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

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

\(\Leftrightarrow2x.2=6x^2+6x+6\)

\(\Leftrightarrow4x=6x^2+6x+6\)

\(\Leftrightarrow6x^2+2x+6=0\)

Ta có \(\Delta=2^2-4.6.6< 0\)

Vậy pt vô nghiệm

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

\(\Leftrightarrow\left[\left(x+1\right)-\left(x-1\right)\right].\left[\left(x+1\right)+\left(x-1\right)\right]=6\left(x^2+x+1\right)\)

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

\(\Leftrightarrow2.2x=6x^2+6x+6\)\(\Leftrightarrow4x=6x^2+6x+6\)

\(\Leftrightarrow6x^2+2x+6=0\)\(\Leftrightarrow3x^2+x+3=0\)( vô nghiệm vì \(1^2< 4.3.3\)hay \(1< 36\)) 

Vậy tập nghiệm của phương trình là \(S=\varnothing\)

( 3x - 1/2 ) + ( 1/2y + 3/5 ) = 0

=> ( 3 x - 1/2 ) = 0

       3x           = 0+1/2

        3x         = 1/2

          x           = 1/2 : 3

          x          = 1/6

=> ( 1/2 y + 3/5 ) = 0

      1/2y               = 0 - 3/5

      1/2 y              = -3/5

           y              = -3/5 : 1/2

           y              = -6/5

\(\left(x-2\right)^8=\left(x-2\right)^6\)

\(\Leftrightarrow\left(x-2\right)^8-\left(x-2\right)^6=0\)

\(\Leftrightarrow\left(x-2\right)^6\left[\left(x-2\right)^2-1\right]=0\)

\(\Leftrightarrow\left(x-2\right)^6\left(x-3\right)\left(x-1\right)=0\)

\(\Rightarrow x=2;x=3;x=1\)

=>x-2=0 hoặc x-2=1

=>x-2=0=>x=2

=>x-2=1=>x=3

\(\left(x-2\right)^8=\left(x-2\right)^6\)

\(\left(x-2\right)^8-\left(x-2\right)^6=0\)

\(\left(x-2\right)^6.\left[\left(x-2\right)^2-1\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x-2\right)^6=0\\\left(x-2\right)^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x-2=0\\\left(x-2\right)^2=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=2\\x-2=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=2\\x=3\end{cases}}}\)

\(\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right)\left(-1\frac{1}{4}\right)...\left(-1\frac{1}{2003}\right)\left(-1\frac{1}{2004}\right)\)

\(=-\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{2004}{2003}.\frac{2005}{2004}\)

\(=-\frac{3.4.5.....2004.2005}{2.3.4.....2003.2004}=\frac{-2005}{2}\)

cả 2 pt đều giải theo kiểu cái đầu nhóm với cái cuối, 2 cái ở giữa nhóm với nhau. sau đó giải theo cách đặt ẩn phụ

1) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)

\(\Leftrightarrow\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24=0\)

\(\Leftrightarrow\left(x^2+5x+2x+10\right)\left(x^2+4x+3x+12\right)-24=0\)

\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)

Đặt \(x^2+7x=a\), nên ta có :

\(\left(a+10\right)\left(a+12\right)-24=0\)

\(\Leftrightarrow\left(x+11-1\right)\left(x+11+1\right)-24=0\)

\(\Leftrightarrow\left[\left(x+11\right)^2-1\right]-24=0\)

\(\Leftrightarrow\left(x+11\right)^2-25=0\)

\(\Leftrightarrow\left(x+11-5\right)\left(x+11+5\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x+16\right)=0\Leftrightarrow\orbr{\begin{cases}x=-6\\x=-16\end{cases}}\)

1/30;-23/57;1031/9;-79993/100;-124;3/25

ủng hộ nha

to noi that: nguoi ta chi thich giup lam bai kho thoi chu bai nay ai di hoc cung lam dc

tai sao ban phai hoi

Phân số tớ kém lắm

a) Đặt: x = a- b; y = b - c ; z = c- a 

Ta có: x + y + z = 0 

=> \(A=x^3+y^3+z^3=3xyz+\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=3xyz\)

=> \(A=3xyz=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

b) Đặt: \(a=x^2-2x\) 

Ta có: \(B=a\left(a-1\right)-6=a^2-a-6=\left(a+2\right)\left(a-3\right)=\left(x^2-2x+2\right)\left(x^2-2x-3\right)\)

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

d) \(D=4\left(x^2+2x-8\right)\left(x^2+7x-8\right)+25x^2\)

Đặt: \(x^2-8=t\)

Ta có: \(D=4\left(t+2x\right)\left(t+7x\right)+25x^2\)

\(=4t^2+36xt+81x^2=\left(2t+9x\right)^2\)

\(=\left(2x^2+9x-16\right)^2\)