\(x^4-3x^3-x^2+2x-4=0\)

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AH
Akai Haruma
Giáo viên
23 tháng 9 2020

Lời giải:

Ta sẽ thử phân tích $x^4-3x^3-x^2+2x-4$ thành nhân tử

Đặt $x^4-3x^3-x^2+2x-4=(x^2+ax+b)(x^2+cx+d)$ với $a,b,c,d$ nguyên.

$\Leftrightarrow x^4-3x^3-x^2+2x-4=x^4+x^3(a+c)+x^2(ac+b+d)+x(ad+bc)+bd$

Đồng nhất hệ số:

\(\left\{\begin{matrix} a+c=-3\\ ac+b+d=-1\\ ad+bc=2\\ bd=-4\end{matrix}\right.\). Từ $bd=-4$ ta xét các TH nguyên của $b,d$ để thay vào tìm $a,c$

Ta tìm được $a=-2;b=-4; c=-1; d=1$

Do đó:

$x^4-3x^3-x^2+2x-4=0$

$\Leftrightarrow (x^2-2x-4)(x^2-x+1)=0$

$\Leftrightarrow x^2-2x-4=0$ (do $x^2-x+1\neq 0$)

$\Leftrightarrow x=1\pm \sqrt{5}$

NV
22 tháng 10 2019

a/ ĐKXĐ: \(0\le x\le4\)

\(\left(x^2-4x\right)\sqrt{-x^2+4x}+x^2-4x+2=0\)

Đặt \(\sqrt{-x^2+4x}=a\ge0\)

\(-a^2.a-a^2+2=0\)

\(\Leftrightarrow a^3+a^2-2=0\)

\(\Leftrightarrow\left(a-1\right)\left(a^2+2a+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a^2+2a+2=0\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{-x^2+4x}=1\Leftrightarrow x^2-4x+1=0\Rightarrow...\)

b/ \(x^4+2x^2+x\sqrt{2x^2+4}-4=0\)

Đặt \(x\sqrt{2x^2+4}=a\Rightarrow x^2\left(2x^2+4\right)=a^2\Rightarrow x^4+2x^2=\frac{a^2}{2}\)

\(\frac{a^2}{2}+a-4=0\Leftrightarrow a^2+2a-8=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x\sqrt{2x^2+4}=2\left(x>0\right)\\x\sqrt{2x^2+4}=-4\left(x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^4+4x^2=4\\2x^4+4x^2=16\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2=\sqrt{3}-1\\x^2=-\sqrt{3}-1\left(l\right)\\x^2=2\\x^2=-4\left(l\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\sqrt{\sqrt{3}-1}\\x=-\sqrt{2}\end{matrix}\right.\)

NV
22 tháng 10 2019

c/ Đặt \(\sqrt[3]{2x^2+3x-10}=a\Rightarrow2x^2+3x=a^3+10\)

\(a^3+10-14=2a\)

\(\Leftrightarrow a^3-2a-4=0\)

\(\Leftrightarrow\left(a-2\right)\left(a^2+2a+2\right)=0\Rightarrow a=2\)

\(\Rightarrow\sqrt[3]{2x^2+3x-10}=2\Rightarrow2x^2+3x-18=0\Rightarrow...\)

d/ \(\Leftrightarrow2\left(3x^2+x+4\right)+\sqrt[3]{3x^2+x+4}-18=0\)

Đặt \(\sqrt[3]{3x^2+x+4}=a\)

\(2a^3+a-18=0\)

\(\Leftrightarrow\left(a-2\right)\left(2a^2+4a+9\right)=0\Rightarrow a=2\)

\(\Rightarrow\sqrt[3]{3x^2+x+4}=2\Rightarrow3x^2+x-4=0\Rightarrow...\)

e/ \(\Leftrightarrow x^2+5x+2-3\sqrt{x^2+5x+2}-2=0\)

Đặt \(\sqrt{x^2+5x+2}=a\ge0\)

\(a^2-3a-2=0\Rightarrow\left[{}\begin{matrix}a=\frac{3+\sqrt{17}}{2}\\a=\frac{3-\sqrt{17}}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2+5x+2}=\frac{3+\sqrt{17}}{2}\Rightarrow x^2+5x-\frac{9+3\sqrt{17}}{2}=0\)

Bài cuối xấu quá, chắc nhầm số liệu

16 tháng 3 2020

\(21,\frac{2}{x-1}\le\frac{5}{2x-1}\left(x\ne1;x\ne\frac{1}{2}\right)\)

\(\Leftrightarrow\frac{2}{x-1}-\frac{5}{2x-1}\le0\)

\(\Leftrightarrow\frac{4x-2-5x+5}{\left(x-1\right)\left(2x-1\right)}\text{≤}0\)

\(\Leftrightarrow\frac{-x+3}{\left(x-1\right)\left(2x-1\right)}\text{≤}0\)

x -x+3 x-1 2x-1 VT -∞ +∞ 1/2 1 3 0 0 0 | | || | | || | | 0 - + + + + + - - - + + + + + + - -

Vậy \(\frac{-x+3}{\left(x-1\right)\left(2x-1\right)}\le0\Leftrightarrow x\in\left(\frac{1}{2};1\right)\cup[3;+\text{∞})\)

23,24 tương tự 21

\(25,2x^2-5x+2< 0\) (1)

Ta có: \(\left\{{}\begin{matrix}2x^2-5x+2=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{2}\end{matrix}\right.\\a=2>0\end{matrix}\right.\) \(\Leftrightarrow\frac{1}{2}< x< 2\)

\(26,-5x^2+4x+12< 0\)

\(\left\{{}\begin{matrix}-5x^2+4x+12=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\frac{6}{5}\end{matrix}\right.\\a=-5< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>2\\x< -\frac{6}{5}\end{matrix}\right.\)

\(27,16x^2+40x+25>0\)

\(\left\{{}\begin{matrix}16x^2+40x+25=0\Leftrightarrow x=-\frac{5}{4}\\a=16>0\end{matrix}\right.\)

\(\Leftrightarrow x\ne-\frac{5}{4}\)

\(28,-2x^2+3x-7\ge0\)

\(\left\{{}\begin{matrix}-2x^2+3x-7=0\left(vo.nghiem\right)\\a=-2< 0\end{matrix}\right.\)

\(\Rightarrow-2x^2+3x-7< 0\) ∀x

=> bpt vô nghiệm

\(29,3x^2-4x+4\ge0\)

\(\left\{{}\begin{matrix}3x^2-4x+4=0\left(vo.nghiem\right)\\a=3>0\end{matrix}\right.\)

=> \(3x^2-4x+4>0\) => bpt vô số nghiệm

\(30,x^2-x-6\le0\)

\(\left\{{}\begin{matrix}x^2-x-6=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\\a=1>0\end{matrix}\right.\)

\(\Rightarrow-2\le x\le3\)

NV
4 tháng 3 2020

a/

\(\frac{3x-4}{x-2}-1>0\Leftrightarrow\frac{2x-2}{x-2}>0\Rightarrow\left[{}\begin{matrix}x>2\\x< 1\end{matrix}\right.\)

b/

\(\frac{2x-5}{2-x}+1\le0\Rightarrow\frac{x-3}{2-x}\le0\Rightarrow\left[{}\begin{matrix}x\ge3\\x< 2\end{matrix}\right.\)

c/

\(\frac{x^2+x-3}{x^2-4}-1\le0\Rightarrow\frac{x+1}{x^2-4}\le0\Rightarrow\frac{x+1}{\left(x-2\right)\left(x+2\right)}\le0\Rightarrow\left[{}\begin{matrix}x< -2\\-1\le x< 2\end{matrix}\right.\)

d/

\(\frac{4x^2-8x+6+x^2-x-6}{2\left(x^2-x-6\right)}>0\Rightarrow\frac{x\left(5x-9\right)}{2\left(x+2\right)\left(x-3\right)}>0\Rightarrow\left[{}\begin{matrix}x>3\\0< x< \frac{9}{5}\\x< -2\end{matrix}\right.\)

e/

\(\frac{x^2+3x+2}{2x+3}-\frac{2x-5}{4}\ge0\Rightarrow\frac{4x^2+12x+8-\left(2x-5\right)\left(2x+3\right)}{4\left(2x+3\right)}\ge0\)

\(\Rightarrow\frac{28x+23}{4\left(2x+3\right)}\ge0\Rightarrow\left[{}\begin{matrix}x\ge-\frac{23}{28}\\x< -\frac{3}{2}\end{matrix}\right.\)

a: =>4x+12<=2x-1

=>2x<=-13

=>x<=-13/2

b: =>x^2-2x+1+4<0

=>(x-1)^2+4<0(loại)

c: =>(x-2+x+3)/(x+3)<0

=>(2x+1)/(x+3)<0

=>-3<x<-1/2