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\(\text{a) x^2 + y^2 = (x+y)^2 - 2xy = a^2 - 2b}\)
\(\text{b) x^3 + y^3 = (x+y)^3 - 3xy(x+y) = a^3 - 3ab}\)
\(\text{c) x^4 + y^4 = (x^2+y^2)^2 - 2x^2y^2 = (a^2-2b)^2 - 2b^2 = a^4 - 4a^2b + 2b^2}\)
\(\text{d) x^5 + y^5 = (x^3+y^3)(x^2+y^2) - x^2y^2(x+y) = a^5 - 5a^3b + 5ab^2}\)
\(x^5+y^5=\left(x^2+y^2\right)\left(x^3+y^3\right)-x^2y^3-x^3y^2\)
\(=\left(x^2+y^2\right)\left(x^3+y^3\right)-\left(xy\right)^2\left(x+y\right)\)
\(=10.26-\left(-3\right)^2.2=...\)
(x+y)5=32
⇔ x5+5x4y+10x3y2+10x2y3+5xy4+y5 = 32
⇔ x5+y5 = 32-5xy(x3+y3)-10x2y2(x+y)
= 32-5.(-3).26-10.(-3)2.2
= 242
Lời giải:
a.
$x^3+y^3=(x+y)^3-3xy(x+y)=9^3-3.9.18=243$
$x^4+y^4=(x^2+y^2)^2-2x^2y^2=[(x+y)^2-2xy]^2-2x^2y^2$
$=[9^2-2.18]^2-2.18^2=1377$
Nếu $x\geq y$ thì:
$x^3-y^3=(x-y)(x^2+xy+y^2)$
$=|x-y|[(x+y)^2-xy]=\sqrt{(x+y)^2-4xy}[(x+y)^2-xy]$
$=\sqrt{9^2-4.18}(9^2-18)=189$
Nếu $x< y$ thì $x^3-y^3=-189$
b.
$A=(x+y)^2-6(x+y)+y-5$
$=(-9)^2-6(-9)+y-5=130+y$
Chưa đủ cơ sở để tính biểu thức.
6: \(-x^2y\left(xy^2-\dfrac{1}{2}xy+\dfrac{3}{4}x^2y^2\right)\)
\(=-x^3y^3+\dfrac{1}{2}x^3y^2-\dfrac{3}{4}x^4y^3\)
7: \(\dfrac{2}{3}x^2y\cdot\left(3xy-x^2+y\right)\)
\(=2x^3y^2-\dfrac{2}{3}x^4y+\dfrac{2}{3}x^2y^2\)
8: \(-\dfrac{1}{2}xy\left(4x^3-5xy+2x\right)\)
\(=-2x^4y+\dfrac{5}{2}x^2y^2-x^2y\)
9: \(2x^2\left(x^2+3x+\dfrac{1}{2}\right)=2x^4+6x^3+x^2\)
10: \(-\dfrac{3}{2}x^4y^2\left(6x^4-\dfrac{10}{9}x^2y^3-y^5\right)\)
\(=-9x^8y^2+\dfrac{5}{3}x^6y^5+\dfrac{3}{2}x^4y^7\)
11: \(\dfrac{2}{3}x^3\left(x+x^2-\dfrac{3}{4}x^5\right)=\dfrac{2}{3}x^3+\dfrac{2}{3}x^5-\dfrac{1}{2}x^8\)
12: \(2xy^2\left(xy+3x^2y-\dfrac{2}{3}xy^3\right)=2x^2y^3+6x^3y^3-\dfrac{4}{3}x^2y^5\)
13: \(3x\left(2x^3-\dfrac{1}{3}x^2-4x\right)=6x^4-x^3-12x^2\)
\(\left\{{}\begin{matrix}x-y=4\\xy=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=y+4\\y\left(y+4\right)=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=y+4\\y^2+4y-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=y+4\\\left[{}\begin{matrix}y=-2+\sqrt{5}\\y=-2-\sqrt{5}\end{matrix}\right.\end{matrix}\right.\)
Với \(y=-2+\sqrt{5}\Rightarrow x=2+\sqrt{5}\)
Với \(y=-2-\sqrt{5}\Rightarrow x=2-\sqrt{5}\)
\(\Rightarrow A=x^2+y^2=\left(-2+\sqrt{5}\right)^2+\left(2+\sqrt{5}\right)^2=\left(2-\sqrt{5}\right)^2+\left(-2-\sqrt{5}\right)^2=18\)
\(B=x^3+y^3\Rightarrow\left[{}\begin{matrix}B=\left(2+\sqrt{5}\right)^3+\left(-2+\sqrt{5}\right)^3=34\sqrt{5}\\B=\left(2-\sqrt{5}\right)^3+\left(-2-\sqrt{5}\right)^3=-34\sqrt{5}\end{matrix}\right.\)
\(\Rightarrow C=x^4+y^4=\left(-2+\sqrt{5}\right)^4+\left(2+\sqrt{5}\right)^4=\left(2-\sqrt{5}\right)^4+\left(-2-\sqrt{5}\right)^4=322\)
2: \(=\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{-\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{-\left(x+y\right)\left(x^2+y^2\right)}{x^2+xy+y^2}\)
\(1,\\ a,=6x^4-15x^3-12x^2\\ b,=x^2+2x+1+x^2+x-3-4x=2x^2-x-2\\ c,=2x^2-3xy+4y^2\\ 2,\\ a,=7x\left(x+2y\right)\\ b,=3\left(x+4\right)-x\left(x+4\right)=\left(3-x\right)\left(x+4\right)\\ c,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ d,=x^2-5x+3x-15=\left(x-5\right)\left(x+3\right)\\ 3,\\ a,\Leftrightarrow3x\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Câu 1
a)\(3x^2\left(2x^2-5x-4\right)=6x^4-15x^3-12x^2\)
b)\(\left(x+1\right)^2+\left(x-2\right)\left(x+3\right)-4x=x^2+2x+1+x^2+3x-2x-6-4x=2x^2-x-5\)
Lời giải:
$x^3+y^3=(x+y)^3-3xy(x+y)=2^3-3xy.2=8-6xy$
$=8-3.2xy=8-3[(x+y)^2-(x^2+y^2)]=8-3(2^2-34)=98$
----------------
$x^4+y^4=(x^2+y^2)^2-2x^2y^2=34^2-\frac{1}{2}(2xy)^2$
$=34^2-\frac{1}{2}[(x+y)^2-(x^2+y^2)]^2=34^2-\frac{1}{2}(2^2-34)^2=706$
a) x2 + y2 = (x+y)2 - 2xy = a2 - 2b
b) x3 + y3 = (x+y)3 - 3xy(x+y) = a3 - 3ab
c) x4 + y4 = (x2+y2)2 - 2x2y2 = (a2-2b)2 - 2b2 = a4 - 4a2b + 2b2
d) x5 + y5 = (x3+y3)(x2+y2) - x2y2(x+y) = a5 - 5a3b + 5ab2
a) \(\left(x-5\right)^2=\left(3+2x\right)^2\)
\(\Rightarrow\left(3+2x\right)^2-\left(x-5\right)^2=0\)
\(\Rightarrow\left(3+2x+x-5\right)\left(3+2x-x+5\right)=0\)
\(\Rightarrow\left(3x-2\right)\left(x+8\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x-2=0\\x+8=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-8\end{matrix}\right.\)
b) \(27x^3-54x^2+36x=9\)
\(\Rightarrow27x^3-54x^2+36x-9=0\)
\(\Rightarrow27x^3-54x^2+36x-8+8-9=0\)
\(\Rightarrow\left(3x-2\right)^3-1=0\)
\(\Rightarrow\left(3x-2-1\right)\left[\left(3x-2\right)^2+3x-2+1\right]=0\)
\(\Rightarrow\left(3x-3\right)\left[\left(3x-2\right)^2+3x-2+\dfrac{1}{4}-\dfrac{1}{4}+1\right]=0\)
\(\Rightarrow\left(3x-3\right)\left[\left(3x-2+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]=0\)
\(\Rightarrow\left(3x-3\right)\left[\left(3x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\right]=0\left(1\right)\)
mà \(\left(3x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}>0,\forall x\)
\(\left(1\right)\Rightarrow3x-3=0\Rightarrow3x=3\Rightarrow x=1\)
(\(x-5\))2 = (3 +2\(x\))2 ⇒ \(\left[{}\begin{matrix}x-5=3+2x\\x-5=-3-2x\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=-8\\x=\dfrac{2}{3}\end{matrix}\right.\) vậy \(x\in\){-8; \(\dfrac{2}{3}\)}
27\(x^3\) - 54\(x^2\) + 36\(x\) = 9
27\(x^3\) - 54\(x^2\) + 36\(x\) - 8 = 1
(3\(x\) - 2)3 = 1 ⇒ 3\(x\) - 2 = 1 ⇒ \(x\) = 1