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Bài 1:
a) \(=\dfrac{8}{15}\left(\dfrac{7}{13}+\dfrac{6}{13}\right)=\dfrac{8}{15}.1=\dfrac{8}{15}\)
b) \(=\dfrac{3.3-7-2.4}{12}=-\dfrac{6}{12}=-\dfrac{1}{2}\)
Bài 2:
\(\dfrac{x}{2,7}=-\dfrac{2}{3,6}\Rightarrow x=\dfrac{\left(-2\right).2,7}{3,6}\Rightarrow x=-\dfrac{3}{2}\)
Bài 3:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=-\dfrac{21}{7}=-3\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).2=-6\\y=\left(-3\right).5=-10\end{matrix}\right.\)
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Đặt t = x² ( t ≥ 0 )
<=> 6/(t + 2) + 12/(t + 8) = 3 - 7/(t + 3)
<=> 6(t + 8)(t + 3) + 12(t + 2)(t + 3) = 3(t + 2)(t + 8)(t + 3) - 7(t + 2)(t + 8)
<=> 6(t² + 11t + 24) + 12(t² + 5t + 6) = 3(t² + 10t + 16)(t + 3) - 7(t² + 10t + 16)
<=> 6t² + 66t +144 + 12t² + 60t + 72 = 3(t^3 + 3t² + 10t² + 30t + 16t + 48) - 7t² - 70t - 112
<=> 6t² + 66t +144 + 12t² + 60t + 72 = 3(t^3 + 13t² + 46t + 48) - 7t² - 70t - 112
<=> 6t² + 66t +144 + 12t² + 60t + 72 = 3t^3 + 39t² + 138t + 144 - 7t² - 70t - 112
<=> 3t^3 + 14t² - 58t - 184 = 0
<=> 3t^3 + 26t² + 46t - 12t² - 104t - 184 = 0
<=> t(3t² + 26t + 46) - 4(3t² + 26t + 46) = 0
<=> ( t - 4 )( 3t² + 26t + 46 ) = 0
<=> t - 4 = 0
<=> 3t² + 26t + 46 = 0
<=> t = 4 > 0 ( chọn )
=> x² = 4
<=> x =+-2
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Xin lỗi bạn mình mới học lớp 5 thôi
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Xin lỗi bạn nhiều
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Ta có :\(\frac{6}{x^2+2}+\frac{12}{x^2+8}=3-\frac{7}{x^2+3}\)
\(\Leftrightarrow\frac{6}{x^2+2}+\frac{12}{x^2+8}+\frac{7}{x^2+3}=3\)
\(\Leftrightarrow\left(\frac{6}{x^2+2}-1\right)+\left(\frac{12}{x^2+8}-1\right)+\left(\frac{7}{x^2+3}-1\right)=0\)
\(\Leftrightarrow\frac{4-x^2}{x^2+2}+\frac{4-x^2}{x^2+8}+\frac{4-x^2}{x^2+3}=0\)
\(\Leftrightarrow\left(4-x^2\right)\left(\frac{1}{x^2+2}+\frac{1}{x^2+8}+\frac{1}{x^2+3}\right)=0\)
Ta thấy : \(\frac{1}{x^2+2}+\frac{1}{x^2+8}+\frac{1}{x^2+3}\ne0\forall x\)
Do đó : \(4-x^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\) ( thỏa mãn )
Vậy : \(x\in\left\{-2,2\right\}\)
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người ta ko bt thì hỏi , đừng có mak ns mấy câu vô văn hóa thế nhá
\(\frac{6}{x^2+2}+\frac{12}{x^2+8}=3-\frac{7}{x^2+3}\)
\(\Leftrightarrow6\left(x^2+8\right)\left(x^3+3\right)+12\left(x^2+2\right)\left(x^2+3\right)=3\left(x^2+2\right)\left(x^2+8\right)\left(x^2+3\right)-7\left(x^2+8\right)\left(x^2+2\right)\)
\(\Leftrightarrow18x^4+126x^2+216=3x^6+32x^4+62x^2+32\)
\(\Leftrightarrow18x^2+126x^2+216-3x^6-32x^4-68x^2-32=0\)
\(\Leftrightarrow-14x^4+58x^2+184-3x^6=0\)
\(\Leftrightarrow x=\pm2\)
\(\Rightarrow x=\pm2\)