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e) = \(\dfrac{3}{2\left(x+3\right)}\) - \(\dfrac{x-6}{2x\left(x+3\right)}\)
= \(\dfrac{3x}{2x\left(x+3\right)}\) - \(\dfrac{x-6}{2x\left(x+3\right)}\) = \(\dfrac{3x-x+6}{2x\left(x+3\right)}\)
= \(\dfrac{2x-6}{2x\left(x+3\right)}\)
= \(\dfrac{2\left(x-3\right)}{2x\left(x+3\right)}\)
c) = \(\dfrac{2\left(a^3-b^3\right)}{3\left(a+b\right)}\) . \(\dfrac{6\left(a+b\right)}{a^2-2ab+b^2}\)
= \(\dfrac{-2\left(a+b\right)\left(a^2-2ab+b^2\right)}{3\left(a+b\right)}\) . \(\dfrac{6\left(a+b\right)}{a^2-2ab+b^2}\)
= \(\dfrac{-2\left(a+b\right)}{1}\) . \(\dfrac{2}{1}\) = -4 (a+b)
Lời giải:
Vì \(2a-b=5\Rightarrow b=2a-5\Rightarrow 2b=4a-10\)
\(\Rightarrow 7a-2b=7a-(4a-10)=3a+10\)
\(\Rightarrow \frac{7a-2b}{3a+10}=\frac{3a+10}{3a+10}=1\)
Lại có:
\(2a-b=5\Rightarrow 2a=b+5\Rightarrow 4a=2b+10\)
\(\Rightarrow 7b-4a=7b-(2b+10)=5b-10\)
\(\Rightarrow \frac{7b-4a}{15b-30}=\frac{5b-10}{15b-30}=\frac{5b-10}{3(5b-10)}=\frac{1}{3}\)
Vậy: \(A=1-\frac{1}{3}=\frac{2}{3}\)
2a-b=5 nên b=2a-5
\(A=\dfrac{7a-2b}{3a+10}-\dfrac{7b-4a}{15b-30}\)
\(=\dfrac{7a-2\left(2a-5\right)}{3a+10}-\dfrac{7\left(2a-5\right)-4a}{15\left(2a-5\right)-30}\)
\(=\dfrac{7a-4a+10}{3a+10}-\dfrac{14a-35-4a}{30a-75-30}\)
\(=1-\dfrac{5\left(2a-7\right)}{15\left(2a-7\right)}=1-\dfrac{1}{3}=\dfrac{2}{3}\)
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
Ta có:
\(Q=\dfrac{2a-b}{3a-b}+\dfrac{5b-a}{3a+b}\)
\(Q=\dfrac{\left(2a-b\right)\left(3a+b\right)}{\left(3a-b\right)\left(3a+b\right)}+\dfrac{\left(5b-a\right)\left(3a-b\right)}{\left(3a-b\right)\left(3a+b\right)}\)
\(Q=\dfrac{\left(2a-b\right)\left(3a+b\right)+\left(5b-a\right)\left(3a-b\right)}{\left(3a-b\right)\left(3a+b\right)}\)
\(Q=\dfrac{3a^2+15ab-6b^2}{9a^2-b^2}\)
Ta lại có:
\(6a^2-15ab+5b^2=0\)
\(\Rightarrow3a^2+15ab-6b^2=9a^2-b^2\left(1\right)\)
Thay (1) vào Q
=> Q = 1
Ta có \(6a^2-15ab+5b^2=0\Leftrightarrow15ab=6a^2+5b^2\)
\(Q=\dfrac{\left(2a-b\right)\left(3a+b\right)+\left(5b-a\right)\left(3a-b\right)}{9a^2-b^2}\)
\(Q=\dfrac{3a^2+15ab-6b^2}{9a^2-b^2}=\dfrac{3a^2+6a^2+5b^2-6b^2}{9a^2-b^2}\)
\(Q=\dfrac{9a^2-b^2}{9a^2-b^2}=1\)
\(\left(\dfrac{a-3}{a}-\dfrac{a}{a-3}+\dfrac{9}{a^2-3a}\right):\dfrac{2a+2}{a}=\left(\dfrac{a-3}{a}-\dfrac{a}{a-3}+\dfrac{9}{a\left(a-3\right)}\right):\dfrac{2a+2}{a}=\left(\dfrac{a^2-6a+9}{a\left(a-3\right)}-\dfrac{a^2}{a\left(a-3\right)}+\dfrac{9}{a\left(a-3\right)}\right):\dfrac{2a+2}{a}=\left(\dfrac{a^2-6a+9-a^2+9}{a\left(a-3\right)}\right):\dfrac{2a+2}{a}=\dfrac{18-6a}{a\left(a-3\right)}:\dfrac{2a+2}{a}=\dfrac{6a-18}{\left(-a\right)\left(3-a\right)}:\dfrac{2a+2}{a}=\dfrac{6}{\left(-a\right)}:\dfrac{2a+2}{a}=\dfrac{6a}{\left(-2a^2\right)+\left(-2a\right)}.DKXD:a\ne0;a\ne3\)