tìm x biết
a) 15 /x+1/=10
b)2 /x-3/+7=15
c) 1+2+3+...+2x =210
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`a) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3`
`<=>x(x^2-25)-(x^3-8)=3`
`<=>x^3-25x-x^3+8=3`
`<=>-25x=-5`
`<=>x=1/5`
`b) (x – 3)^3 – (x – 3)(x^2 + 3x + 9) + 9(x + 1)^2 = 15`
`<=>x^3-9x^2+27x-27-(x^3-27)+9(x^2+2x+1)=15`
`<=>-9x^2+27x+9x^2+18x+9=15`
`<=>45x+9=15`
`<=>45x=6`
`<=>x=6/45=2/15`
`c) (x+5)(x^2 –5x +25) – (x – 7) = x^3`
`<=>x^3-125-x+7=x^3`
`<=>x^3-x-118=x^3`
`<=>-x-118=0`
`<=>-x=118<=>x=-118`
`d) (x+2)(x^2 – 2x + 4) – x(x^2 + 2) = 4 `
`<=>x^3+8-x^3-2x=4`
`<=>8-2x=4`
`<=>2x=4<=>x=2`
\(a,ĐKXĐ:x\ge1\\ 13-\sqrt{x-1}=10\\ \Leftrightarrow\sqrt{x-1}=3\\ \Leftrightarrow x-1=9\\ \Leftrightarrow x=10\\ b,ĐKXĐ:x\in R\\ \sqrt{\left(2x-1\right)^2}-1=3\\ \Leftrightarrow\left|2x-1\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=-4\\2x-1=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
Tìm x biết :
a) 1+3+5+7+...+(2x-1) = 225
b) x+(x+1)+(x+2)+...+(x+2010) = 2029099
c) 2+4+6+8+...+2x=210
\(a,\Leftrightarrow x^2+2x+1-x^2+3x-2x=3\\ \Leftrightarrow3x=2\Leftrightarrow x=\dfrac{3}{2}\\ b,\Leftrightarrow x^2-x-6-x^2+6x-9=15\\ \Leftrightarrow5x=30\Leftrightarrow x=6\\ c,\Leftrightarrow x^3+3x^2+3x+1-x^3-3x^2-2x+3=0\\ \Leftrightarrow x=-4\)
a) \(\left(x+1\right)^2-x\left(x-3\right)=2x+3\Rightarrow x^2+2x+1-x^2+3x=2x+3\)
\(\Rightarrow3x=2\Rightarrow x=\dfrac{2}{3}\)
1/ `|x|=10<=> x=\pm 10`
2/ `|x-8|=0<=>x-8=0<=>x=8`
3/ `7+|x|=12<=>|x|=5<=>x=\pm 5`
4/ `|x+1|=3`
$\Leftrightarrow\left[\begin{array}{1}x+1=3\\x+1=-3\end{array}\right.\\\Leftrightarrow\left[\begin{array}{1}x=3\\x=-4\end{array}\right.$
5/ `15-x=16-(14-42)`
`<=>15-x=16+28`
`<=>15-x=44`
`<=>x=-29`
6/ `210-(x-12)=168`
`<=>210-x+12=168`
`<=>222-x=168`
`<=>x=54`
a) x = 2.
b) x = 7.
c) x= 12.
d) x= 45.
e) x = 18.
f) x = 10.
\(a,\Leftrightarrow4x^2-24x+36-4x^2+1=10\\ \Leftrightarrow-24x=-27\Leftrightarrow x=\dfrac{9}{8}\\ b,\Leftrightarrow x\left(x^2-25\right)=0\\ \Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
\(a,4.\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
\(\Leftrightarrow4.\left(x^2-6x+9\right)-\left(2x^2\right)-1^2=10\)
\(\Leftrightarrow4x^2-24x+36-4x^2+1=10\)
\(\Leftrightarrow-24x+27=10\)
\(\Leftrightarrow-24x=-27\)
\(\Leftrightarrow x=\dfrac{27}{24}\)
Vậy \(x=\dfrac{27}{24}\)
b) 2.| x - 3 | + 7 = 15
2.| x - 3 | = 15 - 7
2.| x - 3 | = 8
| x - 3 | = 8 : 2
| x - 3 | = 4
Ta có 2 trường hợp :
TH 1 : x - 3 = -4 TH 2 : x - 3 = 4
x = -4 +3 x = 4 + 3
x = -1 x = 7
Vậy x là -1 hoặc 7