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Bài 2:
a: =>x=0 hoặc x+3=0
=>x=0 hoặc x=-3
b: =>x-2=0 hoặc 5-x=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1
`@` `\text {Ans}`
`\downarrow`
`a)`
\(5\cdot x^3-5=0\)
`=> 5*x^3 = 0+5`
`=> 5*x^3 = 5`
`=> x^3 = 5 \div 5`
`=> x^3 = 1`
`=> x^3 = 1^3`
`=> x=1`
Vậy, `x=1.`
`b)`
\(( x+1)^2 = 16\)
`=> (x+1)^2 = (+-4)^2`
`=>`\(\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4-1\\x=-4-1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
Vậy, `x \in {3; -5}`
`c)`
\(( x+1)^3 = 27\)
`=> (x+1)^3 = 3^3`
`=> x+1=3`
`=> x=3-1`
`=> x=2`
Vậy, `x=2.`
`d)`
\(( x-1)^3 = 343\)
`=> (x-1)^3 = 7^3`
`=> x-1=7`
`=> x=7+1`
`=> x=8`
Vậy, `x=8.`
`e)`
\((2x - 1^3) = 125\) hay đề là `(2x-1)^3 = 125` vậy ạ?
Mình làm cả 2 TH nhé!
`(2x-1^3)=125`
`=> 2x-1=125`
`=> 2x=125+1`
`=> 2x=126`
`=> x=126 \div 2`
`=> x=63`
TH2:
`(2x-1)^3 = 125`
`=> (2x-1)^3 = 5^3`
`=> 2x-1=5`
`=> 2x=5+1`
`=> 2x=6`
`=> x=6 \div 2`
`=> x=3`
Vậy, `x=3.`
(a) \(5x^3-5=0\Leftrightarrow5x^3=5\Leftrightarrow x^3=1\Leftrightarrow x=1\)
(b) \(\left(x+1\right)^2=16\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
(c) \(\left(x+1\right)^3=27\Leftrightarrow x+1=3\Leftrightarrow x=2\)
(d) \(\left(x-1\right)^3=343\Leftrightarrow x-1=7\Leftrightarrow x=8\)
(e) \(\left(2x-1\right)^3=125\Leftrightarrow2x-1=5\Leftrightarrow2x=6\Leftrightarrow x=3\)
`a)5/8x+2/5=1/5`
`=>5/8x=1/5-2/5`
`=>5/8x=-1/5`
`=>x=-1/5:5/8=-8/25`
`b)5/7:x+11/7=18/7`
`=>5/7:x=18/7-11/7`
`=>5/7:x=1`
`=>x=5/7`
`c)(-1,2).(-3/24)+(0,4-1 4/15):1 2/3`
`=(-6/5).(-1/8)+(2/5-19/15):5/3`
`=3/20+(-13/15)*3/5`
`=3/20-13/25=-37/100`
a)5/8.x+2/5=1/5
5/8.x=1/5 - 2/5
5/8.x=-1/5
x=(-1/5):5/8
x=(-1/5).8/5
x=-8/25. Vậy x=-8/25
b)5/7:x +11/7=18/7
5/7:x=1
x=5/7:1
x=5/7. Vậy x=5/7
a. 2x+\(\dfrac{4}{5}\)=0 hoặc 3x-\(\dfrac{1}{2}\)=0
2x=- 4/5 hoặc 3x=1/2
x=-2/5 hoặc x=\(\dfrac{1}{6}\)
b. x-\(\dfrac{2}{5}\)=0 hoặc x+\(\dfrac{4}{7}\)=0
x=2/5 hoặc x=-\(\dfrac{4}{7}\)
d. x(1+5/8-12/16)=1
\(\dfrac{7}{8}\)x=1=> x=8/7
\(a,\dfrac{1}{4}-\left(2x+\dfrac{1}{2}\right)^2=0\\ \Leftrightarrow\left(2x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}\\ \Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{1}{2}=\dfrac{1}{2}\\2x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{2}-\dfrac{1}{2}\\2x=-\dfrac{1}{2}-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=0\\2x=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\) \(b,\dfrac{1}{2}x+\dfrac{2}{3}x-1=-3\dfrac{1}{3}\\ \Leftrightarrow x\left(\dfrac{1}{2}+\dfrac{2}{3}\right)=-\dfrac{10}{3}+1\\ \Leftrightarrow\left(\dfrac{3+4}{6}\right)x=\dfrac{-10}{3}+\dfrac{3}{3}\\ \Leftrightarrow\dfrac{7}{6}x=\dfrac{-7}{3}\\ \Leftrightarrow x=\left(-\dfrac{7}{3}\right):\dfrac{7}{6}\\ \Leftrightarrow x=-2\)
Vậy \(x=0;x=-\dfrac{1}{2}\) Vậy \(x=-2\)
\(c,\dfrac{x-12}{4}=\dfrac{1}{2}\\ \Leftrightarrow2.\left(x-12\right)=4\\ \Leftrightarrow2x-24=4\\ \Leftrightarrow2x=24+2\\ \Leftrightarrow2x=26\\ \Leftrightarrow x=26:2=13\)
Vậy \(x=13\)
Bài 2 : a, x = -36/9 = -4
b, đề sai
c, <=> -2 =< x =< -3 => x = -1
Bài 1:
a: 2/8=9/36; 2/9=8/36; 8/2=36/9; 9/2=36/8
b: -2/4=9/-18; -2/9=4/-18; 4/-2=-18/9; 9/-2=-18/4
Bài 2:
a: =>x/3=-4/3
hay x=-4
Câu b đề sai rồi bạn
B)\(\frac{11}{2}+x=\frac{3}{2}-7\)
\(\frac{11}{2}+x=\frac{3}{2}-\frac{14}{2}\)
\(\frac{11}{2}+x=\frac{-11}{2}\)
\(x=\frac{-11}{2}-\frac{11}{2}\)
\(x=\frac{-22}{2}\)
\(x=-11\)
A)\(\frac{1}{3}x+\frac{2}{5}\left(x+1\right)=0\)
\(\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
\(x\cdot\left(\frac{1}{3}+\frac{2}{5}\right)=0-\frac{2}{5}\)
\(x\cdot\left(\frac{5}{15}+\frac{6}{15}\right)=-\frac{2}{5}\)
\(x\cdot\frac{11}{15}=\frac{-2}{5}\)
\(x=\frac{-2}{5}:\frac{11}{5}\)
\(x=\frac{-2}{11}\)
nhớ k cho mình nha
a) 1/3 x + 0,4 ( x + 1 ) = 0
<=> 1/3 x + 0,4 . x + 0,4 . 1 = 0
<=> 1/3 x + 0,4 x + 0,4 = 0
<=> x. ( 1/3 + 0,4 ) + 0,4 = 0
<=> x. 11/15 + 0.4 = 0
<=> x. 11/15 = -2/5
<=> x = -2/5 : 11/15 = - 6/11
b) 1 1/2 + x = 3/2 -7
<=> 3/2 + x = -11/2
<=> x = -11/2 - 3/2
<=> x = -7 < GOOD LUCK >