5x+1 * 7y =175x
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điều kiện : \(5x^2+34x+24\ne0\) \(\Leftrightarrow\left(x+6\right)\left(5x+4\right)\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+6\ne0\\5x+4\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne-6\\5x\ne-4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne-6\\x\ne\dfrac{-4}{5}\end{matrix}\right.\)
\(\dfrac{770+175x}{5x^2+34x+24}=\dfrac{35}{12}\Leftrightarrow12\left(770+175x\right)=35\left(5x^2+34x+24\right)\)
\(\Leftrightarrow9240+2100x=175x^2+1190x+840\Leftrightarrow175x^2+1190x+840-9240-2100x=0\)
\(\Leftrightarrow175x^2-910x-8400=0\Leftrightarrow175x^2-1750x+840x-8400=0\)
\(\Leftrightarrow175x\left(x-10\right)+840\left(x-10\right)=\left(175x+840\right)\left(x-10\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}175x+840=0\\x-10=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}175x=-840\\x=10\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-24}{5}\left(tmđk\right)\\x=10\left(tmđk\right)\end{matrix}\right.\)
vậy \(x=\dfrac{-24}{5};x=10\)
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Lấy \(A\left(2;2\right)\) là 1 điểm thuộc \(\Delta_1\)
\(d\left(\Delta_1;\Delta_2\right)=d\left(A;\Delta_2\right)=\dfrac{\left|5.2-7.2+6\right|}{\sqrt{5^2+\left(-7\right)^2}}=\dfrac{\sqrt{74}}{37}\)
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\(b,\left(5x-7y^2\right).\left(5x+7y^2\right)=\left(25x^2-49y^4\right)\)
\(c,\left(5x-7y\right)^2.\left(5x+7y\right)^2=\left(25x^2-70xy+49x^2\right).\left(25x^2+70xy+49x^2\right)\)\(d,\left(\frac{1}{3}x+5y\right).\left(5y-\frac{1}{3}x\right)=25y^2-\frac{1}{9}x^2\) học tốt nhaa) (2x-y)2 = (2x)2 - 2.2x.y + y2 =4x2-4xy+y2
b)(5x-7y2).(5x+7y2) = (5x)2 - (7y2) 2 =25x2 - 49y4
câu c,d mk ko bít làm
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Từ \(\frac{5x-1}{3}=\frac{7y-6}{5}\) Áp dụng TC DTSBN ta có :
\(\frac{5x-1}{3}=\frac{7y-6}{5}=\frac{\left(5x-1\right)+\left(7y-6\right)}{3+5}=\frac{5x+7y-7}{8}=\frac{5x+7y-7}{4x}\)
\(\Rightarrow4x=8\Rightarrow x=2\)
\(\Rightarrow\frac{5.2-1}{3}=\frac{7y-6}{5}\)
\(\Leftrightarrow\frac{7y-6}{5}=3\)
\(\Rightarrow y=3\)
Vậy \(x=2;y=3\)
\(\frac{5x-1}{3}=\frac{7y-6}{5}\Rightarrow5\left(5x-1\right)=3\left(7y-6\right)\Rightarrow25x-5=21y-18\)
\(\Rightarrow21y=25x+13\Rightarrow7y=\frac{25x+13}{3}\)
Xét : \(\frac{5x+7y-7}{4x}=\frac{5x+\frac{25x+13}{3}-7}{4x}=\frac{10x-2}{3x}\)
\(\Rightarrow3x\left(5x-1\right)=3\left(10x-2\right)\Rightarrow15x^2-33x+6=0\)
\(\Rightarrow3\left(x-2\right)\left(5x-1\right)=0\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{5}\end{cases}}\)
Với x=2 , ta có : y=3
Với x =\(\frac{1}{5}\), ta có : y= \(\frac{6}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{5x-1}{3}=\dfrac{7y-6}{5}=\dfrac{5x+7y-7}{8}=\dfrac{5x+7y-7}{4x}\)
+) Xét \(5x+7y-7=0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{5x-1}{3}=0\\\dfrac{7y-6}{5}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}5x-1=0\\7y-6=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=\dfrac{6}{7}\end{matrix}\right.\)
+) Xét \(5x+7y-7\ne0\)
\(\Rightarrow4x=8\Rightarrow x=2\)
Thay \(x=2\) vào \(\dfrac{5x-1}{3}=\dfrac{7y-6}{5}\)
\(\Rightarrow3=\dfrac{7y-6}{5}\)
\(\Rightarrow7y=21\Rightarrow y=3\)
Vậy nếu \(5x+7y-7=0\) thì \(x=\dfrac{1}{5};y=\dfrac{6}{7}\)
nếu \(5x+7y-7\ne0\) thì x = 2, y = 3
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