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a: \(A=\dfrac{10ab^2-5a^2}{16b^2-8ab}=\dfrac{5a\left(2b^2-a\right)}{8b\left(2b-a\right)}=\dfrac{\dfrac{5}{6}\cdot\left(2\cdot\dfrac{1}{49}-\dfrac{1}{6}\right)}{\dfrac{8}{7}\cdot\left(\dfrac{2}{7}-\dfrac{1}{6}\right)}=-\dfrac{37}{48}\)
b: \(A=\dfrac{a^7+1}{a^8\left(a^7+1\right)}=\dfrac{1}{a^8}=\dfrac{1}{0.1^8}=10^8\)
c: \(=\dfrac{2\left(x-2y\right)}{0.2\left(x^2-4y^2\right)}=\dfrac{10}{x+2y}=\dfrac{10}{5}=2\)
d: \(=\dfrac{\left(x-3y\right)\left(x+3y\right)}{1.5\left(x+3y\right)}=\dfrac{x-3y}{1.5}=\dfrac{3}{1.5}=2\)
\(P=27y^3+9y^2+y+\dfrac{1}{27}=\left(3y+3\right)^3\)
Với \(y=\dfrac{2}{3}\) ta có:
\(P=\left(3.\dfrac{2}{3}+3\right)^3=5^3=125\)
\(Q=x^2+4y^2-2x+10+4xy-4y\)
\(=\left(x^2-2x+4xy\right)+4y^2-4y+10\)
\(=\left[x^2-2x\left(1-2y\right)+\left(1-2y\right)^2\right]+4y^2-4y+10-\left(1-2y\right)^2\)\(=\left(x+2y-1\right)^2+4y^2-4y+10-1+4y-4y^2\)\(=\left(x+2y-1\right)^2+9\)
Với \(x+2y=5\) , ta có:
\(Q=\left(5-1\right)^2+9=25\)
a) Ta có: 10(x-y)-8y(y-x)
\(=10\left(x-y\right)+8y\left(x-y\right)\)
\(=2\left(x-y\right)\left(5+4y\right)\)
d) Ta có: \(x^2y-x^3-9y+9x\)
\(=x^2\left(y-x\right)-9\left(y-x\right)\)
\(=\left(y-x\right)\left(x^2-9\right)\)
\(=\left(y-x\right)\left(x-3\right)\left(x+3\right)\)
e) Ta có: \(2x+2y-x^2-xy\)
\(=2\left(x+y\right)-x\left(x+y\right)\)
\(=\left(x+y\right)\left(2-x\right)\)
f) Ta có: \(x^2-25+y^2+2xy\)
\(=\left(x+y\right)^2-5^2\)
\(=\left(x+y-5\right)\left(x+y+5\right)\)
g) Ta có: \(x^2-2x-4y^2-4y\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
h) Ta có: \(x^2\left(x-1\right)+16\left(1-x\right)\)
\(=x^2\left(x-1\right)-16\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-16\right)\)
\(=\left(x-1\right)\left(x-4\right)\left(x+4\right)\)
Bài 4:
a: \(\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2-12=2\)
\(\Leftrightarrow3x-40=2\)
=>3x=42
hay x=14
b: \(\Leftrightarrow x^3+8-x^3-2x=0\)
=>-2x+8=0
=>-2x=-8
hay x=4
c: \(x\left(x-2\right)+\left(x-2\right)=0\)
=>(x-2)(x+1)=0
=>x=2 hoặc x=-1
d: \(5x\left(x-3\right)-x+3=0\)
=>5x(x-3)-(x-3)=0
=>(x-3)(5x-1)=0
=>x=3 hoặc x=1/5
e: \(3x\left(x-5\right)-\left(x-1\right)\left(3x+2\right)=30\)
\(\Leftrightarrow3x^2-15x-3x^2-2x+3x+2=30\)
=>-14x=28
hay x=-2
f: \(\Leftrightarrow\left(x+2\right)\left(x+30-x-5\right)=0\)
=>x+2=0
hay x=-2
a) \(\frac{2x-4y}{0,2x^2-0,8y^2}\) = \(\frac{2(x-2y)}{0,2(x^2-4y^2)}\) = \(\frac{2(x-2y)}{0,2(x-2y)(x+2y)}\)
= \(\frac{10}{x+2y}\) = \(\frac{10}{5}\) = 2 Vì \(x+2y=5\)
b) Ta có : \(\frac{x^2-9y^2}{1,5x+4,5y}\) ĐKXĐ : \(x\ne -3y\)
Ta có : \(3x-9y=1 \) => \(x-3y = \frac{1}{3}\) => x= \(\frac{1}{3} + 3y\) ( Thỏa mãn ĐKXĐ )
Ta có : \(\frac{x^2-9y^2}{1,5x+4,5y}\) = \(\frac{(x-3y)(x+3y)}{1,5(x+3y)}\) = \(\frac{x-3y}{1,5}\) = \(\frac{\frac{1}{3}}{1,5}\) = \(\frac{2}{9}\)