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\(125-x^6=\left(5\right)^3-\left(x^2\right)^3\)
\(=\left(5-x^2\right)\left(25+5x^2+x^4\right)\)
\(49\left(x-4\right)^2-9\left(y+2\right)^2\)
\(=\left[7\left(x-4\right)\right]^2-\left[3\left(y+2\right)\right]^2\)
\(=\left[7x-28\right]^2-\left[3y+6\right]^2\)
\(=\left(7x-28-3y-6\right)\left(7x-28+3y+6\right)\)
\(=\left(7x-3y-34\right)\left(7x-22+3y\right)\)
\(M=x^2+xy+y^2-3x-3y\)
\(\Rightarrow4M=4x^2+4xy+4y^2-12x-12y\)
\(=\left(x^2+4y^2+9+4xy-12y-6x\right)+\left(3x^2-6x+3\right)-12\)
\(=\left(x+2y-3\right)^2+3\left(x-1\right)^2-12\ge-12\)
\(\Rightarrow M\ge-3\)
\(\Rightarrow Min_M=-3\Leftrightarrow x=y=1\)
Bài 2:
a: \(A=1999\cdot2001\)
\(=\left(2000-1\right)\left(2000+1\right)\)
\(=2000^2-1< 2000^2=B\)
Do đó: B lớn hơn
b: \(C=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\)
\(=2^{16}-1< 2^{16}=D\)
Do đó: D lớn hơn
a,(5x-2y)(x2-xy+1)=5x3-5x2+5x-2yx2+2xy2-2y
=5x3-7x2y+2xy2+5x-2y
b,(x-2)(x+2)(\(\dfrac{1}{2}\) x-5)=x2-4.\(\left(\dfrac{1}{2}x-5\right)\)
=\(\dfrac{1}{2}x^3-5x^2-2x+20\)
c,\(\left(x^2-2x+3\right)\left(\dfrac{1}{2}x-5\right)\)
=\(\dfrac{1}{2}x^3-5x^2-1x^2+10x+\dfrac{3}{2}x-15\)
=\(\dfrac{1}{2}x^3-6x^2+\dfrac{23}{2}x-15\)
d,\(\left(x^2-5\right)\left(x+3\right)+\left(x+4\right)\left(x-x^2\right)\)
=\(x^3+3x^2-5x-15+x^2-x^3+4x-4x^2\)
=\(-5x+4x-15\)
=\(-x-15\)
Chúc bạn học tốt(mỏi tay quá)
a) \(x^3-\dfrac{1}{9}x=0\)
\(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)
\(\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\x+\dfrac{1}{3}=0\Leftrightarrow x=-\dfrac{1}{3}\end{matrix}\right.\)
b) \(x\left(x-3\right)+x-3=0\)
\(\Rightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\Rightarrow x=3\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)
c) \(2x-2y-x^2+2xy-y^2=0\) (thêm đề)
\(\Rightarrow2\left(x-y\right)-\left(x-y\right)^2=0\)
\(\Rightarrow\left(x-y\right)\left(2-x+y\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-y=0\Rightarrow x=y\\2-x+y=0\Rightarrow x-y=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=y\left(1\right)\\\left(1\right)\Rightarrow x-x=2\left(loại\right)\end{matrix}\right.\)
d) \(x^2\left(x-3\right)+27-9x=0\)
\(\Rightarrow x^2\left(x-3\right)+\left(x-3\right).9=0\)
\(\Rightarrow\left(x-3\right)\left(x^2+9\right)=0\)
\(\Rightarrow x-3=0\Rightarrow x=3.\)
\(\text{a) }\left(\dfrac{1}{2}a^2x^4+\dfrac{4}{3}\:ax^3-\dfrac{2}{3}ax^2\right):\left(-\dfrac{2}{3}\:ax^2\right)\\ =-3ax^2-2x+1\)
\(\text{b) }4\left(\dfrac{3}{4}x-1\right)+\left(12x^2-3x\right):\left(-3x\right)-\left(2x+1\right)\\ =3x-4-4x+1-2x-1\\ =-3x-4\)
kết quả cuối cùng là: a. -\(\dfrac{3}{4}ax^2-2x+1\)
b. \(\)-\(3x-4\)
1)
a) \(\dfrac{18ab}{27bc}=\dfrac{2a}{3c}\)
b) \(\dfrac{-21b^2y^2}{-28by}=\dfrac{3by}{4}\)
c) \(\dfrac{-49a^3}{14b^3}=\dfrac{-7a^3}{2b^3}\)
d) \(\dfrac{12x^3y^2}{18xy^5}=\dfrac{2x^2}{3y^3}\)
2)
a) \(\dfrac{a^3\left(a-5\right)}{a-5}=a^3\)
b) \(\dfrac{3\left(b+7\right)^4}{8\left(b+7\right)^6}=\dfrac{3}{8\left(b+7\right)^2}\)
c) \(\dfrac{15x\left(x+5\right)^2}{20x^2\left(x+5\right)}=\dfrac{3\left(x+5\right)}{4x}\)
d) \(\dfrac{x^3-4x^2}{y\left(x-4\right)}=\dfrac{x^2\left(x-4\right)}{y\left(x-4\right)}=\dfrac{x^2}{y}\)
e) \(\dfrac{5\left(a-2c\right)^2}{2a^2-4ac}=\dfrac{5\left(a-2c\right)^2}{2a\left(a-2c\right)}=\dfrac{5\left(a-2c\right)}{2a}\)
3)
a) \(\dfrac{ax-3a}{bx-3b}=\dfrac{a\left(x-3\right)}{b\left(x-3\right)}=\dfrac{a}{b}\) (câu này mình sửa lại đề)
b) \(\dfrac{5x+20y}{15x+60y}=\dfrac{5\left(x+4y\right)}{15\left(x+4y\right)}=\dfrac{1}{3}\)
c) \(\dfrac{3b-9c}{5b^2-15bc}=\dfrac{3\left(b-3c\right)}{5b\left(b-3c\right)}=\dfrac{3}{5b}\)
d) \(\dfrac{8a^2+40ab}{ab+5b^2}=\dfrac{8a\left(a+5b\right)}{b\left(a+5b\right)}=\dfrac{8a}{b}\)
4)
a) \(\dfrac{3x^2-12x+12}{x^4-8x}=\dfrac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}\)
\(=\dfrac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)
b) \(\dfrac{7x^2+14x+7}{3x^2+3x}=\dfrac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}\)
\(=\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)
5)
a) \(\dfrac{45x\left(3-x\right)}{15\left(x-3\right)^3}=\dfrac{-45x\left(x-3\right)}{15\left(x-3\right)^3}=\dfrac{-3x}{\left(x-3\right)^2}\)
b) \(\dfrac{36\left(x-2\right)^3}{32-16x}=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=\dfrac{-9\left(x-2\right)^2}{4}\)
c) \(\dfrac{x^2-xy}{5y^2-5xy}=\dfrac{-x\left(y-x\right)}{5y\left(y-x\right)}=\dfrac{-x}{5y}\)
d) \(\dfrac{y^2-x^2}{x^3-3x^2y+3xy^2-y^3}=\dfrac{-\left(y+x\right)\left(x-y\right)}{\left(x-y\right)^3}=\dfrac{-x-y}{\left(x-y\right)^2}\)
1.
a, \(\dfrac{18ab}{27bc}=\dfrac{18ab:9b}{27bc:9b}=\dfrac{2a}{3c}\)
b, \(\dfrac{-21b^2y^2}{-28by}=\dfrac{-21b^2y^2:\left(-7\right)by}{-28by:\left(-7\right)by}=\dfrac{3by}{4}\)
c, \(\dfrac{-49a^3}{14b^3}=\dfrac{-49a^3:7}{14b^3:7}=\dfrac{-7a^3}{2b^3}\)
d, \(\dfrac{12x^3y^2}{18xy^5}=\dfrac{6xy^2\cdot2x^2}{6xy^2\cdot3y^3}=\dfrac{2x^2}{3y^3}\)
2.
a,\(\dfrac{a^3\cdot\left(a-5\right)}{a-5}=\dfrac{a^3}{1}=a^3\)
b,\(\dfrac{3\cdot\left(b+7\right)^4}{8\cdot\left(b+7\right)^6}=\dfrac{3}{8\cdot\left(b+7\right)^2}\)
c,\(\dfrac{15x\cdot\left(x+5\right)^2}{20x^2\cdot\left(x+5\right)}=\dfrac{3\cdot\left(x+5\right)}{4x}\)
d,\(\dfrac{x^3-4x^2}{y\cdot\left(x-4\right)}=\dfrac{x^2}{y}\)
e,\(\dfrac{5\cdot\left(a-2x\right)^2}{2a^2-4ac}=\dfrac{5\cdot\left(a-2x\right)}{2a}\)
3.
a,\(\dfrac{ax-3a}{bx-3b}=\dfrac{a\cdot\left(x-3\right)}{b\cdot\left(x-3\right)}=\dfrac{a}{b}\)
b, \(\dfrac{5x+20y}{15x+60y}=\dfrac{5\cdot\left(x+4y\right)}{15\cdot\left(x+4y\right)}=\dfrac{5}{15}=\dfrac{1}{3}\)
c, \(\dfrac{3b-9c}{5b^2-15bc}=\dfrac{3\cdot\left(b-3c\right)}{5b\cdot\left(b-3c\right)}=\dfrac{3}{5b}\)
d, \(\dfrac{8a^2+40ab}{ab+5b^2}=\dfrac{8a\cdot\left(a+5b\right)}{b\cdot\left(a+5b\right)}=\dfrac{8a}{b}\)
4.
a,\(\dfrac{3x^2-12x+12}{x^4-8x}=\dfrac{3\cdot\left(x^2-4x+4\right)}{x\cdot\left(x^3-8\right)}=\dfrac{3\cdot\left(x-2\right)^2}{x\cdot\left(x-2\right)\cdot\left(x^2+2x+4\right)}=\dfrac{3\cdot\left(x-2\right)}{x\cdot\left(x^2+2x+4\right)}=\dfrac{3\cdot\left(x-2\right)}{x\cdot\left(x+2\right)^2}\)
b, \(\dfrac{7x^2+14x+7}{3x^2+3x}=\dfrac{7\cdot\left(x^2+2x+1\right)}{3x\cdot\left(x+1\right)}=\dfrac{7\cdot\left(x+1\right)^2}{3x\cdot\left(x+1\right)}=\dfrac{7\cdot\left(x+1\right)}{3x}\)
5.
a, \(\dfrac{45x\cdot\left(3-x\right)}{15x\cdot\left(x-3\right)^3}=\dfrac{3\cdot\left(3-x\right)}{\left(x-3\right)^3}=\dfrac{-3\cdot\left(x-3\right)}{\left(x-3\right)^3}=\dfrac{-3}{\left(x-3\right)^2}\)
b, \(\dfrac{36\cdot\left(x-2\right)^3}{36-16x}=\dfrac{36\cdot\left(x-2\right)^3}{16\cdot\left(2-x\right)}=\dfrac{36\cdot\left(-\left(x-2\right)\right)^3}{16\cdot\left(2-x\right)}=\dfrac{-36\cdot\left(2-x\right)^3}{16\cdot\left(2-x\right)}=\dfrac{-9\cdot\left(2-x\right)^2}{4}\)
c, \(\dfrac{x^2-xy}{5y^2-5xy}=\dfrac{x\cdot\left(x-y\right)}{5y\cdot\left(y-x\right)}=\dfrac{-x\cdot\left(y-x\right)}{5y\cdot\left(y-x\right)}=\dfrac{-x}{5y}\)
d, \(\dfrac{y^2-x^2}{x^3-3x^2y+3xy^2+y^3}=\dfrac{\left(x+y\right)\cdot\left(x-y\right)}{\left(x-y\right)^3}=\dfrac{-\left(x+y\right)\cdot\left(y-x\right)}{\left(x-y\right)^3}=\dfrac{-\left(x+y\right)}{\left(x-y\right)^2}\)
bài 1
đặt a = n5 - n = n (n4 - 1) = n (n - 1) (n + 1) (n2 + 1)
n(n + 1) luôn chia hết cho 2 => a luôn chia hết cho 2
ta cần cm a chia hết cho 5 => có 2 trường hợp
th1: n chia hết cho 5 => a chia hết cho 5
th2: n ko chia hết cho 5 => n = 5k + b (với b = 1 ; 2 ; 3 ; 4)
với b = 1 => n - 1 = 5k
với b = 2 => n2 + 1 = (5k+2)2 + 1 = 25k2 + 20k + 5
=> a chia hết cho 5
với b=3 => n2 + 1 = (5k+3)2 +1 = 25k2 + 30k + 10
=> a chia hết cho 5
với b = 4 => n + 1 = 5k + 5
=> a chia hết cho 5
từ các th trên => a luôn chia hết cho 5
2 và 5 nguyên tố cùng nhau => a chia hết cho 00 => a tận cùng là 0
=> đpcm
a: \(N=\left(5x\right)^3-\left(2y\right)^3=1^3-1^3=0\)
b: \(Q=x^3+27y^3=\dfrac{1}{8}+\dfrac{27}{8}=\dfrac{28}{8}=\dfrac{7}{2}\)