Tìm các giá trị của a sao cho mỗi biểu thức có giá trị bằng 2
a) \(\frac{2a^2-3a-2}{a^2-4}\)
b)\(\frac{3a-1}{3a+1}+\frac{a-3}{a+3}\)
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a) \(\frac{2a^2-3a-2}{a^2-4}=2\)
\(\Rightarrow2a^2-3a-2=2\left(a^2-4\right)\)
\(\Rightarrow2a^2-3a-2=2a^2-4\)
\(\Rightarrow-3a-2=-4\)
\(\Rightarrow-3a=-2\Rightarrow a=\frac{2}{3}\)
b) \(\frac{3a-1}{3a+1}+\frac{a-3}{a+3}=2\)
\(\Rightarrow\frac{\left(3a-1\right)\left(a+3\right)+\left(3a+1\right)\left(a-3\right)}{\left(3a+1\right)\left(a+3\right)}=2\)
\(\Rightarrow\frac{6a^2-6}{3a^2+10a+3}=2\)
\(\Rightarrow6a^2-6=2\left(3a^2+10a+3\right)\)
\(\Rightarrow6a^2-6=6a^2+20a+6\)
\(\Rightarrow-6=20a+6\Rightarrow20a=-12\)
\(\Rightarrow a=\frac{-3}{5}\)
ĐKXĐ của phương trình : \(\orbr{\begin{cases}x\ne-\frac{1}{3}\\x\ne-3\end{cases}}\)
\(\frac{3a-1}{3a+1}+\frac{a-3}{a+3}=2\)
\(\Leftrightarrow\left(3a-1\right)\left(a+3\right)+\left(a-3\right)\left(3a+1\right)=2\left(3a+1\right)\left(a+3\right)\)\(\Leftrightarrow3a^2+8a-3+3a^2-8a-3=2\left(3a^2+10a+3\right)\)
\(\Leftrightarrow6a^2-6-6a^2-20a-6=0\)
\(\Leftrightarrow-20a-12=0\Leftrightarrow a=\frac{-12}{20}=-\frac{3}{5}\)(NHẬN)
vậy tập nghiệm của phương trình là : S = { -3/5 }
Tk mk nka !!! Th@nks !!
a, Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\)\(\Rightarrow a=2k\); \(b=3k\); \(c=5k\)
Ta có: \(B=\frac{a+7b-2c}{3a+2b-c}=\frac{2k+7.3k-2.5k}{3.2k+2.3k-5k}=\frac{2k+21k-10k}{6k+6k-5k}=\frac{13k}{7k}=\frac{13}{7}\)
b, Ta có: \(\frac{1}{2a-1}=\frac{2}{3b-1}=\frac{3}{4c-1}\)\(\Rightarrow\frac{2a-1}{1}=\frac{3b-1}{2}=\frac{4c-1}{3}\)
\(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{1}=\frac{3\left(b-\frac{1}{3}\right)}{2}=\frac{4\left(c-\frac{1}{4}\right)}{3}\) \(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{12}=\frac{3\left(b-\frac{1}{3}\right)}{2.12}=\frac{4\left(c-\frac{1}{4}\right)}{3.12}\)
\(\Rightarrow\frac{\left(a-\frac{1}{2}\right)}{6}=\frac{\left(b-\frac{1}{3}\right)}{8}=\frac{\left(c-\frac{1}{4}\right)}{9}\)\(\Rightarrow\frac{3\left(a-\frac{1}{2}\right)}{18}=\frac{2\left(b-\frac{1}{3}\right)}{16}=\frac{\left(c-\frac{1}{4}\right)}{9}\)
\(\Rightarrow\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-\left(c-\frac{1}{4}\right)}{18+16-9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-c+\frac{1}{4}}{25}\)
\(=\frac{\left(3a+2b-c\right)-\left(\frac{3}{2}+\frac{2}{3}-\frac{1}{4}\right)}{25}=\left(4-\frac{23}{12}\right)\div25=\frac{25}{12}\times\frac{1}{25}=\frac{1}{12}\)
Do đó: +) \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)
+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)
+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)
a) \(ĐK:a\ne1;a\ne0\)
\(A=\left[\frac{\left(a-1\right)^2}{3a+\left(a-1\right)^2}-\frac{1-2a^2+4a}{a^3-1}+\frac{1}{a-1}\right]:\frac{a^3+4a}{4a^2}=\left[\frac{a^2-2a+1}{a^2+a+1}-\frac{1-2a^2+4a}{a^3-1}+\frac{a^2+a+1}{a^3-1}\right].\frac{4a^2}{a^3+4a}\)\(=\left[\frac{a^3-3a^2+3a-1}{a^3-1}-\frac{1-2a^2+4a}{a^3-1}+\frac{a^2+a+1}{a^3-1}\right].\frac{4a^2}{a^3+4a}=\frac{a^3-1}{a^3-1}.\frac{4a}{a^2+4}=\frac{4a}{a^2+4}\)
b) Ta có: \(a^2+4\ge4a\)(*)
Thật vậy: (*)\(\Leftrightarrow\left(a-2\right)^2\ge0\)
Khi đó \(\frac{4a}{a^2+4}\le1\)
Vậy MaxA = 1 khi x = 2
a) \(\frac{2a-9}{2a-5}+\frac{3a}{3a-2}=2\)
<=> (2a - 9)(3a - 2) + 3a(2a - 5) = 2(2a - 5)(3a - 2)
<=> 6a2 - 4a - 27a + 16 + 6a2 - 15a = 12a2 - 8a - 30a + 20
<=> 12a2 - 44a + 16 = 12a2 - 38a + 20
<=> 12a2 - 44a + 16 - 12a2 = -38a + 20
<=> -44a + 16 = -36a + 20
<=> -44a + 16 + 36a = 20
<=> -8a + 16 = 20
<=> -8a = 20 - 16
<=> -8a = 4
<=> a = -4/8 = -1/2
b) nhân chéo và làm tương tự