Xin phép sửa đề:

Ta có: \(\frac{3x+1}{\left(x-1\right)^2}-\frac{1}{x+1}=\frac{x+3}{1-x^2}\) \(\left(x\ne\pm1\right)\)

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

\(\Rightarrow3x^2+4x+1-1+2x-x^2=-x^2-2x+3\)

\(\Leftrightarrow3x^2+8x-3=0\)

\(\Leftrightarrow\left(3x^2+9x\right)-\left(x+3\right)=0\)

\(\Leftrightarrow3x\left(x+3\right)-\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\3x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{1}{3}\end{cases}}\)

Vậy tập nghiệm PT \(S=\left(-3;\frac{1}{3}\right)\)

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

\(\Leftrightarrow2x^2-2x+5x-5=0\)

\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)

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

=> x=1, x=-5/2

b/\(x^2+2x^2-8x+5=0\)

\(\Leftrightarrow3x^2-8x+5=0\)

\(\Leftrightarrow3x^2-3x-5x+5=0\)

\(\Leftrightarrow\left(3x^2-3x\right)-\left(5x-5\right)=0\)

\(\Leftrightarrow3x\left(x-1\right)-5\left(x-1\right)=0\)

\(\Leftrightarrow\left(3x-5\right)\left(x-1\right)=0\)

=> x=1, x=5/3

\(\frac{x+1}{x-1}-\frac{x-1}{x-1}=\frac{16}{x^2-1}.\) Đk: x khác 1

\(\Leftrightarrow\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-1=\frac{16}{\left(x-1\right)\left(x+1\right)}.\)

\(\Leftrightarrow\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{16}{\left(x-1\right)\left(x+1\right)}-1=0\)

\(\Leftrightarrow\frac{x^2+2x+1-16}{\left(x-1\right)\left(x+1\right)}-\frac{x^2-1}{\left(\text{​​}x-1\right)\left(x+1\right)}=0\)

\(\Leftrightarrow\frac{x^2+2x+1-16-x^2+1}{\left(x-1\right)\left(x+1\right)}=0\)

\(\Leftrightarrow\frac{2x-14}{\left(x-1\right)\left(x+1\right)}=0\)

\(\Rightarrow2x-14=0\Rightarrow x=7\)

Bonus đk: x khác -1 nữa nha, ;-;

(x + y)² - (x - y)²

= (x² + 2xy + y²) - (x² - 2xy + y²)

= x² + 2xy + y² - x² + 2xy - y²

= (x² - x²) + (y² - y²) + (2xy + 2xy)

= 4xy

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

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

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

\(=4xy\)

Gọi 4 số tự nhiên, liên tiếp đó là n, n+1, n+2, n+3\(\left(n\in N\right)\)

Theo đề bài ra chúng ta có : n(n+1)(n+2)(n+3) + 1 = n.(n+3)(n+1)(n+2)+1

= (n²+3n)(n²+3n+2)+ 1 (*) Đặt n²+3n = t\(\left(t\in N\right)\)thì (*) = t(t+2)+1 = t²+2t+1 = (t+1)²

= (n²+3n+1)² Vì\(n\in N\)nên suy ra : (n²+3n+1)\(\in N\)

=> Vậy n(n+1)(n+2)(n+3) là 1 số chính phương. 

dễ mà bạn ?

khai triển biểu thức ra ta được :

 \(a^2-2ab+b^2=a^2+b^2+2ab-4ab\)

\(\Leftrightarrow a^2-2ab+b^2-a^2-b^2+4ab-2ab=0\)

\(\Leftrightarrow\left(a^2-a^2\right)+\left(b^2-b^2\right)+\left(-2ab-2ab+4ab\right)=0\)

\(\Leftrightarrow0+0+\left(4ab-4ab\right)=0\)

\(\Leftrightarrow0+0+0=0< =>0=0\)*đúng*

thật ra mình mới hc lớp 7 hỏi thử lm bài lớp 8 nhưng k hiểu nên lên đây hỏi.:>

a) 16x^2 - (4x - 5)^2 = 15

<=> 16x^2 - 16x^2 + 40x - 25 = 15

<=> 40x = 40

<=> x = 1

b) (2x + 3)^2 - 4(x - 1)(x + 1) = 49

<=> 4x^2 + 12x + 9 - 4x^2 - 4x + 4x + 4 = 49

<=> 12x + 13 = 49

<=> 12x = 36

<=> x = 3

c) (2x + 1)(1 - 2x) + (1 - 2x)^2 = 18

<=> 1 - 4x^2 + 1 - 4x + 4x^2 = 18

<=> 2 - 4x = 18

<=> -4x = 16

<=> x = -4

d)2(x + 1)^2 - (x - 3)(x + 3) - (x - 4)^2 = 0

<=> 2x^2 + 4x + 2 - x^2 + 3^2 - x^2 + 8x - 16 = 0

<=> 12x - 5 = 0

<=> 12x = 5

<=> x = 5/12

e) (x - 5)^2 - x(x - 4) = 9

<=> x^2 - 10x + 25 - x^2 + 4x = 9

<=> -6x + 25 = 9

<=> -6x = 9 - 25

<=> -6x = -16

<=> x = -16/-6 = 8/3

f) (x - 5)^2 + (x - 4)(1 - x) = 0

<=> x^2 - 10x + 25 + x - x^2 - x - 4 + 4x = 0

<=> -5x + 21 = 0

<=> -5x = -21

<=> x = 21/5

a) A = (x+3)² + (x-3)(x+3) - 2(x+2)(x - 4)

        = (x + 3)(x + 3) + (x - 3)(x + 3) - 2[x(x - 4) + 2(x - 4)]

        = x(x + 3) + 3(x + 3) + x(x + 3) - 3(x + 3) - 2[x² - 4x + 2x - 8]

        = x² + 3x + 3x + 9 + x² + 3x - 3x - 9 - 2(x² - 2x - 8)

        = x² + 3x + 3x + 9 +x² + 3x - 3x - 9 - 2x² + 4x + 16

        = (x² + x² - 2x²) + (3x + 3x + 3x - 3x + 4x) + (9 - 9 + 16) = 10x + 16

Thay x = -1/2 vào biểu thức trên ta có : \(10\cdot\left(-\frac{1}{2}\right)+16=-5+16=11\)

b) \(B=\left(3x+4\right)^2-\left(x-4\right)\left(x+4\right)-10x\)

\(B=9x^2+24x+16-x\left(x+4\right)+4\left(x+4\right)-10x\)

\(B=9x^2+24x+16-x^2-4x+4x+16-10x\)

\(B=\left(9x^2-x^2\right)+\left(24x-4x+4x-10x\right)+\left(16+16\right)\)

\(B=8x^2+14x+32\)

Thay x = -1/10 vào biểu thức trên ta có : \(B=8\cdot\left(-\frac{1}{10}\right)^2+14\cdot\left(-\frac{1}{10}\right)+32=\frac{767}{25}\)

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

\(C=x^2+2x+1-\left(2x-1\right)\left(2x-1\right)+3\left(x^2-4\right)\)

\(C=x^2+2x+1-2x\left(2x-1\right)+1\left(2x-1\right)+3x^2-12\)

\(C=x^2+2x+1-4x^2+2x+2x-1+3x^2-12\)

\(C=\left(x^2-4x^2+3x^2\right)+\left(2x+2x+2x\right)+\left(1-1-12\right)\)

\(C=6x-12\)

Thay x = 1 vào biểu thức ta có : C = 6.1 - 12 = 6 -12 = -6

Còn bài kia làm nốt đi

a) \(9c^2-6c+3\)

\(=\left(9c^2-6c+1\right)+2=\left(3c-1\right)^2+2>0\)

b) \(14m-6m^2-13\)

\(=-6.\left(m^2-\frac{7}{3}m+\frac{13}{6}\right)\)

\(=-6.\left(m^2-2\cdot\frac{7}{6}\cdot m+\frac{49}{36}+\frac{29}{36}\right)\)

\(=-6.\left(m-\frac{7}{6}\right)^2-\frac{29}{6}< 0\)

c) \(a^2-2a+2=\left(a-1\right)^2+1>0\)

d) \(6b-b^2-10=-\left(b^2-6b+9\right)-1=-\left(b-3\right)^2-1< 0\)

a, \(P\left(x\right)=x^7-\left(x+1\right)x^6+\left(x+1\right)x^5-\left(x+1\right)x^4+...+15\)

\(=x^7-x^7-x^6+x^6+x^5-x^5-x^4+...+15=15\)