b: =>|x+2|+|2x-1|<x+1(1)

Trường hợp 1: x<-2

(1) sẽ là -x-2-2x+1<x+1

=>-3x-1<x+1

=>-4x<2

hay x>-1/2(loại)

Trường hợp 2: -2<=x<1/2

(1) sẽ là x+2+1-2x<x+1

=>-x+3<x+1

=>-2x<-2

hay x>1(loại)

Trường hợp 3: x>=1/2

(1) sẽ là x+2+2x-1<x+1

=>3x+1<x+1

=>x<0(loại)

Vậy: BPT vô nghiệm

b: =>|x+2|+|2x-1|<x+1(1)

Trường hợp 1: x<-2

(1) sẽ là -x-2-2x+1<x+1

=>-3x-1<x+1

=>-4x<2

hay x>-1/2(loại)

Trường hợp 2: -2<=x<1/2

(1) sẽ là x+2+1-2x<x+1

=>-x+3<x+1

=>-2x<-2

hay x>1(loại)

Trường hợp 3: x>=1/2

(1) sẽ là x+2+2x-1<x+1

=>3x+1<x+1

=>x<0(loại)

Vậy: BPT vô nghiệm

b: =>|x+2|+|2x-1|<x+1(1)

Trường hợp 1: x<-2

(1) sẽ là -x-2-2x+1<x+1

=>-3x-1<x+1

=>-4x<2

hay x>-1/2(loại)

Trường hợp 2: -2<=x<1/2

(1) sẽ là x+2+1-2x<x+1

=>-x+3<x+1

=>-2x<-2

hay x>1(loại)

Trường hợp 3: x>=1/2

(1) sẽ là x+2+2x-1<x+1

=>3x+1<x+1

=>x<0(loại)

Vậy: BPT vô nghiệm

giống Nguyễn Lê Phước Thịnh nhé

\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)

\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)

\(2.\left(x-2\right)\left(2x-1\right)=x^2-2x.\Leftrightarrow\left(x-2\right)\left(2x-1\right)-x\left(x-2\right)=0.\)

\(\Leftrightarrow\left(x-2\right)\left(2x-1-x\right)=0.\Leftrightarrow\left(x-2\right)\left(x-1\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=1.\end{matrix}\right.\)

\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)

\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)

\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)

\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)

\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)

\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)

\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)

\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)

\(Đk:\) \(x\ne1,x\ne2,x\ne3\)

\(\Rightarrow\dfrac{x+4}{\left(x-2\right)\left(x-1\right)}+\dfrac{x+1}{\left(x-3\right)\left(x-1\right)}=\dfrac{2x+5}{\left(x-3\right)\left(x-1\right)}\)

\(\Rightarrow\dfrac{\left(x+4\right)\cdot\left(x-3\right)+\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)\left(x-3\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(x-3\right)\left(x-1\right)\left(x-2\right)}\)

\(\Rightarrow x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)

\(\Rightarrow0x-14=x-10\)

\(\Rightarrow x=-4\left(tmđk\right)\)

\(\dfrac{2x-1}{x+1}-2< 0.\left(x\ne-1\right).\\ \Leftrightarrow\dfrac{2x-1-2x-2}{x+1}< 0.\Leftrightarrow\dfrac{-3}{x+1}< 0.\)

Mà \(-3< 0.\)

\(\Rightarrow x+1>0.\Leftrightarrow x>-1\left(TMĐK\right).\)

\(\dfrac{x^2-2x+5}{x-2}-x+1\ge0.\left(x\ne2\right).\\ \Leftrightarrow\dfrac{x^2-2x+5-x^2+2x+x-2}{x-2}\ge0.\\ \Leftrightarrow\dfrac{x+3}{x-2}\ge0.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3\ge0.\\x-2\ge0.\end{matrix}\right.\\\left\{{}\begin{matrix}x+3\le0.\\x-2\le0.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-3.\\x\ge2.\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-3.\\x\le2.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge2.\\x\le-3.\end{matrix}\right.\)

Kết hợp ĐKXĐ.

\(\Rightarrow\left[{}\begin{matrix}x>2.\\x\le-3.\end{matrix}\right.\)

\(\dfrac{\left(1+2x\right)\left(x-2\right)}{\left(2x+3\right)\left(1-x\right)}\le0.\left(x\ne1;x\ne\dfrac{-3}{2}\right).\)

Đặt \(\dfrac{\left(1+2x\right)\left(x-2\right)}{\left(2x+3\right)\left(1-x\right)}=f\left(x\right).\)

Ta có bảng sau:

Vậy \(f\left(x\right)\ge0.\Leftrightarrow x\in\left(\dfrac{-3}{2};\dfrac{-1}{2}\right)\cup\)(1;2].

2) −(x2−2x+5)x−2−x+1≥0−(x2−2x+5)x−2−x+1≥0                                                        ⇔$-{}^{}$

Bài 2: 

a) Ta có: \(\Delta=\left(m-1\right)^2-4\cdot1\cdot\left(-m^2-2\right)\)
\(=m^2-2m+1+4m^2+8\)

\(=5m^2-2m+9>0\forall m\)

Do đó, phương trình luôn có hai nghiệm phân biệt với mọi m

Bài 1:

ĐKXĐ \(2x\ne y\)

Đặt \(\dfrac{1}{2x-y}=a;x+3y=b\)

HPT trở thành

\(\left\{{}\begin{matrix}a+b=\dfrac{3}{2}\\4a-5b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{3}{2}-b\\4\left(\dfrac{3}{2}-b\right)-5b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{3}{2}-b\\6-9b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{8}{9}\\a=\dfrac{11}{18}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+3y=\dfrac{8}{9}\\2x-y=\dfrac{18}{11}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=2x-\dfrac{18}{11}\\x+3\left(2x-\dfrac{18}{11}\right)=\dfrac{8}{9}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{82}{99}\\y=\dfrac{2}{99}\end{matrix}\right.\)

a: =>3x-9+5+10x=90

=>13x-4=90

=>13x=94

hay x=94/13

b: \(\Leftrightarrow2x-4-x-1=3x-11\)

=>3x-11=x-5

=>2x=6

hay x=3(nhận)

Mọi người ơi, giúp em với ạ!

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

Bo thi:>

+ đk x > 0 , x khác 1

Bài 1: Giải các phương trình sau:

a) 3(2,2-0,3x)=2,6 + (0,1x-4)

<=> 6.6 - 0.9x = 2,6 + 0,1x - 4

<=> - 0.9x - 0,1x = -6.6 -1,4

<=> -x = -8

<=> x = 8

Vậy x = 8

b) 3,6 -0,5 (2x+1) = x - 0,25(22-4x)

<=> 3,6 - x - 0,5 = x - 5,5 + x

<=> - x - 3,1 = -5,5

<=> - x = -2.4

<=> x = 2.4

Vậy  x = 2.4

=>0,2x+0,4-0,5x=0,25-0,5x+0,25

=>0,2x+0,4=0,5

=>0,2x=0,1

=>x=1/2