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\(2a^2+2b^2=5ab\\ \Leftrightarrow2a^2-5ab+2b^2=0\\ \Leftrightarrow2a^2-4ab-ab+2b^2=0\\ \Leftrightarrow2a\left(a-2b\right)+b\left(a-2b\right)=0\\ \Leftrightarrow\left(2a+b\right)\left(a-2b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=-\dfrac{b}{2}\\a=2b\end{matrix}\right.\)
Với \(a=-\dfrac{b}{2}\Leftrightarrow Q=\dfrac{-\dfrac{b}{2}+b}{-\dfrac{b}{2}-b}=\dfrac{b}{2}:\dfrac{-3b}{2}=\dfrac{b}{-3b}=-\dfrac{1}{3}\)
Với \(a=2b\Leftrightarrow Q=\dfrac{3b}{b}=3\)
Bài 3:
a: Ta có: \(\left(y-5\right)\left(y+8\right)-\left(y+4\right)\left(y-1\right)\)
\(=y^2+8y-5y-40-y^2+y-4y+4\)
=-36
b: Ta có: \(y^4-\left(y^2-1\right)\left(y^2+1\right)\)
\(=y^4-y^4+1\)
=1
Bài 2:
a: \(\left(2a-b\right)\left(4a+b\right)+2a\left(b-3a\right)\)
\(=8a^2+2ab-4ab-b^2+2ab-6a^2\)
\(=2a^2-b^2\)
b: \(\left(3a-2b\right)\left(2a-3b\right)-6a\left(a-b\right)\)
\(=6a^2-9ab-4ab+6b^2-6a^2+6ab\)
\(=6b^2-7ab\)
c: \(5b\left(2x-b\right)-\left(8b-x\right)\left(2x-b\right)\)
\(=10bx-5b^2-16bx+8b^2+2x^2-xb\)
\(=3b^2-7xb+2x^2\)
a, ĐK: \(a\ne0,b\ne0,a+b\ne0\)
\(A=\left[\frac{1}{a^2}+\left(\frac{1}{a}+\frac{1}{b}\right):\frac{a+b}{2}+\frac{1}{b^2}\right].\frac{a^2b^2}{a^3+b^3}:\left(a+b\right)\)
\(=\left[\frac{1}{a^2}+\frac{a+b}{ab}:\frac{a+b}{2}+\frac{1}{b^2}\right].\frac{a^2b^2}{a^3+b^3}:\left(a+b\right)\)
\(=\left[\frac{1}{a^2}+\frac{2}{ab}+\frac{1}{b^2}\right].\frac{a^2b^2}{a^3+b^3}:\left(a+b\right)\)
\(=\frac{\left(a+b\right)^2}{a^2b^2}.\frac{a^2b^2}{\left(a+b\right)\left(a^2-ab+b^2\right)}.\frac{1}{a+b}\)
\(=\frac{1}{a^2-ab+b^2}\)
b, \(a^2-ab+b^2=\left(a-\frac{1}{2}b\right)^2+\frac{3}{4}b^2>0\left(a,b\ne0\right)\)
\(\Rightarrow A=\frac{1}{a^2-ab+b^2}>0\forall a;b\)
\(a,=x+x^2-x^3+x^4-x^5+1+x-x^2+x^3-x^4-x-x^2+x^3-x^4+x^5+1+x-x^2+x^3-x^4\\ =2x-2x^2+2x^3-2x^4\)
Ta có : a2 + 2ab + b2 + b2 - 4b +4 = 0
<=> ( a + b )2 + ( b - 2 )2 = 0
mà: ( a + b )2≥0 ∀a,b
( b - 2 )2 ≥0 ∀b
Dấu "=" xảy ra khi :
a + b =0
b - 2 =0
<=> a + 2 =0 <=> a = -2
b =2
Thay a = -2 ; b =2 vào ta có:
M= 22 +7.2.2 + \(\dfrac{52}{-2-2}\)
M= 4 +28- \(\dfrac{52}{4}\)
M= 4 +28 - 13 = 19
Ta có : 2a2 + 2b2 = 5ab
=> 2a2 + 2b2 - 4ab = 5ab - 4ab
=> 2(a2 + b2 - 2ab) = ab
=> (a - b)2 = ab/2
Lại có 2a2 + 2b2 = 5ab
=> 2a2 + 2b2 + 4ab = 5ab + 4ab
=> 2(a + b)2 = 9ab
=> (a + b)2 = 9ab/2
Ta có P2 = \(\left(\frac{a+b}{a-b}\right)^2=\frac{\left(a+b\right)^2}{\left(a-b\right)^2}=\frac{\frac{9ab}{2}}{\frac{ab}{2}}=9\)
=> P = \(\pm\)3
Vậy P = \(\pm\)3