Tìm giá trị của a sao cho mỗi biểu thức sau có giá trị bằng 2
a. \(\dfrac{2a^2-3a-2}{a^2-4}\)
b. \(\dfrac{3a-1}{3a+1}+\dfrac{a-3}{a+3}\)
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2
29 tháng 9 2023
`a)D` xác định `<=>a-1 ne 0<=>a ne 1`
`b)` Với `a ne 1` có:
`D=([a-1]/[a^2+a+1]-[1-3a+a^2]/[(a-1)(a^2+a+1)]-1/[a-1]).[1-a]/[a^2+1]`
`D=[(a-1)^2-1+3a-a^2-a^2-a-1]/[(a-1)(a^2+a+1)].[-(a-1)]/[a^2+1]`
`D=[a^2-2a+1-1+3a-a^2-a^2-a-1]/[(-a^2-1)(a^2+a+1)]`
`D=[-a^2-1]/[(-a^2-1)(a^2+a+1)]=1/[a^2+a+1]`
`c)` Với `a ne 1` có:
`1/D=1/[1/[a^2+a+1]]=a^2+a+1=(a+1/2)^2+3/4`
Vì `(a+1/2)^2 >= 0 AA a ne 1`
`=>(a+1/2)^2+3/4 >= 3/4 AA a ne 1`
Hay `1/D >= 3/4 AA a ne 1=>1/D _[mi n]=3/4`
Dấu "`=`" xảy ra `<=>a=-1/2` (t/m).
12 tháng 2 2020
a) \(\frac{2a^2-3a-2}{a^2-4}=2\)
\(\Rightarrow2a^2-3a-2=2\left(a^2-4\right)\)
\(\Rightarrow2a^2-3a-2=2a^2-4\)
\(\Rightarrow-3a-2=-4\)
\(\Rightarrow-3a=-2\Rightarrow a=\frac{2}{3}\)
12 tháng 2 2020
b) \(\frac{3a-1}{3a+1}+\frac{a-3}{a+3}=2\)
\(\Rightarrow\frac{\left(3a-1\right)\left(a+3\right)+\left(3a+1\right)\left(a-3\right)}{\left(3a+1\right)\left(a+3\right)}=2\)
\(\Rightarrow\frac{6a^2-6}{3a^2+10a+3}=2\)
\(\Rightarrow6a^2-6=2\left(3a^2+10a+3\right)\)
\(\Rightarrow6a^2-6=6a^2+20a+6\)
\(\Rightarrow-6=20a+6\Rightarrow20a=-12\)
\(\Rightarrow a=\frac{-3}{5}\)
14 tháng 3 2021
Áp dụng bđt Schwarz ta có:
\(P=\dfrac{a^4}{2ab+3ac}+\dfrac{b^4}{2cb+3ab}+\dfrac{c^4}{2ac+3bc}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{5\left(ab+bc+ca\right)}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{5\left(a^2+b^2+c^2\right)}=\dfrac{1}{5}\).
Đẳng thức xảy ra khi và chỉ khi \(a=b=c=\dfrac{\sqrt{3}}{3}\).
Y
10 tháng 7 2023
\(-\left(\dfrac{a-1}{a+1}-\dfrac{a}{a-1}-\dfrac{3a+1}{1-a^2}\right):\dfrac{2a+1}{a^2-1}\left(dk:a\ne1,a\ne-1\right)\)
\(=-\left(\dfrac{a-1}{a+1}-\dfrac{a}{a-1}+\dfrac{3a+1}{a^2-1}\right):\dfrac{2a+1}{\left(a-1\right)\left(a+1\right)}\\ =-\left(\dfrac{\left(a-1\right)^2-a\left(a+1\right)+3a+1}{\left(a-1\right)\left(a+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =-\dfrac{a^2-2a+1-a^2-a+3a+1}{\left(a-1\right)\left(a+1\right)}.\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\)
\(=-\dfrac{2}{2a+1}\)
H
11 tháng 7 2023
\(-\left(\dfrac{a-1}{a+1}-\dfrac{a}{a-1}-\dfrac{3a+1}{1-a^2}\right):\dfrac{2a+1}{a^2-1}\\ =-\left(\dfrac{a-1}{a+1}-\dfrac{a}{a-1}+\dfrac{3a+1}{a^2-1}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =-\left(\dfrac{a-1}{a+1}-\dfrac{a}{a-1}+\dfrac{3a-1}{\left(a-1\right)\left(a+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =-\left(\dfrac{\left(a-1\right)^2}{\left(a+1\right)\left(a-1\right)}-\dfrac{a\left(a+1\right)}{\left(a-1\right)\left(a+1\right)}+\dfrac{3a+1}{\left(x-1\right)\left(x+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =-\left(\dfrac{\left(a-1\right)^2-a\left(a+1\right)+3a+1}{\left(a-1\right)\left(a+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\)\(=-\left(\dfrac{a^2-2a+1-\left(a^2+a\right)+3a+1}{\left(a-1\right)\left(a+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =-\left(\dfrac{a^2-2a+1-a^2-a+3a+1}{\left(a-1\right)\left(a+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =-\left(\dfrac{2}{\left(a-1\right)\left(a+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =\dfrac{-2.\left(a-1\right)\left(a+1\right)}{\left(a-1\right)\left(a+1\right).\left(2a+1\right)}\\ =-\dfrac{2}{2a+1}\)
__
\(-\dfrac{2}{2a+1}=\dfrac{3}{a-1}\\ \Leftrightarrow-2\left(a-1\right)=3\left(2a+1\right)\\ \Leftrightarrow-2a+2-6a-3=0\\ \Leftrightarrow-8a-1=0\\ \Leftrightarrow-8a=1\\ \Leftrightarrow a=-\dfrac{1}{8}\)
DX
23 tháng 12 2022
2.
\(P=\left(\dfrac{a+6}{3\left(a+3\right)}-\dfrac{1}{a+3}\right).\dfrac{27a}{a+2}=\left(\dfrac{a+3}{3\left(a+3\right)}\right).\dfrac{27a}{a+2}=\dfrac{27a}{3\left(a+2\right)}=\dfrac{9a}{a+2}\)
ĐKXĐ là :
\(a\ne0;-3;-2\)
Vs a = 1 ta có:
=> P=3
1.
\(M=\left(\dfrac{2a}{2a+b}-\dfrac{4a^2}{\left(2a+b\right)^2}\right):\left(\dfrac{2a}{\left(2a-b\right)\left(2a+b\right)}-\dfrac{1}{2a-b}\right)=\left(\dfrac{4a^2+2ab-4a^2}{\left(2a+b\right)^2}\right).\left(\dfrac{\left(2a+b\right)\left(2a-b\right)}{b}\right)=\dfrac{2a.\left(2a-b\right)}{\left(2a+b\right)}\)
a.
\(\dfrac{2a^2-3a-2}{a^2-4}=2\)
\(\Leftrightarrow\dfrac{2a^2-4a+a-2}{\left(a-2\right)\left(a+2\right)}=2\)
\(\Leftrightarrow\dfrac{\left(2a^2-4a\right)+\left(a-2\right)}{\left(a-2\right)\left(a+2\right)}=2\)
\(\Leftrightarrow\dfrac{2a\left(a-2\right)+\left(a-2\right)}{\left(a-2\right)\left(a+2\right)}=2\)
\(\Leftrightarrow\dfrac{\left(2a+1\right)\left(a-2\right)}{\left(a-2\right)\left(a+1\right)}=2\)
\(\Leftrightarrow\dfrac{2a+1}{a+1}=2\)
\(\Leftrightarrow\dfrac{2a+1}{a+1}=\dfrac{2\left(a+1\right)}{a+1}\)
\(\Leftrightarrow2a+1=2a+2\)
Suy ra pt vô nghiệm
a) \(\dfrac{2a^{2^{ }}-3a-2}{a^2-4}\)=2
<=> \(\dfrac{2a^{2^{ }}-3a-2}{\left(a-2\right)\left(a+2\right)}\)=2 (1)
ĐKXĐ: a-2 #0 => a#2
a+2#0 -> a#-2
(1) <=> \(\dfrac{2a^{2^{ }}-3a-2}{\left(a-2\right)\left(a+2\right)}\)= \(\dfrac{2\left(a^{^2}-4\right)}{\left(a-2\right)\left(a+2\right)}\)
=> 2a2 - 3a - 2 = 2a2 - 8
<=> 2a2 - 3a - 2 - 2a2 + 8 = 0
<=> -3a + 6 = 0
<=> -3 ( a-2)
<=> -3 = 0 ( vô no )
a-2 = 0 => a = 2
Vậy với A=2 thì biểu thức có giá trị = 2