tìm x biết
b) \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)
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MT
30 tháng 8 2015
\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)
\(\Leftrightarrow\left(x-3\right)^3-\left(x-3\right)^3+6\left(x+1\right)^2+3x^2=-33\)
\(\Leftrightarrow6\left(x+1\right)^2+3x^2=-33\text{ vô lý }\left(\text{vì }6\left(x+1\right)^2\ge0;3x^2\ge0\right)\)
\(\text{Vậy không có x nào thỏa mãn}\)
26 tháng 7 2021
a)(x+3)3-x(3x+1)2+(2x+1)(4x2-2x+1-3x2)=54
\(\Rightarrow\)x3+9x2+27x+27-x(9x2+6x+1)+(2x+1)(x2-2x+1)=54
\(\Rightarrow\)x3+9x2+27x+27-9x3-6x2-x+2x3-4x2+2x+x2-2x+1=54
\(\Rightarrow\)-6x3+26x+28=54
\(\Rightarrow\)-6x3+26x=54-28
\(\Rightarrow\)-6x3+26x=26
\(\Rightarrow\)-6x3+26x-26=0
\(\Rightarrow\)-2(3x3+13x+14)
20 tháng 5 2022
a: \(\Leftrightarrow x^3-27-x\left(x^2-4\right)=1\)
\(\Leftrightarrow x^3-27-x^3+4x=1\)
=>4x-27=1
hay x=7
b: \(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6\left(x+1\right)^2+3x^2=15\)
\(\Leftrightarrow-9x^2+27x+6x^2+12x+6+3x^2=15\)
=>39x+6=15
hay x=3/13
c: \(\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2-12=2\)
\(\Leftrightarrow3x-40=2\)
hay x=14
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):}`
30 tháng 8 2023
6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)
Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)
Phương trình sẽ trở thành là: a^2+a-42=0
=>(a+7)(a-6)=0
=>a=-7(loại) hoặc a=6(nhận)
=>2x^2+3x+9=36
=>2x^2+3x-27=0
=>2x^2+9x-6x-27=0
=>(2x+9)(x-3)=0
=>x=3 hoặc x=-9/2
8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)
=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)
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: \(S=\left\{-\dfrac{7}{9}\right\}\)
13 tháng 10 2023
a: \(y'< 0\)
=>\(\left(x-3\right)^3\cdot\left(x-1\right)^{22}\cdot\left(-3x-6\right)^7< 0\)
=>\(\left(x-3\right)\left(-3x-6\right)< 0\)
=>\(\left(x+2\right)\left(x-3\right)>0\)
=>\(\left[{}\begin{matrix}x>3\\x< -2\end{matrix}\right.\)
y'>0
=>\(\left(x+2\right)\left(x-3\right)< 0\)
=>\(-2< x< 3\)
y'=0
=>\(\left[{}\begin{matrix}x-3=0\\x-1=0\\-3x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\\x=-2\end{matrix}\right.\)
Ta có bảng xét dấu sau:
| x | \(-\infty\) -2 1 3 +\(\infty\) |
| y' | - 0 + 0 + 0 - |
Vậy: Hàm số đồng biến trên các khoảng \(\left(-2;1\right);\left(1;3\right)\)
Hàm số nghịch biến trên các khoảng \(\left(-\infty;-2\right);\left(3;+\infty\right)\)
b: y'<0
=>\(\left(4x-3\right)^3\cdot\left(x^2-1\right)^{21}\left(3x-9\right)^7< 0\)
=>\(\left(4x-3\right)\left(3x-9\right)\left(x^2-1\right)< 0\)
=>\(\left(4x-3\right)\left(x-3\right)\left(x^2-1\right)< 0\)
TH1: \(\left\{{}\begin{matrix}\left(4x-3\right)\left(x-3\right)>0\\x^2-1< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>3\\x< \dfrac{3}{4}\end{matrix}\right.\\-1< x< 1\end{matrix}\right.\Leftrightarrow-1< x< \dfrac{3}{4}\)
TH2: \(\left\{{}\begin{matrix}\left(4x-3\right)\left(x-3\right)< 0\\x^2-1>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{4}< x< 3\\\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow1< x< 3\)
y'>0
=>\(\left(4x-3\right)\left(x-3\right)\left(x^2-1\right)>0\)
TH1: \(\left\{{}\begin{matrix}\left(4x-3\right)\left(x-3\right)>0\\x^2-1>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>3\\x< \dfrac{3}{4}\end{matrix}\right.\\\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -1\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}\left(4x-3\right)\left(x-3\right)< 0\\x^2-1< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{4}< x< 3\\-1< x< 1\end{matrix}\right.\Leftrightarrow\dfrac{3}{4}< x< 1\)
Ta sẽ có bảng xét dấu sau đây:
| x | \(-\infty\) -1 3/4 1 3 +\(\infty\) |
| y' | + 0 - 0 + 0 - 0 + |
Vậy: Hàm số đồng biến trên các khoảng \(\left(-\infty;-1\right);\left(\dfrac{3}{4};1\right);\left(3;+\infty\right)\)
Hàm số nghịch biến trên các khoảng \(\left(-1;\dfrac{3}{4}\right);\left(1;3\right)\)
\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)
<=> \(x^3-9x^2+27x-27\) \(-\left(x^3-3^3\right)+6\left(x^2+2x+1\right)+3x^2=-33\)
<=> \(x^3-9x^2+27x-27-x^3+27+6x^2+12x+6+3x^2=-33\)
<=> \(-6x^2+39x+6=-33\)
<=> \(6x^2-39x-6=33\)
<=> \(6x^2-39x-39=0\)
<=> \(6\left(x^2-\frac{39}{6}x-\frac{39}{6}\right)=0\)
<=> \(x^2-2.x.\frac{39}{12}+\frac{1521}{144}-\frac{273}{16}=0\)
<=> \(\left(x-\frac{39}{12}\right)^2-\frac{273}{16}=0\)
<=> \(\left(x-\frac{39}{12}-\frac{\sqrt{273}}{4}\right)\left(x-\frac{39}{12}+\frac{\sqrt{273}}{4}\right)=0\)
<=> \(\left(x-\frac{13+\sqrt{273}}{4}\right).\left(x-\frac{13-\sqrt{273}}{4}\right)=0\)
<=> \(x=\frac{13+\sqrt{273}}{4}\) ( h ) \(x=\frac{13-\sqrt{273}}{4}\)
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