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TQ
2
H
25 tháng 9 2023
`(x+2)(x^2 -2x+4) -x(x^2-2)=15`
`<=> x^3 +8 - x^3 + 2x-15=0`
`<=> 2x-7=0`
`<=> 2x=7`
`<=>x=7/2`
__
`(x-4)^2 -(x-2)(x+2)=6`
`<=>x^2 - 8x+16- x^2 +4-6=0`
`<=> -8x+14=0`
`<=> -8x=-14`
`<=>x=14/8= 7/4`
__
`x^4 -2x^3 +x^2-2x=0`
`<=>x(x^3-2x^2+x-2)=0`
`<=> x(x^3+x-2x^2-2)=0`
`<=>x(x(x^2+1) -2(x^2+1))=0`
`<=> x(x^2+1)(x-2)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
H9
HT.Phong (9A5)
CTVHS
25 tháng 9 2023
a) \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)
\(\Leftrightarrow\left(x^3+2^3\right)-\left(x^3-2x\right)=15\)
\(\Leftrightarrow x^3+8-x^3+2x=15\)
\(\Leftrightarrow2x+8=15\)
\(\Leftrightarrow2x=15-8\)
\(\Leftrightarrow2x=7\)
\(\Leftrightarrow x=\dfrac{7}{2}\)
b) \(\left(x-4\right)^2-\left(x+2\right)\left(x-2\right)=6\)
\(\Leftrightarrow x^2-8x+16-\left(x^2-4\right)=6\)
\(\Leftrightarrow x^2-8x+16-x^2+4=6\)
\(\Leftrightarrow-8x+20=6\)
\(\Leftrightarrow-8x=6-20\)
\(\Leftrightarrow-8x=-14\)
\(\Leftrightarrow x=\dfrac{7}{4}\)
c) \(x^4-2x^3+x^2-2x=0\)
\(\Leftrightarrow x^3\left(x-2\right)+x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^3+x\right)\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x^2+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
DD
2
GH
27 tháng 7 2023
a
\(x^2\left(2x+15\right)+4\left(2x+15\right)=0\\ \Leftrightarrow\left(2x+15\right)\left(x^2+4\right)=0\\ \Leftrightarrow2x+15=0\left(x^2+4>0\forall x\right)\\ \Leftrightarrow2x=-15\\ \Leftrightarrow x=-\dfrac{15}{2}\)
b
\(5x\left(x-2\right)-3\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\5x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0+2=2\\x=\dfrac{0+3}{5}=\dfrac{3}{5}\end{matrix}\right.\)
c
\(2\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow2\left(x+3\right)-\left(x^2+3x\right)=0\\ \Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\2-x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0-3=-3\\x=2-0=2\end{matrix}\right.\)
27 tháng 7 2023
a: =>(2x+15)(x^2+4)=0
=>2x+15=0
=>2x=-15
=>x=-15/2
b; =>(x-2)(5x-3)=0
=>x=2 hoặc x=3/5
c: =>(x+3)(2-x)=0
=>x=2 hoặc x=-3
22 tháng 9 2021
a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow2x=-7\)
hay \(x=-\dfrac{7}{2}\)
b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)
\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)
\(\Leftrightarrow24x=-13\)
hay \(x=-\dfrac{13}{24}\)
CM
24 tháng 9 2018
a) Rút gọn được VT = 9x + 7. Từ đó tìm được x = 1.
b) Rút gọn được VT = 2x + 8. Từ đó tìm được x = 7 2 .
VK
1
8 tháng 9 2019
a) \(x^3-x=0\)
\(\Leftrightarrow x\left(x^2-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x-1=0\\x+1=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x=1\\x=-1\end{cases}}\)
Vậy \(x\in\left\{0;1;-1\right\}\)
b) \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow8-2x=15\)
\(\Leftrightarrow2x=-7\)
\(\Leftrightarrow x=\frac{-7}{2}\)
Vậy \(x=\frac{-7}{2}\)
HH
7 tháng 8 2017
a)
<=> 10x - 35 + 16x - 10 = 5
<=> 10x + 16x = 5 + 35 + 10
<=> 26x = 50
<=> x = 50/26 = 25/13
DS
1
23 tháng 10 2018
a) \(\left(2x+1\right)^2-4\left(x+2\right)^2=9\)
\(\left(2x+1\right)^2-\left[2\left(x+2\right)\right]^2=9\)
\(\left[2x+1-2\left(x+2\right)\right]\left[2x+1+2\left(x+2\right)\right]=9\)
\(\left(2x+1-2x-4\right)\left(2x+1+2x+4\right)=9\)
\(-3\left(4x+5\right)=9\)
\(4x+5=-3\)
\(4x=-8\)
\(x=-2\)
b) \(x^2-2x-15=0\)
\(x^2-5x+3x-15=0\)
\(x\left(x-5\right)+3\left(x-5\right)=0\)
\(\left(x-5\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}}\)
c) \(2x^2+3x-5=0\)
\(2x^2-2x+5x-5=0\)
\(2x\left(x-1\right)+5\left(x-1\right)=0\)
\(\left(x-1\right)\left(2x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-5}{2}\end{cases}}}\)
TN
10 tháng 7 2018
\(12\left(x-2\right)\left(x+2\right)-3\left(2x+3\right)^2\) \(=52\)
\(12\left(x^2-4\right)-3\left(4x^2+12x+9\right)\) \(=52\)
\(12x^2-48-12x^2-36x-27\) \(=52\)
\(-36x-75=52\)
\(-36x=127\)
\(x=\frac{-127}{36}\)
\(\left(2x+1\right)^2-4\left(x-1\right)\left(x+1\right)\) \(+2x=5\)
\(4x^2+4x+1-4\left(x^2-1\right)\) \(+2x=5\)
\(4x^2+4x-1-4x^2+4+2x=5\)
\(6x+3=5\)
\(6x=2\)
\(x=3\)
\(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)\) \(+6\left(x-1\right)^2=15\)
\(x^3-6x^2+12x-8-\left(x-3\right)\left(x+3\right)^2\) \(+6\left(x^2-2x+1\right)=15\)
\(x^3-6x^2+12x-8-\left(x^2-9\right)\left(x+3\right)\) \(+6x^2-12x+6=15\)
\(x^3-2\) \(-\left(x^3+3x^2-9x-27\right)\)\(=15\)
\(x^3-2-x^3-3x^2+9x+27=15\)
\(-3x^2+9x+25=15\)
\(-3x^2+9x+10=0\)
\(-3\left(x^2-3x-\frac{10}{3}\right)\) \(=0\)
\(x=\frac{9+\sqrt{201}}{6}\)
các câu còn lại tương tự
\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)
\(\Leftrightarrow x^3+8-x^3+2x=15\)
\(\Leftrightarrow2x+8=15\)
\(\Leftrightarrow2x=15-8=7\)
\(\Leftrightarrow x=\frac{7}{2}\)
\(\left(x+2\right)\left(x^2-2x+4\right)-x.\left(x^2-2\right)=15\)