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1 Tinh
89/16+25(9/16÷125/64)÷(-27/8)
= \(\frac{89}{16}+25\left(\frac{9}{16}:\frac{125}{64}\right):-\frac{27}{8}\)
=\(\frac{89}{16}+\frac{25.36}{125}:-\frac{27}{8}\)
=\(\frac{89}{16}-\frac{32}{15}=\frac{823}{240}\)
\(\frac{89}{16}+25.\left(\frac{9}{16}:\frac{125}{64}\right):\left(\frac{-27}{8}\right)\)
=\(\frac{89}{16}+25.\frac{36}{125}:\left(\frac{-27}{8}\right)\)
=\(\frac{89}{16}+\frac{36}{5}:\left(\frac{-27}{8}\right)\)
=\(\frac{89}{16}+\left(\frac{-32}{15}\right)\)
=\(\frac{823}{240}\)
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ta có : \(2x=3y=5z\Leftrightarrow\left(2x\right)^2=\left(3y\right)^2=\left(5z\right)^2\)
\(\Leftrightarrow4x^2=9y^2=25z^2\Leftrightarrow\dfrac{4x^2}{2}=\dfrac{9y^2}{2}=\dfrac{25z^2}{2}\)
\(\Rightarrow\dfrac{4x^2.225-9y^2.100-25z^2.36}{225.2-100.2-36.2}=\dfrac{900x^2-900y^2-900z^2}{450-200-72}\)
\(=\dfrac{900.\left(x^2-y^2-z^2\right)}{178}=\dfrac{900.89}{178}=450\)
ta có : \(\left\{{}\begin{matrix}\dfrac{4x^2}{2}=450\\\dfrac{9y^2}{2}=450\\\dfrac{25z^2}{2}=450\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4x^2=450.2\\9y^2=450.2\\25z^2=450.2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4x^2=900\\9y^2=900\\25z^2=900\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=\dfrac{900}{4}\\y^2=\dfrac{900}{9}\\z^2=\dfrac{900}{25}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2=255\\y^2=100\\z^2=36\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\pm15\\y=\pm10\\z=\pm6\end{matrix}\right.\)
vậy \(x=\pm15;y=\pm10;z=\pm6\)
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câu 1
=> x+1/2+x+1/3+x+1/4-x-1/5-x-1/6=0
=> (x+x+x-x-x)+(1/2+1/3+1/4-1/5-1/6)=0
=> x+43/60=0
=> x = -43/60
câu dưới làm tương tự bạn nhé!
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b) \(\frac{x-11}{89}+\frac{x-13}{87}+\frac{x-15}{85}+\frac{x-17}{83}=4\)
\(=>\left(\frac{x-11}{89}-1\right)+\left(\frac{x-13}{87}-1\right)+\left(\frac{x-15}{85}-1\right)+\left(\frac{x-17}{83}-1\right)=0\)
\(=>\frac{x-100}{89}+\frac{x-100}{87}+\frac{x-100}{85}+\frac{x-100}{83}=0\)
\(=>\left(x-100\right)\left(\frac{1}{89}+\frac{1}{87}+\frac{1}{85}+\frac{1}{83}\right)=0\)
=> x-100 =0 => x=100
Vậy nghiệm là 100
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Có \(\frac{36}{-53}\) = \(\frac{-36}{53}\)
Lại có -58 < -36 => \(\frac{-58}{89}\) < \(\frac{-36}{53}\) hay \(\frac{-58}{89}\) < \(\frac{36}{-53}\)
tuổi con HN là :
50 : ( 1 + 4 ) = 10 ( tuổi )
tuổi bố HN là :
50 - 10 = 40 ( tuổi )
hiệu của hai bố con ko thay đổi nên hiệu vẫn là 30 tuổi
ta có sơ đồ : bố : |----|----|----|
con : |----| hiệu 30 tuổi
tuổi con khi đó là :
30 : ( 3 - 1 ) = 15 ( tuổi )
số năm mà bố gấp 3 tuổi con là :
15 - 10 = 5 ( năm )
ĐS : 5 năm
mình nha
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x+11}{89}+\dfrac{x+13}{87}=\dfrac{x+15}{85}+\dfrac{x+17}{83}\\\Leftrightarrow\left(\dfrac{x+11}{89}+1\right)+\left(\dfrac{x+13}{87}+1\right)=\left(\dfrac{x+15}{85}+1\right)+\left(\dfrac{x+17}{83}+1\right)\\ \Leftrightarrow\dfrac{x+100}{89}+\dfrac{x+100}{87}-\dfrac{x+100}{85}-\dfrac{x+100}{83}=0\\ \Leftrightarrow\left(x+100\right)\left(\dfrac{1}{89}+\dfrac{1}{87}-\dfrac{1}{85}-\dfrac{1}{83}\right)=0\\ \Leftrightarrow x+100=0\left(vì\dfrac{1}{89}+\dfrac{1}{87}-\dfrac{1}{85}-\dfrac{1}{83}\ne0\right)\\ \Leftrightarrow x=-100\)
Ta có: \(2^{5x+2}=8^9\)
\(\Leftrightarrow2^{5x+2}=2^{27}\)
\(\Rightarrow5x+2=27\)
\(\Leftrightarrow5x=25\)
\(\Rightarrow x=5\)