Giúp mình với: 5x -x = 8^2 : 2^4
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2.(x-3)+3x+0.5=\(\dfrac{3}{4}\)
4x+2+4x=272
(1,2-5x).(2\(\dfrac{1}{8}\) +1/2 x)=0
GIÚP MÌNH VỚI !!!!
\(2\left(x-3\right)+3x+0,5=\dfrac{3}{4}\\ \Leftrightarrow2x-6+3x+\dfrac{1}{2}=\dfrac{3}{4}\\ \Leftrightarrow x\left(2+3\right)=\dfrac{3}{4}-\dfrac{1}{2}+6\\ \Leftrightarrow5x=\dfrac{25}{4}\\ \Leftrightarrow x=\dfrac{25}{4}:5=\dfrac{5}{4}\\ ---\\ 4^{x+2}+4^x=272\\ \Leftrightarrow4^x\left(4^2+1\right)=272\\ \Leftrightarrow4^x.17=272\\ \Leftrightarrow4^x=\dfrac{272}{17}=16=4^2\\ Vậy:x=2\\ ----\\ \left(1,2-5x\right)\left(2\dfrac{1}{8}+\dfrac{1}{2}x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}1,2-5x=0\\2,125+0,5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=1,2\\0,5x=-2,125\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1,2}{5}=0,24\\x=\dfrac{-2,125}{0,5}=-4,25\end{matrix}\right.\)
a) \(2\left(x-3\right)+3x+0,5=\dfrac{3}{4}\)
\(\Rightarrow2x-6+3x+\dfrac{1}{2}=\dfrac{3}{4}\)
\(\Rightarrow5x-6=\dfrac{3}{4}-\dfrac{1}{2}\)
\(\Rightarrow5x-6=\dfrac{1}{4}\)
\(\Rightarrow5x=\dfrac{1}{4}+6\)
\(\Rightarrow5x=\dfrac{25}{4}\)
\(\Rightarrow x=\dfrac{25}{4}:5\)
\(\Rightarrow x=\dfrac{5}{4}\)
b) \(4^{x+2}+4^x=272\)
\(\Rightarrow4^x\cdot4^2+4^x\cdot1=272\)
\(\Rightarrow4^x\cdot\left(16+1\right)=272\)
\(\Rightarrow4^x\cdot17=272\)
\(\Rightarrow4^x=16\)
\(\Rightarrow4^x=4^2\)
\(\Rightarrow x=2\)
c) \(\left(1,2-5x\right)\left(2\dfrac{1}{8}+\dfrac{1}{2}x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}1,2-5x=0\\\dfrac{15}{8}+\dfrac{1}{2}x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}5x=1,2\\\dfrac{1}{2}x=-\dfrac{15}{8}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1,2}{5}\\x=-\dfrac{15}{8}:\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{6}{25}\\x=-\dfrac{15}{4}\end{matrix}\right.\)
a: =25x^2-10x+25x^2-1-10x=50x^2-20x-1
b: =x^2-12x+32-x^2+12x-32
=0
** Bổ sung điều kiện $x,y$ là số nguyên.
a/
$(5x-1)(y+1)=4$
Với $x,y$ nguyên thì $5x-1, y+1$ nguyên. Mà tích của chúng bằng 4 nên ta có các trường hợp sau:
TH1: $5x-1=1, y+1=4\Rightarrow x=\frac{2}{5}$ (loại)
TH2: $5x-1=-1, y+1=-4\Rightarrow x=0; y=-5$
TH3: $5x-1=2, y+1=2\Rightarrow x=\frac{3}{5}$ (loại)
TH4: $5x-1=-2, y+1=-2\Rightarrow x=\frac{-1}{5}$ (loại)
TH5: $5x-1=4, y+1=1\Rightarrow x=1; y=0$
TH6: $5x-1=-4; y+1=-1\Rightarrow x=\frac{-3}{5}$ (loại)
Vậy......
b/
$xy-7y+5x=0$
$y(x-7)+5(x-7)=-35$
$(x-7)(y+5)=-35$
Vì $x,y$ nguyên nên $x-7, y+5$ nguyên. $(x-7)(y+5)=-35\Rightarrow x-7$ là ước của $-35$.
Mà $x\geq 3\Rightarrow x-7\geq -4$
$\Rightarrow x-7\in \left\{-1; 1; 5; 7; 35\right\}$
Nếu $x-7=-1\Rightarrow y+5=35$
$\Rightarrow x=6; y=30$
Nếu $x-7=1\Rightarrow y+5=-35$
$\Rightarrow x=8; y=-40$
Nếu $x-7=5\Rightarrow y+5=-7$
$\Rightarrow x=12; y=-12$
Nếu $x-7=7\Rightarrow y+5=-5$
$\Rightarrow x=14; y=-10$
Nếu $x-7=35; y+5=-1$
$\Rightarrow x=42; y=-6$
`(x+3)(x^2-5x+8)=(x+3).x^2`
`<=>(x+3)(x^2-5x+8-x^2)=0`
`<=>(x+3)(8-5x)=0`
`<=>` \(\left[ \begin{array}{l}x+3=0\\8-5x=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac85\\x=-3\end{array} \right.\)
Vậy `S={-3,8/5}`
`(x+3)(x^2-5x+8)=(x+3).x^2`
`<=>(x+3)(x^2-5x+8-x^2)=0`
`<=>(x+3)(-5x+8)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\-5x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{8}{5}\end{matrix}\right.\)
Vậy `S={-3;8/5}`.
\(\Leftrightarrow\frac{6x^2+3}{24}-\frac{10x-4}{24}=\frac{6x^2-6}{24}-\frac{4x-12}{24}\)
\(\Leftrightarrow\frac{6x^2+3-10x+4}{24}=\frac{6x^2-6-4x+12}{24}\)
\(\Leftrightarrow6x^2-10x+7=6x^2-4x+6\)
\(\Leftrightarrow-6x+1=0\)
\(\Rightarrow-6x=-1\)
\(\Leftrightarrow x=\frac{1}{6}\)
Vậy ...
=> 4x = 64 : 16 = 4
=> x = 4 : 4 = 1
k mk nha
5x -x = 8^2 : 2^4
5x-x=64-16
5x-x=48
4x=48
x=12