| x-2| +4=6 timf x
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DV
0
NT
1
T
5 tháng 10 2023
1) \(\left|\dfrac{1}{2}x-\dfrac{1}{6}\right|=\dfrac{1}{3}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{6}=\dfrac{1}{3}\\\dfrac{1}{2}x-\dfrac{1}{6}=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{1}{2}\\\dfrac{1}{2}x=-\dfrac{1}{6}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
\(---\)
2) \(\left|\dfrac{1}{2}x+\dfrac{3}{5}\right|=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x+\dfrac{3}{5}=\dfrac{1}{2}\\\dfrac{1}{2}x+\dfrac{3}{5}=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=-\dfrac{1}{10}\\\dfrac{1}{2}x=-\dfrac{11}{10}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=-\dfrac{11}{5}\end{matrix}\right.\)
\(---\)
3) \(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\left|\dfrac{-3}{4}\right|\)
\(\Rightarrow\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\dfrac{3}{4}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{3}{4}\\\dfrac{3}{4}x-\dfrac{3}{4}=-\dfrac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{3}{2}\\\dfrac{3}{4}x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
\(---\)
4) \(14-\left|\dfrac{3x}{2}-1\right|=9\)
\(\Rightarrow\left|\dfrac{3x}{2}-1\right|=5\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3x}{2}-1=5\\\dfrac{3x}{2}-1=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3x}{2}=6\\\dfrac{3x}{2}=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=12\\3x=-8\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{8}{3}\end{matrix}\right.\)
\(---\)
5) \(17-\left|\dfrac{2}{3}-4x\right|=9\)
\(\Rightarrow\left|\dfrac{2}{3}-4x\right|=8\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}-4x=8\\\dfrac{2}{3}-4x=-8\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4x=-\dfrac{22}{3}\\4x=\dfrac{26}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{6}\\x=\dfrac{13}{6}\end{matrix}\right.\)
\(---\)
6) \(5-\left|2x-3\right|=\dfrac{1}{2}\)
\(\Rightarrow\left|2x-3\right|=\dfrac{9}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=\dfrac{9}{2}\\2x-3=-\dfrac{9}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{15}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
#\(Toru\)
ST
2
FT
0
20 tháng 2 2020
(x^2 +24+14x) (x^2+24+10x) =165x^2
Đặt t = x^2 + 24+12x
(t-2x)(t+2x) = 165x^2
t^2 - 4x^2 =165x^2
t^2 = 169x^2
t = 13x hay t = -13x
Nếu t = 13x thì
x^2 +12x + 24= 13x
x^2 - x + 24 = 0 (Vô nghiệm vì vế trái > 0)
Nếu t = -13x thì:
x^2 +12x+24 = -13x
x^2 +25x +24=0
(x+1)(x+24) = 0
x + 1 =0 hay x+24 = 0
x = -1 hay x= -24
Vậy...
Học tốt!
TL
6
H9
HT.Phong (9A5)
CTVHS
8 tháng 1 2024
\(\dfrac{1}{x-3}+\dfrac{3x^2-8x+10}{x^2-5x+6}-\dfrac{2x-4}{x-2}\left(ĐK:x\ne3;x\ne2\right)\)
\(=\dfrac{1}{x-3}+\dfrac{3x^2-8x+10}{x\left(x-2\right)-3\left(x-2\right)}-\dfrac{2x-4}{x-2}\)
\(=\dfrac{1}{x-3}+\dfrac{3x^2-8x+10}{\left(x-3\right)\left(x-2\right)}-\dfrac{2x-4}{x-2}\)
\(=\dfrac{x-2}{\left(x-2\right)\left(x-3\right)}+\dfrac{3x^2-8x+10}{\left(x-3\right)\left(x-2\right)}-\dfrac{\left(2x-4\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x-2+3x^2-8x+10-\left(2x^2-6x-4x+12\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{3x^2-7x+8-2x^2+10x-12}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2+3x-4}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2+3x-4}{x^2-5x+6}\)
tri tuyet doi cua x-2=2
suy ra x-2=-2
x-2=2
suy ra x=0
x=4
vậy x={0,4}
| x - 2 | + 4 = 6
| x - 2 | = 6 - 4
| x - 2 | = 2
| x - 2 | = 2 <=> | x - 2 | = -2
x = 2 + 2 x = -2 + 2
x = 4 x = 0
Vậy ..