3x,y+9x-y=28
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3: 10x=6y=5z
\(\Leftrightarrow\dfrac{10x}{30}=\dfrac{6y}{30}=\dfrac{5z}{30}\)
hay x/3=y/5=z/6
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x+y-z}{3+5-6}=\dfrac{24}{2}=12\)
Do đó: x=36; y=60; z=72
4: Ta có: 9x=3y=2z
nên \(\dfrac{9x}{18}=\dfrac{3y}{18}=\dfrac{2z}{18}\)
hay x/2=y/6=z/9
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{6}=\dfrac{z}{9}=\dfrac{x-y+z}{2-6+9}=\dfrac{50}{5}=10\)
Do đó: x=20; y=60; z=90

c) \(\dfrac{y^4-1}{y^3+y^2+y+1}\)
\(=\dfrac{\left(y^2+1\right)\left(y^2-1\right)}{y^2\left(y+1\right)+\left(y+1\right)}\)
\(=\dfrac{\left(y^2+1\right)\left(y+1\right)\left(y-1\right)}{\left(y+1\right)\left(y^2+1\right)}\)
\(=y-1\)
d) \(\dfrac{2x^2-9x+7}{-2x^2-x+28}\)
\(=\dfrac{2x^2-2x-7x+7}{-\left(2x^2+8x-7x-28\right)}\)
\(=\dfrac{2x\left(x-1\right)-7\left(x-1\right)}{-\left(2x-7\right)\left(x+4\right)}\)
\(=-\dfrac{\left(2x-7\right)\left(x-1\right)}{\left(2x-7\right)\left(x+4\right)}\)
\(=\dfrac{1-x}{x+4}\)

\(\left(9x^2+3x\right)\left(x-y\right)=3x\left(3+1\right)\left(x-y\right)\)

\(=\left(3x-y\right)\left(9x-3y\right)=3\left(3x-y\right)^2\)
=9x(3x-y)-3y(3x-y)
=(3x-y)(9x-3y)
Chúc bạn học tốt nha ^^

\(a,\)\(2\left(x-y\right)\left(x+y\right)+\left(x-y\right)^2+\left(x+y\right)^2.\)
\(=\left[\left(x-y\right)+\left(x+y\right)\right]^2=\left(x-y+x+y\right)^2=x^2\)
\(b,\)\(\left(2x-3\right)\left(4x^2+6x+9\right)-\left(54+8x\right)\)
\(=8x^2-27-54-8x=8x^2-8x-81\)
\(c,\)\(\left(3x+y\right)\left(9x^2-3xy+y^2\right)-\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)
\(=27x^3+y^3-\left(27x^3-y^3\right)=2y^3\)
\(d,\)\(\left(a+b+c\right)^2-\left(a-c\right)^2-2ab+2bc\)
\(=a^2+b^2+c^2+2ab+2bc+2ac-a^2+2ac-c^2-2ab+2bc\)
\(=b^2+4bc+4ac\)

a) 9x (3x-y) +3y (y-3x)
= 9x(3x-y)-3y(3x-y)
=(3x-y)(9x-3y)
= (3x-y)3(3x-y)
= 3(3x-y)2
b)x3 - 3x2 - 9x + 27
=x3+3x2-6x2-18x+9x+27
= (x3+3x2)-(6x2+18x)+(9x+27)
= x2(x+3)-6x(x+3)+9(x+3)
= (x+3)(x2-6x+9)
=(x+3)(x-3)2
a) \(9x\left(3x-y\right)+3y\left(y-3x\right)=9x\left(3x-y\right)-3y\left(3x-y\right)\)
=\(\left(3x-y\right)\left(9x-3y\right)\)
\(=3\left(3x-y\right)\left(3x-y\right)\)
\(=3\left(3x-y\right)^2\)
b) \(x^3-3x^2-9x+27=\left(x^3-3x^2\right)-\left(9x-27\right)\)
\(=x^2\left(x-3\right)-9\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-9\right)\)
\(=\left(x-3\right)\left(x-3\right)\left(x+3\right)\)
=\(\left(x-3\right)^2\left(x+3\right)\)


a) \(\left(3x-5\right)\left(3x+5\right)=9x^2-25\Leftrightarrow9x^2+15x-15x-25=9x^2-25\Leftrightarrow9x^2-25=9x^2-25\)(đúng)
b) \(x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)\Leftrightarrow x^3-y^3=x^3+x^2y+xy^2-x^2y-xy^2-y^3\Leftrightarrow x^3-y^3=x^3-y^3\)(đúng)
c) \(x^2+y^2=\left(x+y\right)^2-2xy\Leftrightarrow x^2+y^2=x^2+y^2+2xy-2xy\Leftrightarrow x^2+y^2=x^2+y^2\)(đúng)
a: \(\left(3x-5\right)\left(3x+5\right)\)
\(=9x^2+15x-15x-25\)
\(=9x^2-25\)
b: \(\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x^3+x^2y+xy^2-x^2y-xy^2-y^3\)
\(=x^3-y^3\)
c: \(\left(x+y\right)^2-2xy\)
\(=x^2+2xy+y^2-2xy\)
\(=x^2+y^2\)
3x . y + 3x . 3 - y = 28
3x ( y+3) - y = 28
3x ( y+ 3) = 28 + y
3x ( y+3) = 9.3 + 1 + y
x(y+3) = 9+1+y
x(y+3) = 10 +y
x = 10 + y : y + y: 3
x = 10 + 1 + y : 3
x = 11 + y : 3
11+ y : 3 = x
11+ y = 3x
3.4 -1 + y = 3x
3+y = x
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