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tìm x,biết
5-x-16=40+x
4x-10=15-x
15-x=4x-5
x-15=6+4x
-12+x=5x-20
7x-4=20+3x
5x-7=-21-2x
x+15=20-4x
17-x=7-6x
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4x - 3 = 4x - ( 8 - 5 ) = 4x - 8 + 5 = 4x - 4.2 + 5 = 4.( x - 2 ) + 5
k cho mk nha bạn
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Sai rồi kìa! Kém!
\(\dfrac{1}{1.5}=\dfrac{1}{5}\)
\(1-\dfrac{1}{5}=\dfrac{4}{5}\)
\(\Rightarrow\dfrac{4}{5}\ne\dfrac{1}{5}\Rightarrow\dfrac{1}{1.5}\ne1-\dfrac{1}{5}\)
Tốt nhất là nên xóa đi làm lại kẻo người ta cười cho!
Gợi ý: Đến cái đoạn sai...
\(\Leftrightarrow4x\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{33.37}\right)=\dfrac{72}{37}\)
\(\Leftrightarrow4x.\dfrac{1}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{33.37}\right)=...\)
\(\Leftrightarrow4x.\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-....\right)\) (tự làm tiếp)
\(\dfrac{4x}{5}+\dfrac{4x}{45}+\dfrac{4x}{117}+...+\dfrac{4x}{1221}=\dfrac{72}{37}\)
\(\dfrac{4x}{1.5}+\dfrac{4x}{5.9}+\dfrac{4x}{9.13}+...+\dfrac{4x}{33.37}=\dfrac{72}{37}\)
\(\Leftrightarrow4x\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{33.37}\right)=\dfrac{72}{37}\)
\(\Leftrightarrow4x\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{33}-\dfrac{1}{37}\right)=\dfrac{72}{37}\)
\(\Leftrightarrow4x\left(1-\dfrac{1}{37}\right)=\dfrac{72}{37}\)
\(\Leftrightarrow4x.\dfrac{36}{37}=\dfrac{72}{37}\)
\(4x=\dfrac{72}{37}:\dfrac{36}{37}\)
\(4x=2\)
\(x=2:4\)
\(x=\dfrac{1}{2}\)
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4x - 3 = 4x - 3 + 5 - 5
= 4x - 8 + 5
= 4x - 2.4 + 5
= 4.(x-2) + 5
Vì x-2 chia hết cho x-2 nên 5 phải chia hết cho x-2
=> x-2 E Ư(5)
=> x-2 E (-1;-5;1;5)
=> x E (1;-3;3;7)
không còn gì để nói: cảm ơn pn nhiều nha, may mà có bạn, mai mk thi khảo sát rồi
CẢM ƠN BẠN
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`(x - 2)/3 = (x + 1)/4`
`(x - 2) . 4 = (x + 1) . 3`
`<=> 4x - 8 = 3x + 3`
`<=> 4x - 3x = 3 + 8`
`<=> (4 - 3)x = 11`
`=> x = 11`
`=>` `x = 11`
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a) \(x^3+4x=0\)
\(\Rightarrow x\left(x^2+4\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x^2+4=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x^2=-4\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x\in\phi\end{array}\right.\)
Vậy: \(x=0\)
b) \(2\left(5-x\right)=4x-3\)
\(\Rightarrow10-2x=4x-3\)
\(\Rightarrow10+3=4x+2x\)
\(\Rightarrow13=6x\)
\(\Rightarrow x=\frac{13}{6}\)
x3+ 4x=0
<=> x(x2+4)=0
=> x=0 hoặc x2+4=0
Mà: x2+4 >4
=>x=0
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`4^(x+4)-4^x =4080`
`=>4^x *4^4 -4^x =4080`
`=>4^x (256-1)=4080`
`=>4^x *255=4080`
`=>4^x =16`
`=>4^x =4^2`
`=>x=2`
\(4^x=64\\ 4^x=4^3\\ x=3\)
4x = 64
4x = 43
x = 3