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24/9-(x+1/2)^2=2/1/3
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Những câu hỏi liên quan
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
11 tháng 5
a; \(\dfrac{93}{17}\): \(x\) + (- \(\dfrac{21}{17}\)) : \(x\) + \(\dfrac{22}{7}\): \(\dfrac{22}{3}\) = \(\dfrac{5}{14}\)
\(\dfrac{94}{17}\) \(\times\) \(\dfrac{1}{x}\) - \(\dfrac{21}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)
\(\dfrac{72}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)
\(\dfrac{72}{17x}\) = \(\dfrac{5}{14}\) - \(\dfrac{3}{7}\)
\(\dfrac{72}{17x}\) = - \(\dfrac{1}{14}\)
17\(x\) = 72.(-14)
17\(x\) = - 1008
\(x\) = - 1008 : 17
\(x\) = - \(\dfrac{1008}{17}\)
Vậy \(x\) \(=-\dfrac{1008}{17}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
11 tháng 5
b; - \(\dfrac{32}{27}\) - (3\(x\) - \(\dfrac{7}{9}\))3 = - \(\dfrac{24}{27}\)
- \(\dfrac{32}{27}\) + \(\dfrac{24}{27}\) = (3\(x\) - \(\dfrac{7}{9}\))3
(3\(x-\dfrac{7}{9}\))3 = - \(\dfrac{8}{27}\)
(3\(x-\dfrac{7}{9}\))3 = (- \(\dfrac{2}{3}\))3
3\(x-\dfrac{7}{9}\) = - \(\dfrac{2}{3}\)
3\(x\) = - \(\dfrac{2}{3}\) + \(\dfrac{7}{9}\)
3\(x\) = \(\dfrac{1}{9}\)
\(x\) = \(\dfrac{1}{9}\) : 3
\(x\) = \(\dfrac{1}{27}\)
Vậy \(x=\dfrac{1}{27}\)
NV
0
26 tháng 10 2021
a: \(\left(x+1\right)^3+\left(x-2\right)^3=2x^3+2\left(2x-1\right)^2-9\)
\(\Leftrightarrow x^3+3x^2+3x+1+x^3-6x^2+12x-8=2x^3+2\left(4x^2-4x+1\right)-9\)
\(\Leftrightarrow2x^3-3x^2+15x-7=2x^3+8x^2-8x-7\)
\(\Leftrightarrow-11x^2+23x=0\)
\(\Leftrightarrow x\left(-11x+23\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{23}{11}\end{matrix}\right.\)
H9
HT.Phong (9A5)
CTVHS
10 tháng 8 2023
1) \(\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-3-2\right)\left(x-3+2\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
2) \(x^2-2x=24\)
\(\Leftrightarrow x^2-2x-24=0\)
\(\Leftrightarrow x^2+4x-6x-24=0\)
\(\Leftrightarrow x\left(x+4\right)-6\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
16 tháng 12 2023
n: ĐKXĐ: x<>0
\(\left(x+\dfrac{1}{x}\right)^2-3\left(x+\dfrac{1}{x}\right)+2=0\)
=>\(\left(x+\dfrac{1}{x}\right)^2-2\left(x+\dfrac{1}{x}\right)-\left(x+\dfrac{1}{x}\right)+2=0\)
=>\(\left(x+\dfrac{1}{x}-2\right)\left(x+\dfrac{1}{x}-1\right)=0\)
=>\(\dfrac{x^2+1-2x}{x}\cdot\dfrac{x^2+1-x}{x}=0\)
=>\(\left(x^2-2x+1\right)\left(x^2-x+1\right)=0\)
=>\(\left(x-1\right)^2\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]=0\)
=>\(\left(x-1\right)^2=0\)
=>x-1=0
=>x=1
p: \(x^4-4x^3+6x^2-4x+1=0\)
=>\(x^4-x^3-3x^3+3x^2+3x^2-3x-x+1=0\)
=>\(x^3\left(x-1\right)-3x^2\left(x-1\right)+3x\left(x-1\right)-\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x^3-3x^2+3x-1\right)=0\)
=>\(\left(x-1\right)^4=0\)
=>x-1=0
=>x=1
H
21 tháng 2 2023
` 8/23 . 46/24 =1/3 .x`
`=>8/23 . 23/12 =1/3 . x`
`=> 1/3 . x=2/3`
`=>x=2/3 : 1/3`
`=>x=2/3 . 3`
`=> x= 6/3`
`=>x=2`
`----`
`1/5 : x= 1/5-1/7`
`=>1/5 : x= 7/35 - 5/35`
`=> 1/5 :x= 2/35`
`=>x= 1/5 : 2/35`
`=>x=1/5 . 35/2`
`=>x=7/2`
`----`
`4/9 - (x-1/2)^2 =1/3`
`=> (x-1/2)^2 =4/9-1/3`
`=> (x-1/2)^2 =4/9- 3/9`
`=> (x-1/2)^2 =1/9`
`=> (x-1/2)^2 = (+- 1/3)^2`
`@ TH1`
`x-1/2=1/3`
`=>x=1/3+1/2`
`=>x= 2/6 + 3/6`
``=>x= 5/6`
`@ TH2`
`x-1/2=-1/3`
`=>x=-1/3 +1/2`
`=>x= -2/6 + 3/6`
`=>x=1/6`
`----`
`3,2 . x-(4/5+2/3) : 3 2/3 = 7/10`
`=> 3,2 . x-22/15 : 11/3 = 7/10`
`=> 3,2 . x-22/15 = 7/10 . 11/3`
`=> 3,2 . x-22/15 =77/30`
`=> 3,2 .x= 77/30 + 22/15`
`=> 3,2 .x=121/30`
`=>x= 121/30. 5/16`
`=>x= 121/96`
LT
0
\(\frac{24}{9}-\left(x+\frac{1}{2}\right)^2=2\frac{1}{3}\)
\(\Rightarrow\frac{24}{9}-\left(x+\frac{1}{2}\right)^2=\frac{7}{3}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\frac{24}{9}-\frac{7}{3}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\frac{24}{9}-\frac{21}{9}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\frac{3}{9}=\frac{1}{3}\)
Ta thấy: \(\frac{1}{3}\)không là SCP
\(\Rightarrow x=\varnothing\)
Vậy không tìm được x thỏa mãn
\(\frac{24}{9}-\left(x+\frac{1}{2}\right)^2=2\frac{1}{3}\)
\(\frac{8}{3}-\left(x+\frac{1}{2}\right)^2=\frac{7}{3}\)
\(\left(x+\frac{1}{2}\right)^2=\frac{8}{3}-\frac{7}{3}=\frac{1}{3}\)
Mà \(\frac{1}{3}\)không là số chính phương
=> không có x nào thỏa mãn bài trên