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|x-9|=2x+5
Xét 3 TH
TH1: x>9 => x-9=2x+5 =>-9-5=x =>x=-14 (L)
TH2: x<9 => 9-x=2x+5 => 9-5=3x =>x=4/3(t/m)
TH3: x=9 =>0=23(L)
Vậy x= 4/3
Ta có:\(\dfrac{1-2x}{4}-2\le\dfrac{1-5x}{8}+x\\ \)
\(\dfrac{2-4x-16}{8}\le\dfrac{1-5x+8x}{8}\)
\(-4x-14\le1+3x\\ \Leftrightarrow7x+15\ge0\\ \Leftrightarrow x\ge-\dfrac{15}{7}\)
ta có: x4-4x3-2x2+12x+9 < x4-4x3-2x2+15x-3
=> x4-4x3-2x2+15x-3 - (x4-4x3-2x2+12x+9) > 0
=> 3x+6>0
(đề bài có cho điều kiện của x thì chứng minh 3x+6>0 là xong ạ)
Ta có: \(\left(x^2-2x-3\right)^2< x^2\left(x^2-4x-2\right)+3\left(5x-1\right)\)
\(\Leftrightarrow x^4+4x^2+9-4x^3-6x^2+12x< x^4-4x^3-2x^2+15x-3\)
\(\Leftrightarrow3x-12>0\)
\(\Leftrightarrow x-4>0\Rightarrow x>4\)
Vậy x > 4
\(x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
b.\(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
\(ĐK:x\ne\pm2\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)-5\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{12+\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)-5\left(x-2\right)=12+\left(x^2-4\right)\)
\(\Leftrightarrow x^2+3x+2-5x+10=12+x^2-4\)
\(\Leftrightarrow-2x=-4\)
\(\Leftrightarrow x=2\left(ktm\right)\)
Vậy pt vô nghiệm
a)
<=> x (x-2 ) = 0
<=> x =0
x = 2
b)
đkxđ : x khác 2 , x khác -2
<=> \(\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{12}{x^2-4}+\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=0\)
<=> \(\dfrac{x^2+3x+2}{....}-\dfrac{5x-10}{....}-\dfrac{12}{...}+\dfrac{x^2-4}{....}=0\)
<=> \(x^2+3x+2-5x+10-12+x^2-4=0\)
<=> \(2x^2-2x-4=0\)
<=> x =2 (ktm)
Vậy..
<=> \(\dfrac{x+2}{x-2}\)-\(\dfrac{1}{x}\)=\(\dfrac{2}{x\left(x-2\right)}\)
<=> \(\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)
ok, ở đây đã có mẫu chung rồi, em cứ vậy làm tiếp thôi :D
\(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x^2-2x}\) (ĐKXĐ: \(x\ne0;x\ne2\))
\(\Leftrightarrow x\left(x+2\right)-\left(x-2\right)=2\)
\(\Leftrightarrow x^2+2x-x+2=2\)
\(\Leftrightarrow x^2+x+2-2=0\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
\(\Rightarrow S=\left\{-1\right\}\)
a) \(\frac{2-x}{3}< \frac{3-2x}{5}\)
<=> \(10-5x< 9-6x\)
<=> x < - 1
Vậy S = { x| x < -1 }
b)
0 -1
câu a bài 1:(2x+1)(3x-2)=(5x-8)(2x+1)
<=>(2x+1)(3x-2)-(5x-8)(2x+1)=0
<=>(2x+1)(3x-2-5x+8)=0
<=>(2x+1)(6-2x)=0
bước sau tự làm nốt nha !
câu b:gợi ý: tách 4x^2-1thành (2x-1)(2x+1) rồi làm như câu a
Đặng Thị Vân Anh tuy mk k cần nx nhưng dù s cx cảm ơn bn nha :)
\(x^5+y^5-\left(x+y\right)^5\)
\(=x^5+y^5-\left(x^5+5x^4y+10x^3y^2+10x^2y^3+8xy^4+y^5\right)\)
\(=-5xy\left(x^3+2x^2y+2xy^2+y^3\right)\)
\(=-5xy\left[\left(x+y\right)\left(x^2-xy+y^2\right)+2xy\left(x+y\right)\right]\)
\(=-5xy\left(x+y\right)\left(x^2+xy+y^2\right)\)
\(\dfrac{x^2-2x-4}{x^2-2x-3}>1\)
\(\Leftrightarrow\dfrac{x^2-2x-4}{x^2-2x-3}-1>0\)
\(\Leftrightarrow\dfrac{x^2-2x-4-x^2+2x+3}{x^2-3x+x-3}>0\)
\(\Leftrightarrow\dfrac{-1}{\left(x-3\right)\left(x+1\right)}>0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3>0\\x+1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3< 0\\x+1>0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>3\\x< -1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 3\\x>-1\end{matrix}\right.\end{matrix}\right.\)
TH1 : vô lý
Vậy \(-1< x< 3\) thì \(\dfrac{x^2-2x-4}{x^2-2x-3}>1\)
\(\dfrac{x^2-2x-4}{x^2-2x-3}>1\)
\(\Leftrightarrow x^2-2x-4>x^2-2x-3\)
\(\Leftrightarrow x^2-x^2-2x+2x>-3+4\)
\(\Leftrightarrow0x>1\) (vô lí)
Vậy bpt vô nghiệm