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\(\frac{x}{20}=\frac{0,8}{x}\)
x.x = 20 . 0,8
x2 = 16
x2 = 42 = (-4)2
Vậy x = 4 hoặc x = -4
\(1,\frac{x+1}{x-2}=\frac{3}{4}\)
\(\Rightarrow3x-6=4x+4\)
\(\Rightarrow3x-4x=4+6\)
\(\Rightarrow-x=10\Leftrightarrow x=-10\)
\(2,\frac{x-1}{3}=\frac{x+3}{5}\)
\(\Rightarrow5x-5=3x+9\)
\(\Rightarrow5x-3x=9+5\)
\(\Rightarrow2x=14\Leftrightarrow x=7\)
\(3,\frac{2x+3}{24}=\frac{3x-1}{32}\)
\(\Rightarrow64x+96=72x-24\)
\(\Rightarrow72x-64x=24+96\)
\(\Rightarrow8x=120\)
\(\Rightarrow x=15\)
Ta có:x^2\(\ge\)x\(\ge\)0
=>x^2+x\(\ge\)0
=>x^2+x+1\(\ge\)0>1
Vậy h(x) vô nghiệm.
\(\frac{x-3}{5}=\frac{x+4}{-2}\)
=> (x - 3). (-2) = 5(x + 4)
=> -2x + 6 = 5x + 20
=> -2x - 5x = 20 - 6
=> -7x = 14
=> x = 14 : (-7)
=> x = -2
x-3/5=x+4/-2
=> ﴾x ‐ 3﴿. ﴾‐2﴿ = 5﴾x + 4﴿
=> ‐2x + 6 = 5x + 20
=> ‐2x ‐ 5x = 20 ‐ 6 => ‐7x = 14 => x = 14 : ﴾‐7﴿
=> x = ‐2
> =<
\(H\left(x\right)=x+3\)
\(\Rightarrow H\left(x\right)=0\Leftrightarrow x+3=0\Rightarrow x=-3\)
\(T\left(x\right)=12-\dfrac{1}{3}x\)
\(\Rightarrow T\left(x\right)=0\Leftrightarrow12-\dfrac{1}{3}x=0\Rightarrow\dfrac{1}{3}x=12\Rightarrow x=36\)
\(B\left(x\right)=x^2-5x+4=\left(x-1\right)\left(x-4\right)\)
\(\Rightarrow B\left(x\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
\(C\left(x\right)=42x-4x^2=2x\left(21-2x\right)\)
\(\Rightarrow C\left(x\right)=0\Leftrightarrow\left[{}\begin{matrix}2x=0\\21-2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=10\dfrac{1}{2}\end{matrix}\right.\)
MONG CÂU TRẢ LỜI NÀY GIÚP BN
#chúc_bn_học_tốt
a) [2x] = -1\(\Rightarrow-1\le2x< 0\Rightarrow-0,5\le x< 0\)
b) [x + 0,4] = 3\(\Rightarrow3\le x+0,4< 4\Rightarrow2,6\le x< 3,6\)
c)\(\left[\frac{2}{3}x-5\right]=3\Rightarrow3\le\frac{2}{3}x-5< 4\Rightarrow8\le\frac{2}{3}x< 9\Rightarrow12\le x< 13,5\)
Từ bài trên,ta có :\(\left[x\right]=y\Rightarrow y\le x< y+1\left(x\in Q;y\in Z\right)\)
Ta có : \(\frac{x}{6}=\frac{24}{x}\)
\(\Leftrightarrow x\cdot x=6\cdot24\)
\(\Leftrightarrow x^2=144\)
\(\Rightarrow x=\pm12\)
\(\frac{x}{6}=\frac{24}{x}\)
\(=>x^2=24.6\)
=> \(x^2=144\)
=> x = \(12\)