Giải phương trình : \(\left|x^2-1\right|+\left|x\right|=1\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,f'\left(x\right)=3x^2-6x\\ f'\left(x\right)\le0\Leftrightarrow3x^2-6x\le0\\ \Leftrightarrow3x\left(x-2\right)\le0\Leftrightarrow0\le x\le2\)
Lời giải:
a. $f'(x)\leq 0$
$\Leftrightarrow 3x^2-6x\leq 0$
$\Leftrightarrow x(x-2)\leq 0$
$\Leftrightarrow 0\leq x\leq 2$
b.
$f'(x)=x^2-3x+2=0$
$\Leftrightarrow 3x^2-6x=x^2-3x+2=0$
$\Leftrightarrow 3x(x-2)=(x-1)(x-2)=0$
$\Leftrightarrow x-2=0$
$\Leftrightarrow x=2$
c.
$g(x)=f(1-2x)+x^2-x+2022$
$g'(x)=(1-2x)'f(1-2x)'_{1-2x}+2x-1$
$=-2[3(1-2x)^2-6(1-2x)]+2x-1$
$=-24x^2+2x+5$
$g'(x)\geq 0$
$\Leftrightarrow -24x^2+2x+5\geq 0$
$\Leftrightarrow (5-12x)(2x-1)\geq 0$
$\Leftrightarrow \frac{-5}{12}\leq x\leq \frac{1}{2}$
![](https://rs.olm.vn/images/avt/0.png?1311)
`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}
![](https://rs.olm.vn/images/avt/0.png?1311)
a:
ĐKXĐ: \(x\notin\left\{\dfrac{3}{2};1\right\}\)
\(y=\dfrac{\left(x-2\right)^2}{\left(2x-3\right)\left(x-1\right)}=\dfrac{x^2-4x+4}{2x^2-2x-3x+3}\)
=>\(y=\dfrac{x^2-4x+4}{2x^2-5x+3}\)
=>\(y'=\dfrac{\left(x^2-4x+4\right)'\left(2x^2-5x+3\right)-\left(x^2-4x+4\right)\left(2x^2-5x+3\right)'}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{\left(2x-4\right)\left(2x^2-5x+3\right)-\left(2x-5\right)\left(x^2-4x+4\right)}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{4x^3-10x^2+6x-8x^2+20x-12-2x^3+8x^2-8x+5x^2-20x+20}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{2x^3-5x^2-2x+8}{\left(2x^2-5x+3\right)^2}\)
b:
ĐKXĐ: x<>-3
\(y=\left(x+3\right)+\dfrac{4}{x+3}\)
=>\(y'=\left(x+3+\dfrac{4}{x+3}\right)'=1+\left(\dfrac{4}{x+3}\right)'\)
\(=1+\dfrac{4'\left(x+3\right)-4\left(x+3\right)'}{\left(x+3\right)^2}\)
=>\(y'=1+\dfrac{-4}{\left(x+3\right)^2}=\dfrac{\left(x+3\right)^2-4}{\left(x+3\right)^2}\)
y'=0
=>\(\left(x+3\right)^2-4=0\)
=>\(\left(x+3+2\right)\left(x+3-2\right)=0\)
=>(x+5)(x+1)=0
=>x=-5 hoặc x=-1
c:
ĐKXĐ: x<>-2
\(y=\dfrac{\left(5x-1\right)\left(x+1\right)}{x+2}\)
=>\(y=\dfrac{5x^2+5x-x-1}{x+2}=\dfrac{5x^2+4x-1}{x+2}\)
=>\(y'=\dfrac{\left(5x^2+4x-1\right)'\left(x+2\right)-\left(5x^2+4x-1\right)\left(x+2\right)'}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{\left(5x+4\right)\left(x+2\right)-\left(5x^2+4x-1\right)}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{5x^2+10x+4x+8-5x^2-4x+1}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{10x+9}{\left(x+2\right)^2}\)
\(y'\left(-1\right)=\dfrac{10\cdot\left(-1\right)+9}{\left(-1+2\right)^2}=\dfrac{-1}{1}=-1\)
d:
ĐKXĐ: x<>2
\(y=x-2+\dfrac{9}{x-2}\)
=>\(y'=\left(x-2+\dfrac{9}{x-2}\right)'=1+\left(\dfrac{9}{x-2}\right)'\)
\(=1+\dfrac{9'\left(x-2\right)-9\left(x-2\right)'}{\left(x-2\right)^2}\)
=>\(y'=1+\dfrac{-9}{\left(x-2\right)^2}=\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}\)
y'=0
=>\(\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}=0\)
=>\(\left(x-2\right)^2-9=0\)
=>(x-2-3)(x-2+3)=0
=>(x-5)(x+1)=0
=>x=5 hoặc x=-1
![](https://rs.olm.vn/images/avt/0.png?1311)
TH1: \(x\ge2\)
\(\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)=4\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)=4\)
\(\Leftrightarrow x^4-5x^2=0\Rightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-\sqrt{5}\left(loại\right)\\x=\sqrt{5}\end{matrix}\right.\)
TH2: \(x< 2\)
\(-\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)=4\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)=-4\)
\(\Leftrightarrow x^4-5x^2+8=0\)
\(\Leftrightarrow\left(x^2-\dfrac{5}{2}\right)^2+\dfrac{7}{4}=0\) (vô nghiệm)
Vậy \(x=\sqrt{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ĐK:x\ne0;x\ne1\\ PT\Leftrightarrow\left(\dfrac{1}{x}+2\right)\left(2+\dfrac{x+1}{x-1}-x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{x}=-2\\\dfrac{x+1}{x-1}=x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x+1=x^2-x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x^2-2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1+\sqrt{2}\\x=1-\sqrt{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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}
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(S=\left\{-2;\dfrac{1+\sqrt{41}}{2}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-2\right)\left(x-1\right)\left(x-4\right)\left(x-8\right)=4x^2\)
\(\Leftrightarrow[\left(x-2\right)\left(x-4\right)][\left(x-1\right)\left(x-8\right)]=4x^2\)
\(\Leftrightarrow\left(x^2-6x+8\right)\left(x^2-9x+8\right)=4x^2\)
thấy \(x=0;2\) không phải nghiệm của phương trình nên ta chia hai vế của pt cho \(x^2\) ta được \(:\)
\(\Leftrightarrow\left(x+\dfrac{8}{x}-9\right)\left(x+\dfrac{8}{x}-6\right)=4\)
\(Đặt:\) \(x+\dfrac{8}{x}=a\) thì pt trở thành \(:\)
\(\left(a-6\right)\left(a-9\right)=4\)
\(\Leftrightarrow a^2-15a+50=0\)
\(\Leftrightarrow\left(a-5\right)\left(a-10\right)=0\Leftrightarrow\left\{{}\begin{matrix}a=5\\a=10\end{matrix}\right.\)
\(Với\) \(a=5\) thì \(x+\dfrac{8}{x}=5\Leftrightarrow x^2-5x+8=0\left(vônghiem\right)\)
\(Với\) \(a=10\) thì \(x+\dfrac{8}{x}=10\Leftrightarrow x^2-10x+8=0\Leftrightarrow\left\{{}\begin{matrix}x=5-căn17\\x=5+căn17\end{matrix}\right.\)
\(Vậy...\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì $3x^2-x+1>0,x^2+1>0$
$\to \begin{cases}x^2 \geq 4\x<-1\\\end{cases}$
$\to \begin{cases}\left[ \begin{array}{l}x \geq 2\\x \leq -2\end{array} \right.\\x<-1\\\end{cases}$
$\to x \leq -2$
Vậy tập xác định của phương trình là `(-oo,-2]`
đung ko,em học lớp 6