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) \(\dfrac{x^3+8y^3}{2y+x}\)
\(=\dfrac{x^3+\left(2y\right)^3}{x+2y}\)
\(=\dfrac{\left(x+2y\right)\left[x^2+x.2y+\left(2y\right)^2\right]}{x+2y}\)
\(=x^2+2xy+4y^2\)
b) \(\dfrac{a-1}{2\left(a-4\right)}+\dfrac{a}{a-4}\) MTC: \(2\left(a-4\right)\)
\(=\dfrac{a-1}{2\left(a-4\right)}+\dfrac{2a}{2\left(a-4\right)}\)
\(=\dfrac{a-1+2a}{2\left(a-4\right)}\)
\(=\dfrac{3a-1}{2\left(a-4\right)}\)
c) \(\dfrac{x^3+3x^2y+3xy^2+y^3}{2x+2y}\)
\(=\dfrac{\left(x+y\right)^3}{2\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2}\)
d) \(\left(x-5\right)^2+\left(7-x\right)\left(x+2\right)\)
\(=\left(x^2-2.x.5+5^2\right)+\left(7x+14-x^2-2x\right)\)
\(=x^2-10x+25+7x+14-x^2-2x\)
\(=39-5x\)
e) \(\dfrac{3x}{x-2}-\dfrac{2x+1}{2-x}\)
\(=\dfrac{3x}{x-2}+\dfrac{2x+1}{x-2}\)
\(=\dfrac{3x+2x+1}{x-2}\)
\(=\dfrac{5x+1}{x-2}\)
h) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x+6}{4-9x^2}\)
\(=\dfrac{1}{3x-2}-\dfrac{1}{3x+2}+\dfrac{3x+6}{9x^2-4}\)
\(=\dfrac{1}{3x-2}-\dfrac{1}{3x+2}+\dfrac{3x+6}{\left(3x-2\right)\left(3x+2\right)}\) MTC: \(\left(3x-2\right)\left(3x+2\right)\)
\(=\dfrac{3x+2}{\left(3x-2\right)\left(3x+2\right)}-\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}+\dfrac{3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{\left(3x+2\right)-\left(3x-2\right)+\left(3x+6\right)}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+2-3x+2+3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+10}{\left(3x-2\right)\left(3x+2\right)}\)
Bài 1: Chưa đủ dữ kiện để tính. Từ $a+b=2$ bạn chỉ có thể tính $a^2+b^2+2ab$
Bài 2:
\(a^2+b^2-ab-a-b+1=0\)
\(\Leftrightarrow 2a^2+2b^2-2ab-2a-2b+2=0\)
\(\Leftrightarrow (a^2-2ab+b^2)+(a^2-2a+1)+(b^2-2b+1)=0\)
\(\Leftrightarrow (a-b)^2+(a-1)^2+(b-1)^2=0\)
Vì \((a-b)^2\geq 0; (a-1)^2\geq 0;(b-1)^2\geq 0, \forall a,b\in\mathbb{R}\)
\(\Rightarrow (a-b)^2+(a-1)^2+(b-1)^2\geq 0\)
Dấu "=" xảy ra khi \((a-b)^2=(a-1)^2=(b-1)^2=0\Leftrightarrow a=b=1\)
Bài 3:
\(x+y=x^3+y^3=(x+y)(x^2-xy+y^2)\)
\(\Leftrightarrow (x+y)(x^2-xy+y^2-1)=0\)
\(\Rightarrow \left[\begin{matrix} x+y=0\\ x^2-xy+y^2-1=0\end{matrix}\right.\).
Nếu $x+y=0$ \(\Rightarrow x^2+y^2=x+y=0\)
Mà \(x^2\geq 0, y^2\geq 0, \forall x,y\) nên để tổng của chúng bằng $0$ thì \(x^2=y^2=0\Leftrightarrow x=y=0\) (thỏa mãn)
Nếu \(x^2-xy+y^2-1=0\)
\(\Leftrightarrow (x^2+y^2)-xy-1=0\)
\(\Leftrightarrow x+y-xy-1=0\)
\(\Leftrightarrow (x-1)(1-y)=0\) \(\Rightarrow \left[\begin{matrix} x=1\\ y=1\end{matrix}\right.\)
\(x=1\Rightarrow 1+y=1+y^2=1+y^3\)
\(\Leftrightarrow y=y^2=y^3\Rightarrow y=0\) hoặc $y=1$
\(y=1\Rightarrow x+1=x^2+1=x^3+1\)
\(\Leftrightarrow x=x^2=x^3\Rightarrow x=0\) hoặc $x=1$.
Vậy $(x,y)=(0,0); (1,0), (0,1), (1,1)$
b: \(=5:\dfrac{-3}{4}\cdot x^2:x\cdot y^4:y^3=\dfrac{-20}{3}xy\)
c: \(=-9:\dfrac{4}{5}\cdot a^5:a^4\cdot b^4:b^3=-\dfrac{45}{4}ab\)
d: \(=\dfrac{64a^{15}b^6c^9}{4a^6b^2c^8}=16a^9b^4c\)
g: \(=\dfrac{3}{4}:\dfrac{3}{2}\cdot a^5:a^2\cdot b^3:b^2\cdot c^2:c=\dfrac{1}{2}a^3bc\)
Ai biết cách làm thì nhanh tay giải giùm mình nhé!!!!!!!!!!!!
mk đang cần gấp....<3<3<3<3<3<3
a) 8x3+1
=(2x)3+13
=(2x+1)(4x2-+2x+1)
b)27-y3
=33-y3
=(3-y)(9+3y+y2)
c)27a3-8b3
=(3a)3-(2b)3
=(3a-2b)(9a2+6ab+4b2)
c) \(\left(\frac{x}{2}-\frac{y}{3}\right)^3=\left(\frac{x}{2}-\frac{y}{3}\right)\left(\frac{x}{2}-\frac{y}{3}\right)\left(\frac{x}{2}-\frac{y}{3}\right)\)