Tìm x thuộc N biết:
a, \(\left(x-9\right)^4=\left(x-9\right)^7\)
b, \(\left(3x-15\right)^{10}=\left(3x-15\right)^{15}\)
c, \(\left(x-8\right)^3=\left(x-8\right)^6\)
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1 tháng 9 2021
giúp mik mik đang cần gấp
nhưng phả có lời giải đừng cho mỗi đáp án
1 tháng 9 2021
a:Ta có: \(\left(x-9\right)^7=\left(x-9\right)^4\)
\(\Leftrightarrow\left(x-9\right)^4\cdot\left[\left(x-9\right)^3-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-9=0\\x-9=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=10\end{matrix}\right.\)
b: ta có: \(\left(3x-15\right)^{15}=\left(3x-15\right)^{10}\)
\(\Leftrightarrow\left(3x-15\right)^{10}\cdot\left[\left(3x-15\right)^5-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-15=0\\3x-15=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{16}{3}\end{matrix}\right.\)
31 tháng 8 2021
a) \(\left(x-9\right)^4=\left(x-9\right)^7\)
\(\Rightarrow\left[{}\begin{matrix}x-9=1\\x-9=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=10\\x=9\end{matrix}\right.\)
b) \(\left(3x-15\right)^{10}=\left(3x-15\right)^{15}\)
\(\Rightarrow\left[{}\begin{matrix}3x-15=0\\3x-15=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{3}\\x=\dfrac{16}{3}\end{matrix}\right.\)
c) \(\left(x-8\right)^3=\left(x-8\right)^6\)
\(\Rightarrow\left[{}\begin{matrix}x-8=0\\x-8=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=8\\x=9\end{matrix}\right.\)
TA
31 tháng 8 2021
a, `(x-9)^4=(x-9)^7`
`(x-9)^4-(x-9)^7=0`
`(x-9)^4 . [(1-(x-9)^3]=0`
TH1: `(x-9)^4=0`
`x-9=0`
`x=9`
TH2: `1-(x-9)^3=0`
`(x-9)^3=1^3`
`x-9=1`
`x=10`
b, `(3x-15)^10=(3x-15)^15`
`(3x-15)^10 . [1-(3x-15)^5]=0`
TH1: `(3x-15)^10=0`
`3x-15=0`
`x=5`
TH2: `1-(3x-15)^5=0`
`(3x-15)^5=1^5`
`3x-15=1`
`x=16/3` (Loại)
c, `(x-8)^3=(x-8)^6`
`(x-8)^3 .[1-(x-8)^3]=0`
TH1: `(x-8)^3=0`
`x=8`
TH2: `1-(x-8)^3=0`
`x-8=1`
`x=9`
21 tháng 9 2021
\(c,\Rightarrow\left[{}\begin{matrix}-2\left(x+2\right)+\left(4-x\right)=11\left(x< -2\right)\\2\left(x+2\right)+\left(4-x\right)=11\left(-2\le x\le4\right)\\2\left(x+2\right)+\left(x-4\right)=11\left(x>4\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{3}\left(tm\right)\\x=3\left(tm\right)\\x=\dfrac{11}{3}\left(ktm\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{11}{3}\end{matrix}\right.\)
21 tháng 9 2021
\(a,\Rightarrow\left[{}\begin{matrix}x+\dfrac{5}{2}=3x+1\\x+\dfrac{5}{2}=-3x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{7}{8}\end{matrix}\right.\)
16 tháng 2 2021
a) Ta có: \(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x-1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=4\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)
b) Ta có: \(x\left(2x-9\right)=3x\left(x-5\right)\)
\(\Leftrightarrow x\left(2x-9\right)-3x\left(x-5\right)=0\)
\(\Leftrightarrow x\left(2x-9\right)-x\left(3x-15\right)=0\)
\(\Leftrightarrow x\left(2x-9-3x+15\right)=0\)
\(\Leftrightarrow x\left(6-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
Vậy: S={0;6}
c) Ta có: \(3x-15=2x\left(x-5\right)\)
\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{5;\dfrac{3}{2}\right\}\)
d) Ta có: \(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)
\(\Leftrightarrow6\left(5-x\right)=2\left(3x-4\right)\)
\(\Leftrightarrow30-6x=6x-8\)
\(\Leftrightarrow30-6x-6x+8=0\)
\(\Leftrightarrow-12x+38=0\)
\(\Leftrightarrow-12x=-38\)
\(\Leftrightarrow x=\dfrac{19}{6}\)
Vậy: \(S=\left\{\dfrac{19}{6}\right\}\)
e) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)
\(\Leftrightarrow6x+4-3x-1=12x+10\)
\(\Leftrightarrow3x+3-12x-10=0\)
\(\Leftrightarrow-9x-7=0\)
\(\Leftrightarrow-9x=7\)
\(\Leftrightarrow x=-\dfrac{7}{9}\)
Vậy: \(S=\left\{-\dfrac{7}{9}\right\}\)
9 tháng 4 2017
(2x+9)/(x+1)(x+8)-(2x+15)/(x+8)(x+7)+(2x+10)/(x+7)(x+3)=4/3
(x+1+x+8)/(x+1)(x+8)-(x+8+x+7)/(x+8)(x+7)+(x+7+x+3)/(x+7)(x+3)=4/3
1/(x+8)+1/(x+1)-1/(x+7)-1/(x+8)+1/(x+7)+1/(x+3)=4/3
1/(x+1)+1/(x+3)=4/3
(x+3+x+1)/(x+3)(x+1)=4/3
(2x+4)/(x+3)(x+1)=4/3
=>(2x+4).3=(x+3)(x+1).4
6(x+2)=4(x+3)(x+1)
3(x+2)=2(x+3)(x+1)
3x+6=2(x^2+4x+3)
3x+6=2x^2+8x+6
2x^2+8x+6-3x-6=0
2x^2+5x=0
x(2x+5)=0
=> x=0 hoac 2x+5=0
=> x=0 hoac x=-5/2
11 tháng 3 2020
1) \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
<=> \(\frac{21x}{24}-\frac{100\left(x-9\right)}{24}=\frac{80x+6}{24}\)
<=> 21x - 100x + 900 = 80x + 6
<=> -79x - 80x = 6 - 900
<=> -159x = -894
<=> x = 258/53
Vậy S = {258/53}
2) \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x+1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
<=> \(\frac{3\left(4x^2+4x+1\right)}{15}-\frac{5\left(x^2+2x+1\right)}{15}=\frac{7x^2-14x-5}{15}\)
<=> 12x2 + 12x + 3 - 5x2 - 10x - 5 = 7x2 - 14x - 5
<=> 7x2 + 2x - 7x2 + 14x = -5 + 2
<=> 16x = 3
<=> x = 3/16
Vậy S = {3/16}
11 tháng 3 2020
3) 4(3x - 2) - 3(x - 4) = 7x+ 10
<=> 12x - 8 - 3x + 12 = 7x + 10
<=> 9x - 7x = 10 - 4
<=> 2x = 6
<=> x = 3
Vậy S = {3}
4) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
<=> \(\frac{x^2+14x+40}{12}+\frac{3\left(x^2+2x-8\right)}{12}=\frac{4\left(x^2+8x-20\right)}{12}\)
<=> x2 + 14x + 40 + 3x2 + 6x - 24 = 4x2 + 32x - 80
<=> 4x2 + 20x - 4x2 - 32x = -80 - 16
<=> -12x = -96
<=> x = 8
Vậy S = {8}
TT
11 tháng 1 2023
\(8,1-\left(x-6\right)=4\left(2-2x\right)\)
\(\Leftrightarrow1-x+6=8-8x\)
\(\Leftrightarrow-x+8x=8-1-6\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\dfrac{1}{7}\)
\(9,\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
\(10,\left(x+3\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)
2
11 tháng 1 2023
`8)1-(x-5)=4(2-2x)`
`<=>1-x+5=8-6x`
`<=>5x=2<=>x=2/5`
`9)(3x-2)(x+5)=0`
`<=>[(x=2/3),(x=-5):}`
`10)(x+3)(x^2+2)=0`
Mà `x^2+2 > 0 AA x`
`=>x+3=0`
`<=>x=-3`
`11)(5x-1)(x^2-9)=0`
`<=>(5x-1)(x-3)(x+3)=0`
`<=>[(x=1/5),(x=3),(x=-3):}`
`12)x(x-3)+3(x-3)=0`
`<=>(x-3)(x+3)=0`
`<=>[(x=3),(x=-3):}`
`13)x(x-5)-4x+20=0`
`<=>x(x-5)-4(x-5)=0`
`<=>(x-5)(x-4)=0`
`<=>[(x=5),(x=4):}`
`14)x^2+4x-5=0`
`<=>x^2+5x-x-5=0`
`<=>(x+5)(x-1)=0`
`<=>[(x=-5),(x=1):}`
a. x = 9
b. x = 5
c. x = 8
Đề nhìn vô lí quá
a. x = 9
b. x = 5
c. x = 8