Giải các phương trình:
\(a,x\left(x+1\right)\left(x^2+x+1\right)=42\)
\(b,x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\)
Giải các phương trình sau:
a \(\left(x+2\right)\left(x+\text{4}\right)\left(x+6\right)\left(x+8\right)+16=0\)
b \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)
c \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4=0\)
d \(\left(x^2-3x+2\right)\left(x^2+15x+56\right)+8=0\)
b: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24=0\)
\(\Leftrightarrow x^2+7x+6=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)
Giải các phương trình sau:
a, \(\left(x^2+x+1\right)^2=3\left(x^4+x^2+1\right)\)
b, \(x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)
Ta có : \(x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2-x+2x-2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)
\(\Leftrightarrow\left(x^2+x-1+1\right)\left(x^2+x-1-1\right)=24\)
\(\Leftrightarrow\left(x^2+x-1\right)^2-1=24\)
\(\Leftrightarrow\left(x^2+x-1\right)^2=25\)
<=> 2 trường hợp sảy ra là bằng 5 hoặc -5 nhé
a,
\(x^4+2x^3+3x^2+2x+1-3x^4-3x^2-3=0.\)
\(-2x^4+2x^3+2x-2=0\)
\(x^4-x^3-x+1=0\) " chia cả 2 vế cho -2 )
\(x^3\left(x-1\right)-\left(x-1\right)=0\)
\(\left(x-1\right)\left(x^3-1\right)=0\)
\(\hept{\begin{cases}x=1\\x^3=1\end{cases}}\)
Giải các phương trình:
\(a,x\left(x+1\right)\left(x^2+x+1\right)=42\)
\(b,x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\)
Câu a:
\(x(x+1)(x^2+x+1)=42\)
\(\Leftrightarrow (x^2+x)(x^2+x+1)=42\)
Đặt \(x^2+x=a\) thì pt trở thành: \(a(a+1)=42\)
\(\Leftrightarrow a^2+a-42=0\Leftrightarrow (a-6)(a+7)=0\)
\(\Rightarrow \left[\begin{matrix} a=6\\ a=-7\end{matrix}\right.\)
Nếu $a=6$ \(\leftrightarrow x^2+x=6\leftrightarrow x^2+x-6=0\leftrightarrow (x-2)(x+3)=0\)
\(\Rightarrow \left[\begin{matrix} x=2\\ x=-3\end{matrix}\right.\)
Nếu $a=-7$
\(\leftrightarrow x^2+x=-7\Leftrightarrow x^2+x+7=0\)
\(\Leftrightarrow (x+\frac{1}{2})^2+\frac{27}{4}=0\) (vô lý)
Vậy pt có nghiệm \(x=2\) hoặc $x=-3$
Câu b:
\(x(x+1)(x+2)(x+3)=24\)
\(\Leftrightarrow [x(x+3)][(x+1)(x+2)]=24\)
\(\Leftrightarrow (x^2+3x)(x^2+3x+2)=24\)
Đặt \(x^2+3x=a\) thì pt trở thành: \(a(a+2)=24\)
\(\Leftrightarrow a^2+2a+1=25\Leftrightarrow (a+1)^2=25\)
\(\Rightarrow a+1=\pm 5\Rightarrow \left[\begin{matrix} a=4\\ a=-6\end{matrix}\right.\)
Nếu $a=4$ \(\leftrightarrow x^2+3x=4\leftrightarrow x^2+3x-4=0\)
\(\leftrightarrow (x-1)(x+4)=0\Rightarrow x=1\) hoặc $x=-4$
Nếu \(a=-6\leftrightarrow x^2+3x=-6\leftrightarrow x^2+3x+6=0\leftrightarrow (x+\frac{3}{2})^2+\frac{15}{4}=0\)(vô lý)
Do đó pt có nghiệm $x=1$ hoặc $x=-4$
Giải các bất phương trình sau :
\(a.4\left(x-3\right)^2-\left(2x-1\right)^2\ge12\)
\(b.\left(x-4\right)\left(x+4\right)\ge\left(x+3\right)^2+5\)
c. \(\left(3x-1\right)^2-9\left(x+2\right)\left(x-2\right)< 5x\)
\(a,4\left(x-3\right)^2-\left(2x-1\right)^2\ge12\)
\(\Leftrightarrow4x^2-24x+36-4x^2-4x+1\ge12\)
\(\Leftrightarrow-28x+37\ge12\)
\(\Leftrightarrow-28x\ge12-37\)
\(\Leftrightarrow-28x\ge-25\)
\(\Leftrightarrow x\le\dfrac{25}{28}\)
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 \(S=\left\{x\left|x>\dfrac{37}{11}\right|\right\}\)
Bài 3: Giải các hệ phương trình sau:
a)\(\left\{{}\begin{matrix}2\left(x-2\right)+3\left(1+y\right)=-2\\3\left(x-2\right)-2\left(1+y\right)=-3\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
a: =>2x-4+3+3y=-2 và 3x-6-2-2y=-3
=>2x+3y=-2+4-3=2-3=-1 và 3x-2y=-3+6+2=5
=>x=1; y=-1
b: =>x^2-x+xy-y=x^2+x-xy-y+2xy
=>-x-y=x-y và y^2+y-yx-x=y^2-2y+xy-2x-2xy
=>x=0 và y-x=-2y-2x
=>x=0 và y=0
1. giải phương trình tích:
a) \(\left(x+3\right)\left(x^2+2021\right)=0\)
\(\)2. giải các phương trình sau bằng cách đưa về phương trình tích:
b) \(x\left(x-3\right)+3\left(x-3\right)=0\)
c) \(\left(x^2-9\right)+\left(x+3\right)\left(3-2x\right)=0\)
d) \(3x^2+3x=0\)
e) \(x^2-4x+4=4\)
`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`
`=>` $\left[ \begin{array}{l}x=0\\x=4\end{array} \right.$
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}
Bài 1:
a) Ta có: \(\left(x+3\right)\left(x^2+2021\right)=0\)
mà \(x^2+2021>0\forall x\)
nên x+3=0
hay x=-3
Vậy: S={-3}
Bài 2:
b) Ta có: \(x\left(x-3\right)+3\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Vậy: S={3;-3}
Giải các phương trình:
a) (\(x^2-4x\))\(^2\) = 4\(\left(x^2-4x\right)\)
b) \(\left(x+2\right)^2-x+1=\left(x-1\right)\left(x+1\right)\)
a) (x2 - 4x)2 = 4(x2 - 4x)
<=> (x2 - 4x)(x2 - 4x - 4) = 0
<=> x(x - 4)(x2 - 4x - 4) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\\left(x-2\right)^2=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=\pm\sqrt{8}+2\end{matrix}\right.\)
b) (x + 2)2 - x + 1 = (x - 1)(x + 1)
<=> x2 + 4x + 4 - x + 1 = x2 - 1
<=> 3x + 5 = -1
<=> x = -2
giải các phương trình sau:
a)\(3\left(x^2+x\right)^2-7\left(x^2+x\right)+4=0\)0
b)\(x\left(x+1\right)\left(x^2+x+1\right)=42\)
c)\(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
d)\(\left(5x^2-2x+10\right)^2=\left(3x^2+10x-8\right)^2\)
a) đặt \(\left(x^2+x\right)\)là \(y\)
ta có: \(3y^2-7y+4\)\(=0\)
<=>\(\left(3y-4\right)\left(y-1\right)=0\)
còn lại bạn tự xử nhé