Ta có: 4x=5y => x/5=y/4=>x2/25=y2/16
ta có:
x2/25=y2/16=x2-y2/25-16=1/9
x^2/25=1/9=>x^2=25/9=>x=5/3
y^2/16=1/9=>y^2=16/9=>y=4/3
tích của chúng bằng:5/3*4/3=20/9
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a) \(\left(2x+3y\right)^2=\left(2x\right)^2+2\cdot2x\cdot3y+\left(3y\right)^2=4x^2+12xy+9y^2\)
b) \(\left(x+\dfrac{1}{4}\right)^2=x^2+2\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2=x^2+\dfrac{1}{2}x+\dfrac{1}{16}\)
c) \(\left(x^2+\dfrac{2}{5}y\right)\left(x^2-\dfrac{2}{5}y\right)=\left(x^2\right)^2-\left(\dfrac{2}{5}y\right)^2=x^4-\dfrac{4}{25}y^2\)
d) \(\left(2x+y^2\right)^3=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y^2+3\cdot2x\cdot\left(y^2\right)^2+\left(y^2\right)^3=8x^3+12x^2y^2+6xy^4+y^6\)
e) \(\left(3x^2-2y\right)^2=\left(3x^2\right)^2-2\cdot3x^2\cdot2y+\left(2y\right)^2=9x^4-12x^2y+4y^2\)
f) \(\left(x+4\right)\left(x^2-4x+16\right)=x^3+4^3=x^3+64\)
g) \(\left(x^2-\dfrac{1}{3}\right)\cdot\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3=x^6-\dfrac{1}{27}\)
a) (a - 2b)x(a + 2b)
b) x2-(y-3)2
=> (x-y+3)(x+y-3)
c) (2a + b - a)(2a + b + a)
=> (a+b)(3a+b)
d) (4(x - 1))2 - (5(x + y))2
⇔ (4x - 4 - 5x - 5y)(4x - 4 + 5x + 5y)
⇔ -(x + 5y + 4)(9x + 5y + -4)
e) (x + 5)2
f) (5x - 2y)2
h) (x - 5)(x2 + 5x + 25)
k) (x + 5)3
a: \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)=\dfrac{1}{27}x^3+8y^3\)
b: \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=x^6-\dfrac{1}{27}\)
c: \(\left(y-5\right)\left(y^2+5y+25\right)=y^3-125\)
\(a,=5\left(x-y\right)+a\left(x-y\right)=\left(5+a\right)\left(x-y\right)\\ b,=a\left(x+y\right)+b\left(x+y\right)=\left(a+b\right)\left(x+y\right)\\ c,=x\left(x+1\right)+a\left(x+1\right)=\left(x+a\right)\left(x+1\right)\\ d,Sửa:x^2y+xy^2-3x-3y=xy\left(x+y\right)-3\left(x+y\right)=\left(xy-3\right)\left(x+y\right)\\ e,=xy\left(x+1\right)-\left(x+1\right)=\left(xy-1\right)\left(x+1\right)\\ f,=x^2-4=\left(x-2\right)\left(x+2\right)\\ g,=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\\ h,=\left(x+5\right)^2-y^2=\left(x-y+5\right)\left(x+y+5\right)\\ i,=\left(x-4\right)^2-24y^2=\left(x-2\sqrt{6}y-4\right)\left(x+2\sqrt{6}y+4\right)\)
a: \(=5x\left(xy^2+3x+6y^2\right)\)
b: \(=\left(x-2\right)\left(x+3\right)-\left(x-2\right)\left(x+2\right)=\left(x-2\right)\left(x+3-x-2\right)=\left(x-2\right)\)
c: \(=\left(x-3\right)\left(x-4\right)\)
d: \(=x\left(x^2-2xy+y^2-9\right)\)
=x(x-y-3)(x-y+3)
e: \(=\left(x+y\right)^2-25=\left(x+y+5\right)\left(x+y-5\right)\)
f: \(=\left(x-4\right)\left(x+3\right)\)
Bài 3
a) x² + 10x + 25
= x² + 2.x.5 + 5²
= (x + 5)²
b) 8x - 16 - x²
= -(x² - 8x + 16)
= -(x² - 2.x.4 + 4²)
= -(x - 4)²
c) x³ + 3x² + 3x + 1
= x³ + 3.x².1 + 3.x.1² + 1³
= (x + 1)³
d) (x + y)² - 9x²
= (x + y)² - (3x)²
= (x + y - 3x)(x + y + 3x)
= (y - 2x)(4x + y)
e) (x + 5)² - (2x - 1)²
= (x + 5 - 2x + 1)(x + 5 + 2x - 1)
= (6 - x)(3x + 4)
Bài 4
a) x² - 9 = 0
x² = 9
x = 3 hoặc x = -3
b) (x - 4)² - 36 = 0
(x - 4 - 6)(x - 4 + 6) = 0
(x - 10)(x + 2) = 0
x - 10 = 0 hoặc x + 2 = 0
*) x - 10 = 0
x = 10
*) x + 2 = 0
x = -2
Vậy x = -2; x = 10
c) x² - 10x = -25
x² - 10x + 25 = 0
(x - 5)² = 0
x - 5 = 0
x = 5
d) x² + 5x + 6 = 0
x² + 2x + 3x + 6 = 0
(x² + 2x) + (3x + 6) = 0
x(x + 2) + 3(x + 2) = 0
(x + 2)(x + 3) = 0
x + 2 = 0 hoặc x + 3 = 0
*) x + 2 = 0
x = -2
*) x + 3 = 0
x = -3
Vậy x = -3; x = -2