Thực hiện các phép tính sau:
a) x + 4 x + 5 : x + 5 x + 6 : x + 6 x + 4 với x ≠ − 6 ; x ≠ − 5 và x ≠ − 4
b) 3 x 2 7 y 2 : 6 x 7 y : 2 x 3 y với x ≠ 0 và y ≠ 0
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a) \({x^2} + \dfrac{1}{4}{x^2} - 5{x^2} = (1 + \dfrac{1}{4} - 5){x^2} = - \dfrac{{15}}{4}{x^2}\);
b) \({y^4} + 6{y^4} - \dfrac{2}{5}{y^4} = (1 + 6 - \dfrac{2}{5}){y^4} = \dfrac{{33}}{5}{y^4}\).
a/ \(\left(2x+3\right)\left(x-5\right)-\left(x-1\right)^2+x\left(7-x\right)\)
\(=2x^2-2x-15-x^2+2x-1+7x-x^2\)
\(=7x-16\)
\(a,\left(x^3+5x^2-2x+1\right)\left(x-7\right)\\ =x^4-7x^3+5x^3-35x^2-2x^2+14x+x-7\\ =x^4-2x^3-37x^2+15x-7\\ b,\left(2x^2-3xy+y^2\right)\left(x+y\right)\\ =2x^3+2x^2y-3x^2y-3xy^2+xy^2+y^3\\ =2x^3-x^2y-2xy^2+y^3\\ c,\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\\ =x^3-5x^2+x-2x^2+10x--x^3-11x\\ =x^3-7x^2\\ d,x\left(1-3x\right)\left(4-3x\right)-\left(x-4\right)\left(3x+5\right)\\ =x\left(4-15x+9x^2\right)-\left(3x^2-7x-20\right)\\ =4x-15x^2+9x^3-3x^2+7x+20\\ =9x^3-18x^2+11x+20\)
\(a,\dfrac{x+2}{x-1}-\dfrac{x-3}{x-1}-\dfrac{x-4}{1-x}\\ =\dfrac{x+2}{x-1}-\dfrac{x-3}{x-1}+\dfrac{x-4}{x-1}\\ =\dfrac{x+2-x+3+x-4}{x-1}\\ =\dfrac{x+1}{x-1}\)
\(b,\dfrac{1}{x+5}-\dfrac{1}{x-5}+\dfrac{2x}{x^2-25}\\ =\dfrac{1}{x+5}-\dfrac{1}{x-5}+\dfrac{2x}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{x-5-x-5+2x}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{2x-10}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{2\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{2}{x+5}\)
\(c,x+\dfrac{2y^2}{x+y}-y\\ =\dfrac{x\left(x+y\right)+2y^2-y\left(x+y\right)}{x+y}\\ =\dfrac{x^2+xy+2y^2-xy-y^2}{x+y}\\ =\dfrac{x^2+y^2}{x+y}\)
Bài 2:
1: \(A=\left(x+2\right)\left(x^2-2x+4\right)+2\left(x+1\right)\left(1-x\right)\)
\(=\left(x+2\right)\left(x^2-x\cdot2+2^2\right)-2\left(x+1\right)\left(x-1\right)\)
\(=x^3+2^3-2\left(x^2-1\right)\)
\(=x^3+8-2x^2+2=x^3-2x^2+10\)
\(B=\left(2x-y\right)^2-2\left(4x^2-y^2\right)+\left(2x+y\right)^2+4\left(y+2\right)\)
\(=\left(2x-y\right)^2-2\cdot\left(2x-y\right)\left(2x+y\right)+\left(2x+y\right)^2+4\left(y+2\right)\)
\(=\left(2x-y-2x-y\right)^2+4\left(y+2\right)\)
\(=\left(-2y\right)^2+4\left(y+2\right)\)
\(=4y^2+4y+8\)
2: Khi x=2 thì \(A=2^3-2\cdot2^2+10=8-8+10=10\)
3: \(B=4y^2+4y+8\)
\(=4y^2+4y+1+7\)
\(=\left(2y+1\right)^2+7>=7>0\forall y\)
=>B luôn dương với mọi y
Bài 1:
5: \(x^2\left(x-y+1\right)+\left(x^2-1\right)\left(x+y\right)\)
\(=x^3-x^2y+x^2+x^3+x^2y-x-y\)
\(=2x^3-x+x^2-y\)
6: \(\left(3x-5\right)\left(2x+11\right)-6\left(x+7\right)^2\)
\(=6x^2+33x-10x-55-6\left(x^2+14x+49\right)\)
\(=6x^2+23x-55-6x^2-84x-294\)
=-61x-349
Bài 1:
b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)
Bài 2:
a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)
d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)
\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)
e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)
\(a,=\dfrac{5x}{4y^3}\times\left(\dfrac{-20y}{x^4}\right)=\dfrac{-100xy}{4x^4y^3}=\dfrac{-25}{x^3y^2}\\ b,=\dfrac{\left(x-4\right)\left(x+4\right)}{\left(x+4\right)}\times\dfrac{x}{2\left(x-4\right)}=\dfrac{x}{2}\)
\(c,=\dfrac{2\left(x+3\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\times\dfrac{2\left(x-2\right)}{\left(x+3\right)^3}=\dfrac{4}{\left(x+3\right)^2.\left(x^2+2x+4\right)}\)
a) \(\dfrac{5x}{4y^3}:\left(-\dfrac{x^4}{20y}\right)=\dfrac{5x}{4y^3}\cdot\left(-\dfrac{20y}{x^4}\right)=\dfrac{5\cdot-5}{y^2\cdot x^3}=\dfrac{-25}{x^3y^2}\)
b) \(\dfrac{x^2-16}{x+4}:\dfrac{2x-8}{x}=\left(x-4\right)\cdot\dfrac{x}{2\left(x-4\right)}=\dfrac{x}{2}\)
c) \(\dfrac{2x+6}{x^3-8}:\dfrac{\left(x+3\right)^3}{2x-4}=\dfrac{2\left(x+3\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\cdot\dfrac{2\left(x-2\right)}{\left(x+3\right)^3}=\dfrac{4}{\left(x^2+2x+4\right)\left(x+3\right)^2}\)
a)
160 : (4 × 8) = 160 : 32 = 5
160 : 4 : 8 = 40 : 8 = 5
Vậy 160 : (4 × 8) = 160 : 4 : 8.
96 : (3 × 8) = 96 : 24 = 4
96 : 3 : 8 = 32 : 8 = 4
Vậy 96 : (3 × 8) = 96 : 3 : 8.
105 : (5 × 7) = 105 : 35 = 3
105 : 5 : 7 = 21 : 7 = 3
Vậy 105 : (5 × 7) = 105 : 5 : 7.
Nhận xét: a : (b × c) = a : b : c
b)
270 : (9 × 6) = 270 : 9 : 6 = 30 : 6 = 5
420 : (7 × 3) = 420 : 7 : 3 = 60 : 3 = 20
144 : (2 × 8) = 144 : 2 : 8 = 72 : 8 = 9
a) ( x + 4 ) 2 ( x + 5 ) 2 b) x 2 3 y 2