im chưa học\

Ta có : x² + 3x 

= x² + \(2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2\)

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

Ta có : (2x - 3)(4x² - 6x + 9) 

= (2x -3)(2x + 3)²

= (4x² - 9)(2x + 3)

a. (2x+3y)²= (2x)²+2.2x.3y+(3y)²

=4x²+12xy+9y²

b. 2(\(\dfrac{1}{2}\)x²+y)(x²-2y)

=(x²+2y)(x²-2y)

=x⁴-4y²

c, (x+y+z)²= [(x+y)+z]²

=(x+y)²+2(x+y)z+z²

=x²+2xy+y²+2xz+2yz+z²

=x²+y²+z²+2xy+2yz+2xz

a) \(\left(2x+1\right)^3\)

\(=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1+1\)

\(=8x^3+12x^2+6x+1\)

b) \(\left(x-3\right)^3\)

\(=x^3-3.x^2.3+3.x.3^2-3^3\)

\(=x^3-9x^2+27x-27\)

Bài 2: 

a: \(x^3+15x^2+75x+125=\left(x+5\right)^3\)

b: \(1-15y+75y^2-125y^3=\left(1-5y\right)^3\)

c: \(8x^3+4x^2y+\dfrac{3}{2}xy^2+8y^3=\left(2x+2y\right)^3\)

\(2x-2y=by+cz-cz-ax=by-ax\)

\(\Rightarrow2x-2y=by-ax\)

\(\Rightarrow2x+ax=2y+by\)

\(\Rightarrow x\left(a+2\right)=y\left(b+2\right)\)

\(\Rightarrow a+2=\dfrac{y\left(b+2\right)}{x}\)

\(2z-2y=ax+by-cz-ax=by-cz\)

\(\Rightarrow2z+cz=2y+by\)

\(\Rightarrow z\left(c+2\right)=y\left(b+2\right)\)

\(\Rightarrow c+2=\dfrac{y\left(b+2\right)}{z}\)

\(A=\dfrac{2}{a+2}+\dfrac{2}{b+2}+\dfrac{2}{c+2}=\dfrac{2}{\dfrac{y\left(b+2\right)}{x}}+\dfrac{2}{b+2}+\dfrac{2}{\dfrac{y\left(b+2\right)}{z}}=\dfrac{2x}{y\left(b+2\right)}+\dfrac{2}{b+2}+\dfrac{2z}{y\left(b+2\right)}=\dfrac{2x}{y\left(b+2\right)}+\dfrac{2y}{y\left(b+2\right)}+\dfrac{2z}{y\left(b+2\right)}=\dfrac{2x+2y+2z}{y\left(b+2\right)}=\dfrac{by+cz+cz+ax+ax+by}{by+2y}=\dfrac{2\left(ax+by+cz\right)}{by+cz+ax}=2\)

a) \(\left(2x^3-y^2\right)^3\)

\(=\left(2x^3\right)^3-3\cdot\left(2x^3\right)^2\cdot y^2+3\cdot2x^3\cdot\left(y^2\right)^{^2}-\left(y^2\right)^3\)

\(=8x^9-3\cdot4x^6y^2+3\cdot2x^3y^4-y^6\)

\(=8x^9-12x^6y^2+6x^3y^4-y^6\)

b) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)

\(=x^3-\left(3y\right)^3\)

\(=x^3-27y^3\)

c) \(\left(x+2y+z\right)\left(x+2y-z\right)\)

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

\(=x^2+4xy+4y^2-z^2\)

d) \(\left(2x^3y-0,5x^2\right)^3\)

\(=\left(2x^3y-\dfrac{1}{2}x^2\right)^3\)

\(=8x^9y^3-6x^8y^2+\dfrac{3}{2}x^7y-\dfrac{1}{8}x^6\)

e) \(\left(x^2-3\right)\left(x^4+3x^2+9\right)\)

\(=\left(x^2-3\right)\left(4x^2+9\right)\)

\(=4x^4+9x^2-12x^2-27\)

\(=4x^4-3x^2-27\)

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

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

\(=8x^3-1\)

\(a,\left(2x^3-y^2\right)^3=8x^9-12x^6y^2+6x^3y^4-y^6\)\(b,\left(x-3y\right)\left(x^2+3xy+9y^2\right)=x^3-27y^3\)

\(c,\left(x+2y+z\right)\left(x+2y-z\right)=\left(x+2y\right)^2-z^2=x^2+4xy+4y^2-z^2\)\(d,\left(2x^3y-0,5x^2\right)^3=8x^9y^3-6x^4y^2x^2+3x^3yx^4-0,125x^6=8x^9y^3-6x^6y^2+3x^7y-0,125x^6\)

a

Để biểu thức có nghĩa thì \(x-2\ne0\Rightarrow x\ne2\)

b

Để biểu thức có nghĩa thì \(2x+1\ne0\Rightarrow x\ne-\dfrac{1}{2}\)

c

Ủa câu c là (x-1)/(x^2+1) đúng không bạn:v

Để biểu thức có nghĩa thì \(x^2+1\ne0\)

Vì \(x^2\ge0\forall x\Rightarrow x^2+1>0\forall x\)

Vậy biểu thức có nghĩa với mọi giá trị x.

d

Để biểu thức có nghĩa thì \(xy-3y\ne0\Leftrightarrow y\left(x-3\right)\ne0\)

\(\Leftrightarrow\left\{{}\begin{matrix}y\ne0\\x-3\ne0\Rightarrow x\ne3\end{matrix}\right.\)

Vậy để biểu thức có nghĩa thì đồng thời \(y\ne0,x\ne3\)

a) \(\dfrac{5}{x-2}\) 

Có nghĩa khi:

\(x-2\ne0\)

\(\Rightarrow x\ne2\)

b) \(\dfrac{x-y}{2x+1}\)

Có nghĩa khi:

\(2x+1\ne0\)

\(\Rightarrow2x\ne-1\)

\(\Rightarrow x\ne-\dfrac{1}{2}\)

c) \(\dfrac{x-1}{x^2+1}\)

Có nghĩa khi:

\(x^2+1\ne0\)

\(\Rightarrow x^2\ne-1\) (luôn đúng)

Vậy biểu thức được xác định với mọi x

d) \(\dfrac{ax+by+c}{xy-3y}=\dfrac{ax+by+c}{y\left(x-3\right)}\)

Có nghĩa khi:

\(y\left(x-3\right)\ne0\)

\(\Leftrightarrow\left[{}\begin{matrix}y\ne0\\x-3\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}y\ne0\\x\ne3\end{matrix}\right.\)

\(a,=x^3+3x^2+3x+1\\ b,=8x^3+36x^2+54x+27\\ c,=x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}\\ d,=x^6-6x^4+12x^2-8\\ e,=8x^3-36x^2y+54xy^2-27y^3\)

Cộng vế với vế:

\(\Rightarrow x+y+z=2ax+2by+2cz\)

\(\Rightarrow x+y+z-2x=2ax+2by+2cx-2\left(by+cz\right)=2ax\)

\(\Rightarrow2ax=y+z-x\)

\(\Rightarrow a=\dfrac{y+z-x}{2x}\Rightarrow1+a=\dfrac{x+y+z}{2x}\)

Tương tự ta có: \(1+b=\dfrac{x+y+z}{2y}\) ; \(1+c=\dfrac{x+y+z}{2z}\)

\(\Rightarrow\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}=\dfrac{2x+2y+2z}{x+y+z}=2\)