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24 tháng 2 2021
`a,(x+3)(x^2+2021)=0`
`x^2+2021>=2021>0`
`=>x+3=0`
`=>x=-3`
`2,x(x-3)+3(x-3)=0`
`=>(x-3)(x+3)=0`
`=>x=+-3`
`b,x^2-9+(x+3)(3-2x)=0`
`=>(x-3)(x+3)+(x+3)(3-2x)=0`
`=>(x+3)(-x)=0`
`=>` $\left[ \begin{array}{l}x=0\\x=-3\end{array} \right.$
`d,3x^2+3x=0`
`=>3x(x+1)=0`
`=>` $\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.$
`e,x^2-4x+4=4`
`=>x^2-4x=0`
`=>x(x-4)=0`
`=>` $\left[ \begin{array}{l}x=0\\x=4\end{array} \right.$
24 tháng 2 2021
1) a) \(\left(x+3\right).\left(x^2+2021\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2021=0\end{matrix}\right.\\\left[{}\begin{matrix}x=-3\left(nhận\right)\\x^2=-2021\left(loại\right)\end{matrix}\right. \)
=> S={-3}
22 tháng 3 2021
ĐKXĐ: \(x\notin\left\{-1;-2;-3;-4\right\}\)
Ta có: \(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+4}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{x+4}{\left(x+1\right)\left(x+4\right)}-\dfrac{x+1}{\left(x+1\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{x+4-x-1}{\left(x+1\right)\left(x+4\right)}=\dfrac{x^2+5x+4}{6\left(x+1\right)\left(x+4\right)}\)
\(\Leftrightarrow\dfrac{18}{6\left(x+1\right)\left(x+4\right)}=\dfrac{x^2+5x+4}{6\left(x+1\right)\left(x+4\right)}\)
Suy ra: \(x^2+5x+4=18\)
\(\Leftrightarrow x^2+5x-14=0\)
\(\Leftrightarrow x^2+7x-2x-14=0\)
\(\Leftrightarrow x\left(x+7\right)-2\left(x+7\right)=0\)
\(\Leftrightarrow\left(x+7\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)
Vậy: S={-7;2}
LD
31 tháng 3 2022
\(a,4\left(x-3\right)^2-\left(2x-1\right)^2\ge12\)
\(\Leftrightarrow4x^2-24x+36-4x^2-4x+1\ge12\)
\(\Leftrightarrow-28x+37\ge12\)
\(\Leftrightarrow-28x\ge12-37\)
\(\Leftrightarrow-28x\ge-25\)
\(\Leftrightarrow x\le\dfrac{25}{28}\)
Vậy \(S=\left\{x\left|x\le\dfrac{25}{28}\right|\right\}\)
b, \(\left(x-4\right)\left(x+4\right)\ge\left(x+3\right)^2+5\)
\(\Leftrightarrow x^2-16\ge x^2+6x+9+5\)
\(\Leftrightarrow x^2-x^2-6x\ge9+5+16\)
\(\Leftrightarrow-6x\ge30\)
\(\Leftrightarrow x\le-5\)
Vậy \(S=\left\{x\left|x\le-5\right|\right\}\)
\(c,\left(3x-1\right)^2-9\left(x+2\right)\left(x-2\right)< 5x\)
\(\Leftrightarrow9x^2-6x-1-9x^2+36< 5x\)
\(\Leftrightarrow9x^2-9x^2-6x-5x+36+1< 0\)
\(\Leftrightarrow-11x+37< 0\)
\(\Leftrightarrow-11x< -37\)
\(\Leftrightarrow x>\dfrac{37}{11}\)
vậy \(S=\left\{x\left|x>\dfrac{37}{11}\right|\right\}\)
XO
1 tháng 2 2023
1) |x| + x2 - x = x + 10 (1)
Nếu x < 0 thì
|x| = - x
Khi đó (1) <=> x2 - 3x - 10 = 0
Có \(\Delta=\left(-3\right)^2-4.\left(-10\right).1=49>0\)
=> Phương trình 2 nghiệm : \(x_1=\dfrac{3+\sqrt{49}}{2}=5\left(\text{loại}\right);x_2=\dfrac{3-\sqrt{49}}{2}=-2\)
Nếu \(x\ge0\Leftrightarrow\left|x\right|=x\)
Phương trình (1) <=> x2 - x - 10 = 0
\(\Delta=\left(-1\right)^2-4.\left(-10\right).1=41>0\)
=> Phương trình 2 nghiệm \(x_1=\dfrac{1+\sqrt{41}}{2};x_2=\dfrac{1-\sqrt{41}}{2}\left(\text{loại}\right)\)
Vậy tập nghiệm phương trình \(S=\left\{-2;\dfrac{1+\sqrt{41}}{2}\right\}\)
TD
3 tháng 2 2022
f. 5 – (x – 6) = 4(3 – 2x)
<=>5-x+6=12-8x
<=>7x=1
<=>x=\(\dfrac{1}{7}\)
g. 7 – (2x + 4) = – (x + 4)
<=>7-2x-4=-x-4
<=>x=7
h. 2x(x+2)\(^2\)−8x\(^2\)=2(x−2)(x\(^2\)+2x+4)
<=>\(2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)
<=>\(2x^3+8x^2+8x-8x^2=2\left(x^3-8\right)\)
<=>\(2x^3+8x=2x^3-16\)
<=>\(8x=-16\)
<=>\(x=-2\)
i. (x−2\(^3\))+(3x−1)(3x+1)=(x+1)\(^3\)
<=>\(x-8+9x^2-1=x^3+3x^2+3x+1\)
<=>\(6x^2-2x-10=0\)
<=>\(3x^2-x-5=0\)
<=>\(\left[{}\begin{matrix}x=\dfrac{1+\sqrt{61}}{6}\\x=\dfrac{1-\sqrt{61}}{6}\end{matrix}\right.\)
k. (x + 1)(2x – 3) = (2x – 1)(x + 5)
<=>\(2x^2-x-3=2x^2+9x-5\)
<=>10x=2
<=>\(x=\dfrac{1}{5}\)
TD
3 tháng 2 2022
f. 5 – (x – 6) = 4(3 – 2x)
<=>5-x+6=12-8x
<=>7x=1
<=>x=\(\dfrac{1}{7}\)
g. 7 – (2x + 4) = – (x + 4)
<=>7-2x-4=-x-4
<=>x=7
h. \(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)
<=>\(2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)
<=>\(2x^3+8x^2+8x-8x^2=2x^3-16\)
<=>\(8x=-16\)
<=>x=-2
i.\(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)
<=>\(x^3-6x^2+12x+8+9x^2-1=x^3+3x^2+3x+1\)
<=>\(9x+6=0\)
<=>x=\(\dfrac{-2}{3}\)
k. (x + 1)(2x – 3) = (2x – 1)(x + 5)
<=>\(2x^2-x-3=2x^2+9x-5\)
<=>10x=2
<=>x=\(\dfrac{1}{5}\)
LT
0
BC
1
3 tháng 1 2021
Nhận thấy x = 0 không là nghiệm của phương trình.
Xét x \(\neq\) 0.
Chia cả hai vế cảu pt cho x2 ta được:
\(\left(x+1+\dfrac{1}{x}\right)^2=3\left(x^2+1+\dfrac{1}{x^2}\right)\). (*)
Đặt \(x+\frac{1}{x}=t\).
\((*)\Leftrightarrow (a+1)^2=3(a^2-1)\)
\(\Leftrightarrow a^2-a-2=0\Leftrightarrow\left(a+1\right)\left(a-2\right)=0\Leftrightarrow\left[{}\begin{matrix}a=-1\\a=2\end{matrix}\right.\).
+) \(a=-1\Leftrightarrow x+\dfrac{1}{x}=-1\Leftrightarrow x^2+x+1=0\) (vô nghiệm).
+) \(a=2\Leftrightarrow x+\dfrac{1}{x}=2\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\) (thoả mãn \(x\neq 0\)).
Vậy...
- Ta có: \(\left|x-3\right|^2+\left|x-4\right|^3=1\)( ** )
* Với \(x\le3\)thay vào phương trình ( ** ), ta có:
\(\left(3-x\right)^2+\left(4-x\right)^3=1\)
\(\Leftrightarrow9-6x+x^2+64+12x^2-48x-x^3=1\)
\(\Leftrightarrow-x^3+13x^2-54x+72=0\)
\(\Leftrightarrow x^3-13x^2+54x-72=0\)
\(\Leftrightarrow\left(x^3-3x^2\right)-\left(10x^2-30x\right)+\left(24x-72\right)=0\)
\(\Leftrightarrow x^2.\left(x-3\right)-10x.\left(x-3\right)+24.\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left[\left(x^2-4x\right)-\left(6x-24\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right).\left[x.\left(x-4\right)-6.\left(x-4\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right).\left(x-4\right).\left(x-6\right)=0\)
\(\Leftrightarrow\)\(x-3=0\)hoặc \(x-4=0\)hoặc \(x-6=0\)
+ \(x-3=0\)\(\Leftrightarrow\)\(x=3\left(TM\right)\)
+ \(x-4=0\)\(\Leftrightarrow\)\(x=4\left(L\right)\)
+ \(x-6=0\)\(\Leftrightarrow\)\(x=6\left(L\right)\)
* Với \(3< x\le4\)thay vào phương trình ( ** ), ta có:
\(\left(x-3\right)^2+\left(4-x\right)^3=1\)
\(\Leftrightarrow x^2-6x+9+64+12x^2-48x-x^3=1\)
\(\Leftrightarrow-x^3+13x^2-54x+72=0\)
\(\Leftrightarrow x^3-13x^2+54x-72=0\)
\(\Leftrightarrow\left(x^3-3x^2\right)-\left(10x^2-30x\right)+\left(24x-72\right)=0\)
\(\Leftrightarrow x^2.\left(x-3\right)-10x.\left(x-3\right)+24.\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left[\left(x^2-4x\right)-\left(6x-24\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right).\left[x.\left(x-4\right)-6.\left(x-4\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right).\left(x-4\right).\left(x-6\right)=0\)
\(\Leftrightarrow\)\(x-3=0\)hoặc \(x-4=0\)hoặc \(x-6=0\)
+ \(x-3=0\)\(\Leftrightarrow\)\(x=3\left(L\right)\)
+ \(x-4=0\)\(\Leftrightarrow\)\(x=4\left(TM\right)\)
+ \(x-6=0\)\(\Leftrightarrow\)\(x=6\left(L\right)\)
* Với \(x>4\)thay vào phương trình ( ** ), ta có:
\(\left(x-3\right)^2+\left(x-4\right)^3=1\)
\(\Leftrightarrow x^2-6x+9+x^3-12x^2+48x-64=1\)
\(\Leftrightarrow x^3-11x^2+42x-56=1\)
\(\Leftrightarrow\left(x^3-4x^2\right)-\left(7x^2-28x\right)+\left(14x-56\right)=1\)
\(\Leftrightarrow x^2.\left(x-4\right)-7x.\left(x-4\right)+14.\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right).\left(x^2-7x+14\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x^2-7x+14=0\end{cases}}\)
+ \(x-4=0\)\(\Leftrightarrow\)\(x=4\left(L\right)\)
+ \(x^2-7x+14=0\)\(\Leftrightarrow\)\(4x^2-28x+56=0\)
\(\Leftrightarrow\left(4x^2-28x+49\right)+7=0\)
\(\Leftrightarrow\left(2x-7\right)^2+7=0\)
- Vì \(\left(2x-7\right)^2\ge0\forall x\)\(\Rightarrow\)\(\left(2x-7\right)^2+7\ge7>0\forall x\)mà \(\left(2x-7\right)^2+7=0\)
\(\Rightarrow\)\(\left(2x-7\right)^2+7=0\)( vô nghiệm )
Vậy \(S=\left\{3,4\right\}\)
\(\left|x-3\right|^2+\left|x-4\right|^3=1\)
<=> \(\left(x-3\right)^2+\left|x-4\right|^3=1\)
Theo mình nghĩ bài này chỉ cần xét 2 trường hợp thôi!