+7}{n+1}=1+\frac{7}{n+1}\)ĐK : \(n\ne-1\)

\(\Rightarrow n+1\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

Mình đang gấp 

Mình đang chuẩn bị đi học rồi, còn 3 phút nữa 

Bài 3: 

Diện tích là:

\(600^2\cdot\dfrac{1}{4}\cdot\dfrac{3}{4}=67500\left(m^2\right)\)