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KV
23 tháng 6 2023
Bài 7
Ta có:
VT = a³ + b³ = (a + b)(a² - ab + b²)
= (a + b)(a² - 2ab + b² + ab)
= (a + b)[(a - b)² + ab]
= VP
Vậy a³ + b³ = (a + b)[(a - b)² + ab]
KV
23 tháng 6 2023
Bài 8
(x + 2)³ + (x - 2)³ + x³ - 3x(x + 2)(x - 2)
= x³ + 6x² + 12x + 8 + x³ - 6x² + 12x - 8 + x³ - 3x(x² - 4)
= 3x³ + 24x - 3x³ + 12x
= 36x
24 tháng 6 2023
22:
A=9x^2+12x+4+4x^2-28x+49-2(6x^2+15x+4x+10)
=13x^2-16x+53-12x^2-38x-20
=x^2-54x+33
=(-19)^2-54*(-19)+33=1420
24 tháng 6 2023
Bài 4 :
\(A=5\left(x+3\right)\left(x-3\right)+\left(2x+3\right)^2+\left(x-6\right)^2\)
\(A=5\left(x^2-9\right)+\left(4x^2+12x+9\right)+\left(x^2-12x+36\right)\)
\(A=5x^2-45+4x^2+12x+9+x^2-12x+36\)
\(A=10x^2\)
Với x = -1/5 => A = \(10.\left(\dfrac{-1}{5}\right)^2=\dfrac{2}{5}\)
H9
HT.Phong (9A5)
CTVHS
26 tháng 6 2023
Bài 3:
1) \(x^2-25=x^2-5^2=\left(x+5\right)\left(x-5\right)\)
2) \(9x^2-\dfrac{1}{16}y^2=\left(3x\right)^2-\left(\dfrac{1}{4}y\right)^2=\left(3x+\dfrac{1}{4}y\right)\left(3x-\dfrac{1}{4}y\right)\)
3) \(x^6-y^4=\left(x^3\right)^2-\left(y^2\right)^2=\left(x^3+y^2\right)\left(x^3-y^2\right)\)
4) \(\left(2x-5\right)^2-64=\left(2x-5\right)^2-8^2=\left[\left(2x-5\right)+8\right]\left[\left(2x-5\right)-8\right]=\left(2x+3\right)\left(2x-13\right)\)
5) \(81-\left(3x+2\right)^2=9^2-\left(3x+2\right)=\left[9-\left(3x+2\right)\right]\left[9+\left(3x+2\right)\right]=\left(7-3x\right)\left(11+3x\right)\)
26 tháng 6 2023
17:x^2-9=(x-3)(x+3)
18: 4x^2-25=(2x-5)(2x+5)
19: =(x^2-y^2)(x^2+y^2)
=(x-y)(x+y)(x^2+y^2)
20: 9x^2+6xy+y^2=(3x+y)^2
21 6x-9-x^2
=-(x^2-6x+9)
=-(x-3)^2
22: x^2+4xy+4y^2
=x^2+2*x*2y+(2y)^2
=(x+2y)^2
23: =(x+y+x-y)(x+y-x+y)
=2x*2y=4xy
25: =(3x+1+x+1)(3x+1-x-1)
=(4x+2)*2x
=4x(2x+1)
27: =(2x-y)(4x^2+2xy+y^2)-6xy(2x-y)
=(2x-y)(2x-y)^2
=(2x-y)^3
GH
27 tháng 6 2023
5
\(=4^2-x^2=\left(4-x\right)\left(4+x\right)\)
6
\(=4^2-\left(3x+1\right)^2=\left(4-3x-1\right)\left(4+3x+1\right)=\left(3-3x\right)\left(5+3x\right)\\ =3\left(1-x\right)\left(5+3x\right)\)
7
\(=\left(2x+5\right)^2-\left(3x\right)^2=\left(2x+5+3x\right)\left(2x+5-3x\right)\\ =\left(5x+5\right)\left(5-x\right)\\ =5\left(x+1\right)\left(5-x\right)\)
8
\(=\left(2x-1-3x+1\right)\left(2x-1+3x-1\right)=\left(-x\right)\left(5x-2\right)\)
9
\(=\left(2x\right)^2-2.2x.y+y^2=\left(2x-y\right)^2\)
10
\(=\left(x+1\right)^2-\left(3y\right)^2=\left(x+1-3y\right)\left(x+1+3y\right)\)
11
\(=\left(x^2y^2\right)^2+2.2x^2y^2+2^2=\left(x^2y^2+2\right)\)
12
\(=\left(y-2\right)^2-x^2=\left(y-2-x\right)\left(y-2+x\right)\)
13
\(=1-\left(3\sqrt{3}x\right)^3=\left(1-3\sqrt{3}x\right)\left[1^2+3\sqrt{3}.x+\left(3\sqrt{3}.x\right)^2\right]=\left(1-3\sqrt{3}x\right)\left(1+3\sqrt{3}x+27x^2\right)\)
H9
HT.Phong (9A5)
CTVHS
27 tháng 6 2023
Bài 5:
a) \(M=\left(2x-1\right)^2+2\left(2x-1\right)\left(3x+1\right)+\left(3x+1\right)^2\)
\(M=\left[\left(2x-1\right)+\left(3x+1\right)\right]^2\)
\(M=\left(5x\right)^2\)
Thay \(x=-\dfrac{1}{5}\) vào biểu thức M ta có:
\(\left(5\cdot-\dfrac{1}{5}\right)^2=\left(-1\right)^2=1\)
Vậy: ...
b) \(N=\left(3x-1\right)^2-2\left(9x^2-1\right)+\left(3x+1\right)^2\)
\(N=\left(3x-1\right)^2-2\left(3x+1\right)\left(3x-1\right)+\left(3x+1\right)^2\)
\(N=\left[\left(3x-1\right)-\left(3x+1\right)\right]^2\)
\(N=\left(-2\right)^2=4\)
Vậy: ...
KV
28 tháng 6 2023
a) (x + 2)³ + 1
= (x + 2)³ + 1³
= [(x + 2) + 1][(x + 2)² - (x + 2).1 + 1²]
= (x + 3)(x² + 4x + 4 - x - 2 + 1)
= (x + 3)(x² + 3x + 3)
b) x³ + 6x² + 12x + 9
= x³ + 3x² + 3x² + 9x + 3x + 9
= (x³ + 3x²) + (3x² + 9x) + (3x + 9)
= x²(x + 3) + 3x(x + 3) + 3(x + 3)
= (x + 3)(x² + 3x + 3)
c) x³ + 6x² + 12x + 7
= x³ + x² + 5x² + 5x + 7x + 7
= (x³ + x²) + (5x² + 5x) + (7x + 7)
= x²(x + 1) + 5x(x + 1) + 7(x + 1)
= (x + 1)(x² + 5x + 7)
d) 2x³ + 6x² + 12x + 8
= 2(x³ + 3x² + 6x + 4)
= 2(x³ + x² + 2x² + 2x + 4x + 4)
= 2[(x³ + x²) + (2x² + 2x) + (4x + 4)]
= 2[x²(x + 1) + 2x(x + 1) + 4(x + 1)]
= 2(x + 1)(x² + 2x + 4)
Bài 1:
a: \(\dfrac{2x^3+3x^2-2x-1}{2x+3}\)
\(=\dfrac{2x^3+3x^2-2x-3+2}{2x+3}\)
\(=\dfrac{x^2\left(2x+3\right)-\left(2x+3\right)+2}{2x+3}\)
\(=x^2-1+\dfrac{2}{2x+3}\)
vậy: \(2x^3+3x^2-2x-1=\left(x^2-1\right)\left(2x+3\right)+2\)
b: \(3x\left(x-2\right)+5\left(2-x\right)=0\)
=>3x(x-2)-5(x-2)=0
=>(x-2)(3x-5)=0
=>\(\left[{}\begin{matrix}x-2=0\\3x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\)
Bài 2:
\(A=\left(\dfrac{2xy}{x^2-y^2}+\dfrac{x-y}{2x+2y}\right)\cdot\dfrac{2x}{x+y}+\dfrac{y}{x-y}\)
\(=\left(\dfrac{2xy}{\left(x-y\right)\left(x+y\right)}+\dfrac{x-y}{2\left(x+y\right)}\right)\cdot\dfrac{2x}{x+y}+\dfrac{y}{x-y}\)
\(=\dfrac{4xy+\left(x-y\right)^2}{2\left(x+y\right)\left(x-y\right)}\cdot\dfrac{2x}{x+y}+\dfrac{y}{x-y}\)
\(=\dfrac{\left(x+y\right)^2\cdot2x}{2\left(x+y\right)^2\cdot\left(x-y\right)}+\dfrac{y}{x-y}\)
\(=\dfrac{x}{x-y}+\dfrac{y}{x-y}=\dfrac{x+y}{x-y}\)
b: \(\dfrac{x^2+y^2}{xy}=\dfrac{25}{12}\)
=>\(12\left(x^2+y^2\right)-25xy=0\)
=>\(12x^2-16xy-9xy+12y^2=0\)
=>\(4x\left(3x-4y\right)-3y\left(3x-4y\right)=0\)
=>\(\left(3x-4y\right)\left(4x-3y\right)=0\)
=>\(\left[{}\begin{matrix}3x-4y=0\\4x-3y=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}y\\x=\dfrac{3}{4}y\end{matrix}\right.\)
x<y<0 nên \(x=\dfrac{4}{3}y\)
\(A=\dfrac{x+y}{x-y}=\dfrac{\dfrac{4}{3}y+y}{\dfrac{4}{3}y-y}=\dfrac{7}{3}:\dfrac{1}{3}=7\)