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NH
1
H9
HT.Phong (9A5)
CTVHS
22 tháng 4 2023
a) \(\dfrac{3,5}{15}=\dfrac{-2}{x}\)
\(\Rightarrow x=\dfrac{15.-2}{3,5}\)
\(\Rightarrow x=-8,57\)
b) \(2\left(3x-2\right)-3\left(x-2\right)-=-1\)
\(\Rightarrow6x-4-3x+6=-1\)
\(\Rightarrow6x-3x=-1+4-6\)
\(\Rightarrow3x=-3\)
\(\Rightarrow x=-\dfrac{3}{3}=-1\)
18 tháng 5 2021
a) Ta có: \(\left(2x-3\right)-\left(x-5\right)=\left(x+2\right)-\left(x-1\right)\)
\(\Leftrightarrow2x-3-x+5=x+2-x+1\)
\(\Leftrightarrow x+2=3\)
hay x=1
Vậy: x=1
b) Ta có: \(2\left(x-1\right)-5\left(x+2\right)=-10\)
\(\Leftrightarrow2x-2-5x-10=-10\)
\(\Leftrightarrow-3x=-10+10+2=2\)
hay \(x=-\dfrac{2}{3}\)
Vậy: \(x=-\dfrac{2}{3}\)
PT
18 tháng 5 2021
a, (2x - 3) - (x - 5) = (x + 2) - (x - 1)
2x - 3 - x + 5 = x + 2 - x + 1
(2x - x) + (-3 + 5) = (x - x) + (2 + 1)
x + 2 = 3
x = 1
YT
3
H9
HT.Phong (9A5)
CTVHS
9 tháng 8 2023
a) \(1=\left(2x+0,5\right)^{600}\)
\(\Rightarrow1^{600}=\left(2x+0,5\right)^{600}\)
\(\Rightarrow\left[{}\begin{matrix}2x+0,5=1\\2x+0,5=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=0,5\\2x=-1,5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0,25\\x=-0,75\end{matrix}\right.\)
b) \(\left(x-0,125\right)^2=0,25\)
\(\Rightarrow\left(x-0,125\right)^2=0,5^2\)
\(\Rightarrow\left[{}\begin{matrix}x-0,125=0,5\\x-0,125=-0,5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0,625\\x=-0,375\end{matrix}\right.\)
c) \(\left(x-3\right)^{11}=\left(x-3\right)^{41}\)
\(\Rightarrow\left(x-3\right)^{11}-\left(x-3\right)^{41}=0\)
\(\Rightarrow\left(x-3\right)^{11}\left[1-\left(x-3\right)^{30}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^{11}=0\\\left(x-3\right)^{30}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-3=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
9 tháng 8 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
`1 = (2x + 0,5)^600`
`=> (2x+0,5)^600 = (+-1)^600`
`=> \text {TH1: } 2x + 0,5 = 1`
`=> 2x = 1 - 0,5`
`=> 2x = 0,5`
`=> x = 0,5 \div 2`
`=> x = 0,25`
`\text {TH2: } 2x + 0,5 = -1`
`=> 2x = -1 - 0,5`
`=> 2x = -1,5`
`=> x = -1,5 \div 2`
`=> x = -0,75`
Vậy, `x \in {-0,75; 0,25}.`
`b)`
`(x - 0,125)^2 = 0,25`
`=> (x - 0,125)^2 = (+-0,5)^2`
`=> `\(\left[{}\begin{matrix}x-0,125=0,5\\x-0,125=-0,5\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0,5+0,125\\x=-0,5+0,125\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0,625\\x=-0,375\end{matrix}\right.\)
Vậy, `x \in {-0,375; 0,625}.`
`c)`
`(x - 3)^11 = (x - 3)^41`
`=> (x - 3)^11 - (x - 3)^41 = 0`
`=> (x - 3)^11 * [ 1 - (x - 3)^30] = 0`
`=>`\(\left[{}\begin{matrix}\left(x-3\right)^{11}=0\\1-\left(x-3\right)^{30}=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^{30}=1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=3\\x-3=1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
Vậy, `x \in {3; 4}.`
NP
0
12 tháng 9 2021
a) \(\dfrac{1}{2}+\dfrac{2}{3}x=\dfrac{1}{4}\\ \Rightarrow\dfrac{2}{3}x=-\dfrac{1}{4}\\ \Rightarrow x=-\dfrac{3}{8}\)
b) \(2\dfrac{2}{3}:x=1\dfrac{7}{9}:0,02\\ \Rightarrow2\dfrac{2}{3}:x=\dfrac{800}{9}\\ \Rightarrow x=\dfrac{3}{100}\)
c) \(x^x-x+1=1\\ \Rightarrow x^x-x=0\\ \Rightarrow x^x=x\\ \Rightarrow x=1\)
d) \(5-\left|3x-1\right|=3\\ \Rightarrow\left|3x-1\right|=2\\ \Rightarrow\left[{}\begin{matrix}3x-1=-2\\3x-1=2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=1\end{matrix}\right.\)
H
18 tháng 10 2023
\(\dfrac{1}{2}-3x+\left|x-1\right|=0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}-0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}\\ \Rightarrow\left|x-1\right|=\dfrac{1}{2}-3x\\ \Rightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{2}-3x\\x-1=-\dfrac{1}{2}+3x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x+3x=\dfrac{1}{2}+1\\x-3x=-\dfrac{1}{2}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}4x=\dfrac{3}{2}\\2x=\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{8}\\x=\dfrac{1}{4}\end{matrix}\right.\)
__
\(\dfrac{1}{2}\left|2x-1\right|+\left|2x-1\right|=x+1\\ \Rightarrow\left|2x-1\right|\cdot\left(\dfrac{1}{2}+1\right)=x+1\\ \Rightarrow\left|2x-1\right|\cdot\dfrac{3}{2}=x+1\\ \Rightarrow\left|2x-1\right|=x+1:\dfrac{3}{2}\\ \Rightarrow\left|2x-1\right|=x+\dfrac{2}{3}\\ \Rightarrow\left[{}\begin{matrix}2x-1=x+\dfrac{2}{3}\\2x-1=-x-\dfrac{2}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-x=\dfrac{2}{3}+1\\2x+x=-\dfrac{2}{3}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\3x=\dfrac{1}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{1}{9}\end{matrix}\right.\)
LP
3
18 tháng 8 2020
+) \(2x\left(x-4\right)-x\left(2x+3\right)+22=0\)
\(\Leftrightarrow2x^2-8x-2x^2-3x+22=0\)
\(\Leftrightarrow-11x+22=0\)
\(\Leftrightarrow-11\left(x-2\right)=0\)
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
+) \(\left(2x+3\right)\left(3x+2\right)+2\left(1-3x\right)\left(x+\frac{1}{2}\right)=1\)
\(\Leftrightarrow6x^2+4x+9x+6+\left(2-6x\right)\left(x+\frac{1}{2}\right)=1\)
\(\Leftrightarrow6x^2+13x+6+2x+1-6x^2-3x=1\)
\(\Leftrightarrow12x+7=1\)
\(\Leftrightarrow x=\frac{-1}{2}\)
18 tháng 8 2020
2x( x - 4 ) - x( 2x + 3 ) + 22 = 0
<=> 2x2 - 8x - 2x2 - 3x + 22 = 0
<=> -11x + 22 = 0
<=> -11x = -22
<=> x = 2
( 2x + 3 )( 3x + 2 ) + 2( 1 - 3x )( x + 1/2 ) = 1
<=> 6x2 + 13x + 6 + 2( -3x2 - 1/2x + 1/2 ) = 1
<=> 6x2 + 13x + 6 - 6x2 - x + 1 = 1
<=> 12x + 7 = 1
<=> 12x = -6
<=> x = -6/12 = -1/2
