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Ta có :
\(\frac{7x+2}{5x+7}=\frac{7x-1}{5x+1}\)
\(=>\left(7x+2\right)\left(5x+1\right)=\left(7x-1\right)\left(5x+7\right)\)
\(=>35x^2+7x+10x+2=35x^2+49x-5x-7\)
\(=>35x^2+17x+2=35x^2+44x-7\)
\(=>17x+2=44x-7\)
\(=>44x-17x=2+7\)
\(=>27x=9\)
\(=>x=\frac{9}{27}=\frac{1}{3}\)
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a: \(\Leftrightarrow6x=30\)
hay x=5
b: \(\Leftrightarrow6x=25+12-1=36\)
hay x=6
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a: \(\Leftrightarrow8x=108+12=120\)
hay x=15
b: \(\Leftrightarrow6x=60\)
hay x=10
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6-7x=-48
7x=-48+6
7x=-42
x=-42:7
x=-6
k nha
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Ta có :
6x = 3x+1-5x+6x-5x+4x-9x+7x
=> 6x+1 = (3x+7x) - (5x+5x) + (6x+4x) - 9x
=> 6x+9x+1 = 10x - 10x + 10x
=> 15x+1 = 10x
=> 10x-15x = 1
=> -5x = 1
=> x = 1 : (-5)
=> x = -1
Vậy x = -1
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7x - 5x = 28
x.7 - x.5 = 28
x.(7-5) = 28
x.2 = 28
x = 28 : 2
x = 14
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`|7x+1|-|5x+6|=0`
`<=> |7x+1|=|5x+6|`
\(\Leftrightarrow\left[{}\begin{matrix}7x+1=5x+6\\7x+1=-5x-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{12}\end{matrix}\right.\)
Vậy...
|7x+1| - |5x+6|=0
⇔ |7x+1| = |5x+6|
⇔\(\left[{}\begin{matrix}7x+1=5x+6 \\7x+1=-5x-6\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}2x=5\\12x=-7\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-7}{12}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Trả lời :
Bn tham khảo link này nhé :
Câu hỏi của Hoa Thiên Cốt - Toán lớp 8 - Học toán với OnlineMath
Chúc bn hc tốt <3
\(a,x^2+4x+x+4\)
\(\Rightarrow x\left(x+4\right)+\left(x+4\right)\)
\(\Rightarrow\left(x+4\right)\left(x+1\right)\)
\(b,2x^2+11x-4x-22\)
\(\Rightarrow x\left(2x-11\right)-2\left(2x-11\right)\)
\(\Rightarrow\left(2x-11\right)\left(x-2\right)\)
bạn tự làm các câu tiếp theo nhé nếu câu nào k biết thì cứ hỏi mk
\(-12x=48\)
\(x=4\)
-5x-7x=48
-12x=48
X=48:(-12)
X=(-4)