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![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)
\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2abxy+b^2y^2\)
\(\Leftrightarrow a^2y^2+b^2x^2-2abxy=0\)
\(\Leftrightarrow\left(ay-bx\right)^2=0\)
\(\Leftrightarrow ay=bx\)
\(\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}\)
\(\Rightarrowđpcm\)
Bài 2:
Ta có: \(VT=\left(5a-3b+8c\right)\left(5a-3b-8c\right)\)
\(=\left(5a-3b\right)^2-64c^2\)
\(=25a^2-30ab+9b^2-64c^2\)
\(=25a^2-30ab+9b^2-16a^2+16b^2\left(a^2-b^2=4c^2\right)\)
\(=9a^2-30ab+25b^2=\left(3a-5b\right)^2=VP\)
\(\Rightarrowđpcm\)
![](https://rs.olm.vn/images/avt/0.png?1311)
5A+4B=0
=>10/2x+1+12/2x-1=0
=>10(2x-1)+12(2x+1)=0
=>20x-10+24x+12=0
=>44x+2=0
=>x=-1/22(nhận)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{4a^2-9b^2}{a^2b^2}\div\dfrac{2ax+3bx}{2ab}\)
\(=\dfrac{\left(2a-3b\right)\left(2a+3b\right)}{a^2b^2}\times\dfrac{2ab}{x\left(2a+3b\right)}\)
\(=\dfrac{2ab\left(2a-3b\right)\left(2a+3b\right)}{a^2b^2x\left(2a+3b\right)}=\dfrac{4a-6b}{xab}\)
\(=\dfrac{2x}{\left(5-2b\right)\left(5+2b\right)}\times\dfrac{5+2b}{1}\)
\(=\dfrac{2x\left(5+2b\right)}{\left(5-2b\right)\left(5+2b\right)}=\dfrac{2x}{5-2b}\)
\(=\dfrac{\left(2-a\right)^2b}{2ab\left(2-a\right)}+\dfrac{1}{2}\)
\(=\dfrac{2b-ab}{2ab}+\dfrac{1}{2}\)
\(=\dfrac{2b-ab}{2ab}+\dfrac{ab}{2ab}=\dfrac{2b}{2ab}=\dfrac{1}{a}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$5a^2+2b^2=11ab$
$\Leftrightarrow 5a^2+2b^2-11ab=0$
$\Leftrightarrow (5a^2-10ab)-(ab-2b^2)=0$
$\Leftrightarrow 5a(a-2b)-b(a-2b)=0$
$\Leftrightarrow (a-2b)(5a-b)=0$
Do $a>2b>0$ nên $a-2b>0$. Do dó $5a-b=0$
$\Leftrightarrow b=5a$. Khi đó:
$A=\frac{4a^2-5b^2}{a^2+2ab}=\frac{4a^2-5(5a)^2}{a^2+2a.5a}=\frac{-121a^2}{11a^2}=-11$
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=\dfrac{x^2+2-2x\left(x-2\right)+\left(x-1\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}\)
\(=\dfrac{x^2+2-2x^2+4x+x^2-1}{\left(x-2\right)\left(x+1\right)}=\dfrac{4x+1}{\left(x-2\right)\left(x+1\right)}\)
Khi x=5 thì \(A=\dfrac{4\cdot5+1}{\left(5-2\right)\left(5+1\right)}=\dfrac{21}{3\cdot6}=\dfrac{7}{6}\)
b: P=A:B
\(=\dfrac{4x+1}{\left(x-2\right)\left(x+1\right)}\cdot\dfrac{x-2}{1}=\dfrac{4x+1}{x+1}\)
c: P^2=P+2
=>P^2-P-2=0
=>(P-2)(P+1)=0
=>P=2 hoặc P=-1
=>4x+1=2x+2 hoặc 4x+1=-x-1
=>2x=1 hoặc 5x=-2
=>x=-2/5(nhận) hoặc x=1/2(nhận)
\(\dfrac{4a-4b}{5a+5b}.x=\dfrac{a^2-b^2}{a^2+2ab+b^2}\)
\(\Leftrightarrow\dfrac{4\left(a-b\right)}{5\left(a+b\right)}.x=\dfrac{\left(a-b\right)\left(a+b\right)}{\left(a+b\right)^2}\)
\(\Leftrightarrow x=\dfrac{a-b}{a+b}.\dfrac{5\left(a+b\right)}{4\left(a-b\right)}\)
\(\Leftrightarrow x=\dfrac{5}{4}\)
Vậy \(x=\dfrac{5}{4}\)
Ta có: \(\dfrac{4\left(a-b\right)}{5\left(a+b\right)}x=\dfrac{\left(a-b\right)\left(a+b\right)}{\left(a+b\right)^2}\)
\(\Rightarrow\)\(\dfrac{4\left(a-b\right)}{5\left(a+b\right)}x=\dfrac{a-b}{a+b}\)\(\Rightarrow\dfrac{4}{5}x=\dfrac{a-b}{a+b}.\dfrac{a+b}{a-b}=1\)
\(\Rightarrow\)x = \(1:\dfrac{4}{5}=\dfrac{5}{4}\)
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