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a) \(A=y\left(x^2-y^2\right)\left(x^2+y^2\right)-y\left(x^4-y^4\right)=y\left(x^4-y^4\right)-y\left(x^4-y^4\right)=0\)
b) \(B=\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3\left(1-x\right)x=x^3-3x^2+3x-1-x^3-x^2-x+x^2+x+1-3x+3x^2=0\)
a: Ta có: \(A=y\left(x^2-y^2\right)\left(x^2+y^2\right)-y\left(x^4-y^4\right)\)
\(=y\left(x^4-y^4\right)-y\left(x^4-y^4\right)\)
=0
b: Ta có: \(B=\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(1-x\right)\)
\(=x^3-3x^2+3x-1-x^3+1-3x+3x^2\)
=0
Bài 3:
\(\left(x-3\right)\left(x-1\right)\left(x+1\right)\left(x+3\right)+15\)
\(=\left(x^2-9\right)\left(x^2-1\right)+15\)
\(=x^4-10x^2+9+15\)
\(=x^4-10x^2+24\)
\(=\left(x^2-4\right)\left(x^2-6\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x^2-6\right)\)
a: \(A=x^2+y^2=\left(x+y\right)^2-2xy=15^2-2\cdot50=115\)
c: \(x-y=\sqrt{\left(x+y\right)^2-4xy}=\sqrt{15^2-4\cdot50}=5\)
\(C=x^2-y^2=\left(x+y\right)\left(x-y\right)=15\cdot5=75\)
\(\text{a) x^2 + y^2 = (x+y)^2 - 2xy = a^2 - 2b}\)
\(\text{b) x^3 + y^3 = (x+y)^3 - 3xy(x+y) = a^3 - 3ab}\)
\(\text{c) x^4 + y^4 = (x^2+y^2)^2 - 2x^2y^2 = (a^2-2b)^2 - 2b^2 = a^4 - 4a^2b + 2b^2}\)
\(\text{d) x^5 + y^5 = (x^3+y^3)(x^2+y^2) - x^2y^2(x+y) = a^5 - 5a^3b + 5ab^2}\)
\(\left\{{}\begin{matrix}x-y=4\\xy=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=y+4\\y\left(y+4\right)=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=y+4\\y^2+4y-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=y+4\\\left[{}\begin{matrix}y=-2+\sqrt{5}\\y=-2-\sqrt{5}\end{matrix}\right.\end{matrix}\right.\)
Với \(y=-2+\sqrt{5}\Rightarrow x=2+\sqrt{5}\)
Với \(y=-2-\sqrt{5}\Rightarrow x=2-\sqrt{5}\)
\(\Rightarrow A=x^2+y^2=\left(-2+\sqrt{5}\right)^2+\left(2+\sqrt{5}\right)^2=\left(2-\sqrt{5}\right)^2+\left(-2-\sqrt{5}\right)^2=18\)
\(B=x^3+y^3\Rightarrow\left[{}\begin{matrix}B=\left(2+\sqrt{5}\right)^3+\left(-2+\sqrt{5}\right)^3=34\sqrt{5}\\B=\left(2-\sqrt{5}\right)^3+\left(-2-\sqrt{5}\right)^3=-34\sqrt{5}\end{matrix}\right.\)
\(\Rightarrow C=x^4+y^4=\left(-2+\sqrt{5}\right)^4+\left(2+\sqrt{5}\right)^4=\left(2-\sqrt{5}\right)^4+\left(-2-\sqrt{5}\right)^4=322\)
a: \(A=x^2+y^2=\left(x+y\right)^2-2xy=15^2-2\cdot50=125\)
b:\(B=x^4+y^4\)
\(=\left(x^2+y^2\right)^2-2x^2y^2\)
\(=125^2-2\cdot2500\)
=10625
c: \(x-y=\sqrt{\left(x+y\right)^2-4xy}=\sqrt{15^2-4\cdot50}=5\)
\(C=x^2-y^2=\left(x-y\right)\left(x+y\right)=15\cdot5=75\)
a.
\(1-4x^2=\left(1-2x\right)\left(1+2x\right)\)
b.
\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
c.
\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)
\(=\left(x+3\right)^3\)
d.
\(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)
e.
\(x^2-y^2-5x+5y=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
f.
\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
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
(x - 5)² = (3 + 2x)²
(x - 5)² - (3 + 2x)² = 0
[(x - 5) - (3 + 2x)][(x - 5) + (3 + 2x)] = 0
(x - 5 - 3 - 2x)(x - 5 + 3 + 2x) = 0
(-x - 8)(3x - 2) = 0
-x - 8 = 0 hoặc 3x - 2 = 0
*) -x - 8 = 0
-x = 8
x = -8
*) 3x - 2 = 0
3x = 2
x = 2/3
Vậy x = -8; x = 2/3
--------------------
27x³ - 54x² + 36x = 9
27x³ - 54x² + 36x - 9 = 0
27x³ - 27x² - 27x² + 27x + 9x - 9 = 0
(27x³ - 27x²) - (27x² - 27x) + (9x - 9) = 0
27x²(x - 1) - 27x(x - 1) + 9(x - 1) = 0
(x - 1)(27x² - 27x + 9) = 0
x - 1 = 0 hoặc 27x² - 27x + 9 = 0
*) x - 1 = 0
x = 1
*) 27x² - 27x + 9 = 0
Ta có:
27x² - 27x + 9
= 27(x² - x + 1/3)
= 27(x² - 2.x.1/2 + 1/4 + 1/12)
= 27[(x - 1/2)² + 1/12] > 0 với mọi x ∈ R
⇒ 27x² - 27x + 9 = 0 (vô lí)
Vậy x = 1
A = x² + y²
= x² - 2xy + y² + 2xy
= (x - y)² + 2xy
= 4² + 2.1
= 16 + 2
= 18
B = x³ - y³
= (x - y)(x² + xy + y²)
= (x - y)(x² - 2xy + y² + xy + 2xy)
= (x - y)[(x - y)² + 3xy]
= 4.(4² + 3.1)
= 4.(16 + 3)
= 4.19
= 76
C = x⁴ + y⁴
= (x²)² + (y²)²
= (x²)² + 2x²y² + (y²)² - 2x²y²
= (x² + y²)² - 2x²y²
= (x² - 2x²y² + y² + 2x²y²)² - 2x²y²
= [(x - y)² + 2x²y²]² - 2x²y²
= (4² + 2.1²)² - 2.1²
= (16 + 2)² - 2
= 18² - 2
= 324 - 2
= 322
a: Ta có: \(y\left(x^2-y^2\right)\cdot\left(x^2+y^2\right)-y\left(x^4-y^4\right)\)
\(=y\left(x^4-y^4\right)-y\left(x^4-y^4\right)\)
=0
b: Ta có: \(\left(2x+\dfrac{1}{3}\right)\left(4x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)-\left(8x^3-\dfrac{1}{27}\right)\)
\(=8x^3+\dfrac{1}{27}-8x^3+\dfrac{1}{27}\)
\(=\dfrac{2}{27}\)
c: Ta có: \(\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(1-x\right)\)
\(=x^3-3x^2+3x-1-x^3+1-3x+3x^2\)
=0
Ta có: \(x^2+y^2=1\Leftrightarrow\left(x^2+y^2\right)^2=1\) (1)
Thay (1) vào \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\) ta được:
\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\Leftrightarrow\frac{x^4b+y^4a}{ab}=\frac{x^4+2x^2y^2+y^4}{a+b}\)
\(\Leftrightarrow\left(x^4b+y^4a\right)\left(a+b\right)=\left(x^4+2x^2y^2+y^4\right)ab\)
\(\Leftrightarrow x^4ab+x^4b^2+y^4a^2+y^4ab=x^4ab+2x^2y^2ab+y^4ab\)
\(\Leftrightarrow x^4b^2+y^4a^2=2x^2y^2ab\)
\(\Leftrightarrow\left(x^2b\right)^2-2x^2y^2ab+\left(y^2a\right)^2=0\)
\(\Leftrightarrow\left(x^2b-y^2a\right)^2=0\)
\(\Leftrightarrow x^2b-y^2a=0\)
\(\Leftrightarrow x^2b=y^2a\)
\(\Rightarrow\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)
\(\Rightarrow\left(\frac{x^2}{a}\right)^{1009}=\left(\frac{y^2}{b}\right)^{1009}=\left(\frac{1}{a+b}\right)^{1009}\)
\(\Rightarrow\frac{x^{2018}}{a^{1009}}=\frac{y^{2018}}{b^{1009}}=\frac{1}{\left(a+b\right)^{1009}}\)
\(\Rightarrow\frac{x^{2018}}{a^{1009}}+\frac{y^{2018}}{b^{1009}}=\frac{1}{\left(a+b\right)^{1009}}+\frac{1}{\left(a+b\right)^{1009}}=\frac{2}{\left(a+b\right)^{1009}}\left(đpcm\right)\)
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