Đặt \(a+\dfrac{1}{36a}=x\)

pt đã cho trở thành 9x² - 6x + 1 = 0

\(\Leftrightarrow9\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)=0\)

\(\Leftrightarrow9\left(x-\dfrac{1}{3}\right)^2=0\)

\(\Leftrightarrow x-\dfrac{1}{3}=0\)

\(\Leftrightarrow x=\dfrac{1}{3}=a+\dfrac{1}{36a}=\dfrac{36a^2+1}{36a}\)

\(\Leftrightarrow12a=36a^2+1\)

\(\Leftrightarrow36a^2-12a+1=0\)

\(\Leftrightarrow\left(6a-1\right)^2=0\)

\(\Leftrightarrow6a-1=0\)

\(\Leftrightarrow a=\dfrac{1}{6}\) \(\Rightarrow a=6\)

6

4a²b² + 36a²b³ + 6ab⁴

= 2ab²(2a + 18ab + 3b²)

4a²b³ - 6a³b²

= 2a²b²(2b - 3a)

\(PT\Leftrightarrow\dfrac{5}{2}\sqrt{2x+1}-\sqrt{\dfrac{\dfrac{2x+1}{2}}{2}}=\dfrac{3}{2}\\ \Leftrightarrow\dfrac{5}{2}\sqrt{2x+1}-\dfrac{1}{2}\sqrt{2x+1}=\dfrac{3}{2}\\ \Leftrightarrow2\sqrt{2x+1}=\dfrac{3}{2}\\ \Leftrightarrow\sqrt{2x+1}=\dfrac{3}{4}\\ \Leftrightarrow2x+1=\dfrac{9}{16}\\ \Leftrightarrow2x=-\dfrac{7}{16}\\ \Leftrightarrow x=-\dfrac{7}{32}\\ \Leftrightarrow a=-\dfrac{7}{32}\\ \Leftrightarrow1-36a=1+36\cdot\dfrac{7}{32}=...\)

a: =(5a-a+b)(5a+a-b)

=(4a+b)(5a-b)

b: =(2a-a-b)(2a+a+b)

=(a-b)(3a+b)

c: =(7a-2a+b)(7a+2a-b)

=(5a+b)(9a-b)

d: =(6a-3a+2b)(6a+3a-2b)

=(3a+2b)(9a-2b)

e: =(9a-5a+3b)(9a+5a-3b)

=(4a+3b)(14a-3b)

Lời giải:

$25a^2-(a-b)^2=(5a)^2-(a-b)^2=[5a-(a-b)][5a+(a-b)]=(4a+b)(6a-b)$

$4a^2-(a+b)^2=(2a)^2-(a+b)^2=[2a-(a+b)][2a+(a+b)]=(a-b)(3a+b)$

$49a^2-(2a-b)^2=(7a)^2-(2a-b)^2=[7a-(2a-b)][7a+(2a-b)]=(5a+b)(9a-b)$

$36a^2-(3a-2b)^2=(6a)^2-(3a-2b)^2=[6a-(3a-2b)][6a+(3a-2b)]$

$=(3a+2b)(9a-2b)$

$81a^2-(5a-3b)^2=(9a)^2-(5a-3b)^2=[9a-(5a-3b)][9a+(5a-3b)]$

$=(4a+3b)(14a-3b)$

\(7ab\cdot\sqrt{\dfrac{36a^4}{49b^2}}\\ =>7ab\cdot\sqrt{\left(\dfrac{6a^2}{7b}\right)^2}\\ =>7ab\cdot\dfrac{6a^2}{7b}\\ =>\dfrac{7ab\cdot6a^2}{7b}\\ =>6a^3\)

\(A=7ab.\sqrt{\dfrac{36a^4}{49b^2}}\)

\(=7ab.\dfrac{6a^2}{\left|7b\right|}\)

\(=7ab.\dfrac{6a^2}{7b}\left(vib>0\right)\)

\(=6a^3\)

4a²b² + 36a²b³ + 6ab⁴

= 2ab²(2a + 18ab + 3b²)

3n(m - 3) + 5m(m - 3)

= (3n + 5m)(m - 3)

2a(x - y) - (y - x)

= (x - y)(2a + 1)

4a²b³ - 6a³b²

= 2a²b²(2b - 3a)

Theo đề +áp dụng cô si ,ta có:

\(1\ge2a+3b\ge2\sqrt{6ab}\\ \Rightarrow ab\le\frac{1}{24}\)(1)

ÁP dụng cô si cho 2 số ko âm ,ta có:

\(4a^2+9b^2\ge12ab\)(2)

Thay (1),(2) vào ,ta có:

\(36a^2b^2\left(4a^2+9b^2\right)\le36\cdot\frac{1}{24^2}\cdot12\cdot\frac{1}{24}=\frac{1}{32}\)

đến đây thì xong oy

Học tốt nha

^-^

ngược dấu kìa 

phải làm thế này mới đúng 

Áp dụng BĐT xy \(\le\)\(\frac{\left(x+y\right)^2}{4}\). Dấu " = " xảy ra \(\Leftrightarrow\)x = y

Ta có : \(36a^2b^2\left(4a^2+9b^2\right)=3ab.12ab.\left(4a^2+9b^2\right)\)

\(\le\frac{1}{2}.\left(2a.3b\right).\frac{4a^2+9b^2+12ab}{4}\le\frac{1}{2}.\frac{\left(2a+3b\right)^2}{4}.\frac{\left(2a+3b\right)^2}{4}\le\frac{1}{32}\)

Dấu " = " xảy ra \(\Leftrightarrow\)\(a=\frac{1}{4};b=\frac{1}{6}\)

\(\dfrac{4a^2-9b^2}{a^2b^2}\div\dfrac{2ax+3bx}{2ab}\)

\(=\dfrac{\left(2a-3b\right)\left(2a+3b\right)}{a^2b^2}\times\dfrac{2ab}{x\left(2a+3b\right)}\)

\(=\dfrac{2ab\left(2a-3b\right)\left(2a+3b\right)}{a^2b^2x\left(2a+3b\right)}=\dfrac{4a-6b}{xab}\)

$\frac{2x}{25-4b^2}:\frac{1}{5+2b}$

\(=\dfrac{2x}{\left(5-2b\right)\left(5+2b\right)}\times\dfrac{5+2b}{1}\)

\(=\dfrac{2x\left(5+2b\right)}{\left(5-2b\right)\left(5+2b\right)}=\dfrac{2x}{5-2b}\)

$\frac{{\left(2-a\right)}^2}{2ab}.\frac{b}{\left(2-a\right)}+\frac{1}{2}$

\(=\dfrac{\left(2-a\right)^2b}{2ab\left(2-a\right)}+\dfrac{1}{2}\)

\(=\dfrac{2b-ab}{2ab}+\dfrac{1}{2}\)

\(=\dfrac{2b-ab}{2ab}+\dfrac{ab}{2ab}=\dfrac{2b}{2ab}=\dfrac{1}{a}\)

$\frac{2b+2}{2b-b^2}:\frac{b+1}{b}+\frac{2b+2}{3b-6}$