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=>4a^2-5ab+b^2=0
=>(a-b)(4a-b)=0
=>a=b hoặc b=4a(loại)
=>P=b^2/3b^2=1/3
Ta có:
\(4a^2+b^2=5ab\Leftrightarrow4a^2+b^2-4ab-ab=0\)
\(\Leftrightarrow4a\left(a-b\right)-b\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(4a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\4a-b=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=b\left(ktm\right)\\4a=b\left(tm\right)\end{matrix}\right.\)
\(\Rightarrow4a=b\)
\(\Rightarrow\dfrac{5ab}{3a^2+2b^2}=\dfrac{5a.4a}{3a^2+2.\left(4a\right)^2}=\dfrac{20a^2}{3a^2+32a^2}\)
\(=\dfrac{20a^2}{35a^2}=\dfrac{4}{7}\)
\(4a^2+b^2=5ab\)
\(\Rightarrow4a\left(a-b\right)-b\left(a-b\right)=0\)
\(\Rightarrow\left(a-b\right)\left(4a-b\right)=0\)
\(\Rightarrow b=4a\left(do.a\ne b\right)\)
\(\dfrac{5ab}{3a^2+2b^2}=\dfrac{20a^2}{3a^2+32a^2}=\dfrac{4}{7}\)
a,\(5ab-45a^3b\)
=\(5ab\left(1-9a^2\right)\)
=\(5ab\left(1-3a\right)\left(1+3a\right)\)
b,\(3a-6ab+5-10b\)
=\(\left(3a-6ab\right)+\left(5-10b\right)\)
=\(3a\left(1-2b\right)+5\left(1-2b\right)\)
=\(\left(1-2b\right)\left(3a+5\right)\)
c,\(a^2-7ab-2a+14b\)
=\(\left(a^2-7ab\right)-\left(2a-14b\right)\)
=\(a\left(a-7b\right)-2\left(a-7b\right)\)
=\(\left(a-7b\right)\left(a-2\right)\)
d,\(4a^2-8b+4a-8ab\)
=\(\left(4a^2-8ab\right)+\left(4a-8b\right)\)
=\(4a\left(a-2b\right)+4\left(a-2b\right)\)
=\(\left(a-2b\right)\left(4a+4\right)\)
=\(4\left(a-2b\right)\left(a+1\right)\)
e,\(a^2-5a+15b-9b^2\)
=\(\left(a^2-9b^2\right)-\left(5a-15b\right)\)
=\(\left(a-3b\right)\left(a+3b\right)-5\left(a-3b\right)\)
=\(\left(a-3b\right)\left(a+3b-5\right)\)
Bài 1:
a: \(4a^2-6b=2\left(2a^2-3b\right)\)
b: \(m^3n-2m^2n^2-mn\)
\(=mn\left(m^2-2mn-1\right)\)
Bài 1:
a) \(4a^2-6b=2\left(a^2-3b\right)\)
b) \(=mn\left(m^2-2mn-1\right)\)
Bài 2:
a) \(=4\left(u-2\right)^2+v\left(u-2\right)=\left(u-2\right)\left(4u-8+v\right)\)
b) \(=a\left(a-b\right)^3-b\left(a-b\right)^2-b^2\left(a-b\right)=\left(a-b\right)\left[a\left(a-b\right)^2-b\left(a-b\right)-b^2\right]=\left(a-b\right)\left(a^3-2a^2b+ab^2-ab+b^2-b^2\right)=\left(a-b\right)\left(a^3-2a^2b+ab^2-ab\right)\)
`a)x^4+2x^2y+y^2`
`=(x^2+y)^2`
`b)(2a+b)^2-(2b+a)^2`
`=(2a+b-2b-a)(2a+b+2b+a)`
`=(a-b)(3a+3b)`
`=3(a-b)(a+b)`
`c)8a^3-27b^3-2a(4a^2-9b^2)`
`=(2a-3b)(4a^2+6ab+9b^2)-2a(2a-3b)(2a+3b)`
`=(2a-3b)(4a^2+6ab+9b^2-3a^2-6ab)`
`=9b^2(2a-3b)`
a) Ta có: \(x^4+2x^2y+y^2\)
\(=\left(x^2\right)^2+2\cdot x^2\cdot y+y^2\)
\(=\left(x^2+y\right)^2\)
b) Ta có: \(\left(2a+b\right)^2-\left(2b+a\right)^2\)
\(=\left(2a+b-2b-a\right)\left(2a+b+2b+a\right)\)
\(=\left(a-b\right)\left(3a+3b\right)\)
\(=3\left(a+b\right)\left(a-b\right)\)
d) (8a3 – 27b3) – 2a(4a2 – 9b2)
= (2a – 3b)(4a2 + 6ab + 9b2) – 2a(2a – 3b)(2a + 3b)
= (2a – 3b)(4a2 + 6ab + 9b2 – 4a2 – 6ab) = 9b2(2a – 3b)