\(B=\left(\frac{x}{x+1}+\frac{x-1}{x}\right)\div\left(\frac{x}{x+1}-\frac{x-1}{x}\right)\)

ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne-1\end{cases}}\)

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

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

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

Để B = 1 => 2x² - 1 = 1

=> 2x² - 1 - 1 = 0

=> 2x² - 2 = 0

=> 2( x² - 1 ) = 0

=> 2( x - 1 )( x + 1 ) = 0

=> x - 1 = 0 hoặc x + 1 = 0

=> x = 1 ( tm ) hoặc x = -1 ( ktm )

Vậy x = 1 thì B = 1

a, \(3x\left(2x-3\right)-6x\left(1+x\right)=5\)

\(\Leftrightarrow6x^2-9x-6x-6x^2=5\)

\(\Leftrightarrow-15x=5\Leftrightarrow x=-\frac{1}{3}\)

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

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

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

\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-6x^2+12x-12\right)-x^3-1=0\)

\(\Leftrightarrow-x^3+9x^2-9x+12=0\)

a,\(3x\left(2x-3\right)-6x\left(1+x\right)=5\)

\(\Leftrightarrow6x^2-9x-6x-6x^2=5\)

\(\Leftrightarrow-15x=5\)

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

Bài làm 

\(x^3-2x^2+3x-4x^2+8x-12\)

\(=x^3-6x^2+11x-12\)

a, \(5\left(2x-1\right)-2\left(5x+3\right)=10x-5-10x-6=-11\)

b, \(3\left(x+y\right)-3\left(x-2xy+y\right)\)

\(=3\left[\left(x+y\right)-\left(x-2xy+y\right)\right]=3\left(x+y-x+2xy-y\right)\)

\(=3.2xy=6xy\)

ta có

\(1+2+..+2^{60}=2^{x-1}-1\)

nhân 2 cho cả hai vế ta được

\(2+2^2..+2^{60}+2^{61}=2.2^{x-1}-2\)

\(\Leftrightarrow\left(1+2+2^2..+2^{60}\right)+2^{61}-1=2.2^{x-1}-2\)

\(\Leftrightarrow2^{x-1}-1+2^{61}-1=2.2^{x-1}-2\)

\(\Leftrightarrow2^{61}=2^{x-1}\Leftrightarrow x=62\)

chỉnh lại đi bạn ơi . ko phải môn toán