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Nhìn nản lắm.

Câu 3: 

a: \(49^2=2401\)

b: \(51^2=2601\)

c: \(99\cdot100=9900\)

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

\(=9x^2-6x+1-2x^2+x-6x+3\)

\(=7x^2-11x+4\)

mọi người trả lời giúp mình với mình cần gấp

\(a,=x^2-4-x^2-2x-1=-2x-5\\ b,=8x^3-1-8x^3-1=-2\\ 3,\\ a,\Rightarrow x^3+8-x^3+2x=15\\ \Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\\ b,\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\\ \Rightarrow7x=14\Rightarrow x=2\)

Bài 2:

a) \(=x^2-4-x^2-2x-1=-2x-5\)

b) \(=8x^3-1-8x^3-1=-2\)

Bài 3:

a) \(\Rightarrow x^3+8-x^3+2x=15\)

\(\Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\)

b) \(\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\)

\(\Rightarrow7x=14\Rightarrow x=2\)

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

\(=4x^2+4x+1+4x^2-4x+1-2\left(x^2-6x+9\right)\)

\(=8x^2+2-2x^2+12x-18\)

\(=6x^2+12x-16\)

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

\(=x^2-2x+1-9x^2-12x-4\)

\(=-8x^2-14x-3\)

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

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

\(=\left(6x+1-6x+1\right)^2=2^2=4\)

a: \(\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{x+y}\cdot\left(\frac{1}{x}+\frac{1}{y}\right)\)

\(=\frac{x^2+y^2}{x^2y^2}+\frac{2}{x+y}\cdot\frac{x+y}{xy}=\frac{x^2+y^2}{x^2y^2}+\frac{2}{xy}=\frac{x^2+y^2+2xy}{x^2y^2}=\frac{\left(x+y\right)^2}{x^2y^2}\)

\(\left\lbrack\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{x+y}\cdot\left(\frac{1}{x}+\frac{1}{y}\right)\right\rbrack:\frac{x^3+y^3}{x^2y^2}\)

\(=\frac{\left(x+y\right)^2}{x^2y^2}\cdot\frac{x^2y^2}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\frac{x+y}{x^2-xy+y^2}\)

b: \(\frac{1}{\left(2x-y\right)^2}+\frac{2}{4x^2-y^2}+\frac{1}{\left(2x+y\right)^2}\)

\(=\frac{\left(2x+y\right)^2+2\left(2x-y\right)\left(2x+y\right)+\left(2x-y\right)^2}{\left(2x-y\right)^2\cdot\left(2x+y\right)^2}\)

\(=\frac{\left(2x+y+2x-y\right)^2}{\left(2x+y\right)^2\cdot\left(2x-y\right)^2}=\frac{16x^2}{\left(2x+y\right)^2\cdot\left(2x-y\right)^2}\)

\(\left\lbrack\frac{1}{\left(2x-y\right)^2}+\frac{2}{4x^2-y^2}+\frac{1}{\left(2x+y\right)^2}\right\rbrack\cdot\frac{4x^2+4xy+y^2}{16x}\)

\(=\frac{16x^2}{\left(2x+y\right)^2\cdot\left(2x-y\right)^2}\cdot\frac{\left(2x+y\right)^2}{16x}=\frac{x}{\left(2x-y\right)^2}\)

a) Ta có: \(\left(x-\dfrac{1}{1-x}\right):\dfrac{x^2-x+1}{x^2-2x+1}\)

\(=\left(x+\dfrac{1}{x-1}\right):\dfrac{x^2-x+1}{\left(x-1\right)^2}\)

\(=\dfrac{x^2-x+1}{x-1}\cdot\dfrac{\left(x-1\right)^2}{x^2-x+1}\)

\(=x-1\)

b) Ta có: \(\left(1+\dfrac{x}{y}+\dfrac{x^2}{y^2}\right)\left(1-\dfrac{x}{y}\right)\cdot\dfrac{y^2}{x^3-y^3}\)

\(=\left(\dfrac{y^2}{y^2}+\dfrac{xy}{y^2}+\dfrac{x^2}{y^2}\right)\cdot\left(\dfrac{y-x}{y}\right)\cdot\dfrac{y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)

\(=\dfrac{x^2+xy+y^2}{y^2}\cdot\dfrac{-\left(x-y\right)}{y}\cdot\dfrac{y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)

\(=\dfrac{-1}{y}\)