\(\dfrac{2}{36a^2b^2-1};\dfrac{1}{6ab+1^2};\dfrac{1}{6ab-1^2}\)
\(\dfrac{x}{x^3-27};\dfrac{2x}{x^2-6x+9};\dfrac{1}{x^2+3x+9x}\)
\(\dfrac{x^2-x}{x^2-1};\dfrac{3x}{x^3+2x^2+x};2x\)
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Những câu hỏi liên quan
13 tháng 3 2017
Đặt \(a+\dfrac{1}{36a}=x\)
pt đã cho trở thành 9x2 - 6x + 1 = 0
\(\Leftrightarrow9\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)=0\)
\(\Leftrightarrow9\left(x-\dfrac{1}{3}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{1}{3}=0\)
\(\Leftrightarrow x=\dfrac{1}{3}=a+\dfrac{1}{36a}=\dfrac{36a^2+1}{36a}\)
\(\Leftrightarrow12a=36a^2+1\)
\(\Leftrightarrow36a^2-12a+1=0\)
\(\Leftrightarrow\left(6a-1\right)^2=0\)
\(\Leftrightarrow6a-1=0\)
\(\Leftrightarrow a=\dfrac{1}{6}\) \(\Rightarrow a=6\)
DN
1
PA
15 tháng 9 2016
4a2b2 + 36a2b3 + 6ab4
= 2ab2(2a + 18ab + 3b2)
4a2b3 - 6a3b2
= 2a2b2(2b - 3a)
15 tháng 11 2021
\(PT\Leftrightarrow\dfrac{5}{2}\sqrt{2x+1}-\sqrt{\dfrac{\dfrac{2x+1}{2}}{2}}=\dfrac{3}{2}\\ \Leftrightarrow\dfrac{5}{2}\sqrt{2x+1}-\dfrac{1}{2}\sqrt{2x+1}=\dfrac{3}{2}\\ \Leftrightarrow2\sqrt{2x+1}=\dfrac{3}{2}\\ \Leftrightarrow\sqrt{2x+1}=\dfrac{3}{4}\\ \Leftrightarrow2x+1=\dfrac{9}{16}\\ \Leftrightarrow2x=-\dfrac{7}{16}\\ \Leftrightarrow x=-\dfrac{7}{32}\\ \Leftrightarrow a=-\dfrac{7}{32}\\ \Leftrightarrow1-36a=1+36\cdot\dfrac{7}{32}=...\)
10 tháng 9 2023
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)
AH
Akai Haruma
Giáo viên
10 tháng 9 2023
Lời giải:
$25a^2-(a-b)^2=(5a)^2-(a-b)^2=[5a-(a-b)][5a+(a-b)]=(4a+b)(6a-b)$
$4a^2-(a+b)^2=(2a)^2-(a+b)^2=[2a-(a+b)][2a+(a+b)]=(a-b)(3a+b)$
$49a^2-(2a-b)^2=(7a)^2-(2a-b)^2=[7a-(2a-b)][7a+(2a-b)]=(5a+b)(9a-b)$
$36a^2-(3a-2b)^2=(6a)^2-(3a-2b)^2=[6a-(3a-2b)][6a+(3a-2b)]$
$=(3a+2b)(9a-2b)$
$81a^2-(5a-3b)^2=(9a)^2-(5a-3b)^2=[9a-(5a-3b)][9a+(5a-3b)]$
$=(4a+3b)(14a-3b)$
NC
2
NV
19 tháng 7 2023
\(7ab\cdot\sqrt{\dfrac{36a^4}{49b^2}}\\ =>7ab\cdot\sqrt{\left(\dfrac{6a^2}{7b}\right)^2}\\ =>7ab\cdot\dfrac{6a^2}{7b}\\ =>\dfrac{7ab\cdot6a^2}{7b}\\ =>6a^3\)
19 tháng 7 2023
\(A=7ab.\sqrt{\dfrac{36a^4}{49b^2}}\)
\(=7ab.\dfrac{6a^2}{\left|7b\right|}\)
\(=7ab.\dfrac{6a^2}{7b}\left(vib>0\right)\)
\(=6a^3\)
BH
1 tháng 5 2019
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
^-^
HT
24 tháng 11 2018
\(\dfrac{4a^2-9b^2}{a^2b^2}\div\dfrac{2ax+3bx}{2ab}\)
\(=\dfrac{\left(2a-3b\right)\left(2a+3b\right)}{a^2b^2}\times\dfrac{2ab}{x\left(2a+3b\right)}\)
\(=\dfrac{2ab\left(2a-3b\right)\left(2a+3b\right)}{a^2b^2x\left(2a+3b\right)}=\dfrac{4a-6b}{xab}\)
\(=\dfrac{2x}{\left(5-2b\right)\left(5+2b\right)}\times\dfrac{5+2b}{1}\)
\(=\dfrac{2x\left(5+2b\right)}{\left(5-2b\right)\left(5+2b\right)}=\dfrac{2x}{5-2b}\)
\(=\dfrac{\left(2-a\right)^2b}{2ab\left(2-a\right)}+\dfrac{1}{2}\)
\(=\dfrac{2b-ab}{2ab}+\dfrac{1}{2}\)
\(=\dfrac{2b-ab}{2ab}+\dfrac{ab}{2ab}=\dfrac{2b}{2ab}=\dfrac{1}{a}\)
\(\dfrac{2}{36a^2b^2-1}=\dfrac{2}{\left(6ab-1\right)\left(6ab+1\right)}\\ \dfrac{1}{6ab+1}=\dfrac{6ab-1}{\left(6ab-1\right)\left(6ab+1\right)};\dfrac{1}{6ab-1}=\dfrac{6ab+1}{\left(6ab-1\right)\left(6ab+1\right)}\)
\(\dfrac{x}{x^3-27}=\dfrac{x\left(x-3\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\\ \dfrac{2x}{x^2-6x+9}=\dfrac{2x\left(x^2+3x+9\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\\ \dfrac{1}{x^2+3x+9}=\dfrac{\left(x-3\right)^2}{\left(x-3\right)^2\left(x^2+3x+9\right)}\)
\(\dfrac{x^2-x}{x^2-1}=\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x}{x+1}=\dfrac{x\left(x+1\right)}{\left(x+1\right)^2}\\ \dfrac{3x}{x^3+2x^2+x}=\dfrac{3x}{x\left(x^2+2x+1\right)}=\dfrac{3}{\left(x+1\right)^2}\\ 2x=\dfrac{2x\left(x+1\right)^2}{\left(x+1\right)^2}\)