a) 1\(\dfrac{2}{3}\).           b)\(\dfrac{1}{7}\).             c) 1               d )0

a: =>x+5>0

hay x>-5

b: =>2x+1<0

hay x<-1/2

c: =>(x-1)(x-4)>0

=>x>4 hoặc x<1

a) x>-5 ĐKXĐ x\(\ne\)-5

b)x<\(-\dfrac{1}{2}\)

c)x>4 hoặc x<1

d)ĐKXĐ x\(\ne\)1, ko tìm đc x

a) ĐKXĐ: \(x\notin\left\{0;-5\right\}\)

Ta có: \(B=\dfrac{x^2+2x}{2x+10}+\dfrac{x-5}{x}-\dfrac{5x-50}{2x^2+10x}\)

\(=\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{x-5}{x}-\dfrac{5x-50}{2x\left(x+5\right)}\)

\(=\dfrac{x^3+2x^2}{2x\left(x+5\right)}+\dfrac{2\left(x+5\right)\left(x-5\right)}{2x\left(x+5\right)}-\dfrac{5x-50}{2x\left(x+5\right)}\)

\(=\dfrac{x^3+2x^2+2x^2-50-5x+50}{2x\left(x+5\right)}\)

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

\(=\dfrac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}\)

\(=\dfrac{x^2+5x-x-5}{2\left(x+5\right)}\)

\(=\dfrac{x\left(x+5\right)-\left(x+5\right)}{2\left(x+5\right)}\)

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

\(=\dfrac{x-1}{2}\)

b) Để B=0 thì \(\dfrac{x-1}{2}=0\)

\(\Leftrightarrow x-1=0\)

hay x=1(nhận)

Vậy: Để B=0 thì x=1

Để \(B=\dfrac{1}{4}\) thì \(\dfrac{x-1}{2}=\dfrac{1}{4}\)

\(\Leftrightarrow4\left(x-1\right)=2\)

\(\Leftrightarrow4x-4=2\)

\(\Leftrightarrow4x=6\)

hay \(x=\dfrac{3}{2}\)(nhận)

Vậy: Để \(B=\dfrac{1}{4}\) thì \(x=\dfrac{3}{2}\)

c) Thay x=3 vào biểu thức \(B=\dfrac{x-1}{2}\), ta được:

\(B=\dfrac{3-1}{2}=\dfrac{2}{2}=1\)

Vậy: Khi x=3 thì B=1

d) Để B<0 thì \(\dfrac{x-1}{2}< 0\)

\(\Leftrightarrow x-1< 0\)

\(\Leftrightarrow x< 1\)

Kết hợp ĐKXĐ, ta được: 

\(\left\{{}\begin{matrix}x< 1\\x\notin\left\{0;-5\right\}\end{matrix}\right.\)

Vậy: Để B<0 thì \(\left\{{}\begin{matrix}x< 1\\x\notin\left\{0;-5\right\}\end{matrix}\right.\)

Để B>0 thì \(\dfrac{x-1}{2}>0\)

\(\Leftrightarrow x-1>0\)

hay x>1

Kết hợp ĐKXĐ, ta được: x>1

Vậy: Để B>0 thì x>1

a) Ta có: \(\dfrac{4}{5}-3\left|x\right|=\dfrac{1}{5}\)

\(\Leftrightarrow3\left|x\right|=\dfrac{4}{5}-\dfrac{1}{5}=\dfrac{3}{5}\)

\(\Leftrightarrow\left|x\right|=\dfrac{1}{5}\)

hay \(x\in\left\{\dfrac{1}{5};-\dfrac{1}{5}\right\}\)

b) Ta có: \(4x-\dfrac{1}{2}x+\dfrac{3}{5}x=\dfrac{4}{5}\)

nên \(\dfrac{41}{10}x=\dfrac{4}{5}\)

hay \(x=\dfrac{8}{41}\)

c) Ta có: \(\left(2x-8\right)\left(10-5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-8=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=8\\5x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

d) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}\)

\(\Leftrightarrow\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}-\dfrac{3}{4}=\dfrac{14}{4}-\dfrac{3}{4}=\dfrac{11}{4}\)

\(\Leftrightarrow\left|2x-1\right|=11\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=11\\2x-1=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=12\\2x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\)

a) ĐKXĐ: \(x\ne-10;x\ne0;x\ne-5\)

b) \(P=\dfrac{x^2+2x}{2x+20}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)

\(=\dfrac{x^2+2x}{2\left(x+10\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)

\(=\dfrac{x\left(x^2+2x\right)\left(x+5\right)}{2x\left(x+10\right)\left(x+5\right)}+\dfrac{2\left(x-5\right)\left(x+10\right)}{2x\left(x+10\right)\left(x+5\right)}+\dfrac{\left(50-5x\right)\left(x+10\right)}{2x\left(x+5\right)\left(x+10\right)}\)

\(=\dfrac{x^4+7x^3+10x^2+2x^2+10x-100+500-5x^2}{2x\left(x+10\right)\left(x+5\right)}\)

\(=\dfrac{x^4+7x^3+7x^2+10x+400}{2x\left(x+10\right)\left(x+5\right)}\)

c) \(P=0\Rightarrow x^4+7x^3+7x^2+10x+400=0\Leftrightarrow...\)

Số xấu thì câu c, d làm cũng như không. Bạn xem lại đề.

a) Ta có: \(A=\left(\dfrac{2x}{2x^2-5x+3}-\dfrac{5}{2x-3}\right):\left(3+\dfrac{2}{1-x}\right)\)

\(=\dfrac{2x-5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}:\dfrac{3\left(x-1\right)-2}{x-1}\)

\(=\dfrac{2x-5x+5}{2x-3}\cdot\dfrac{1}{3x-3-2}\)

\(=\dfrac{-3x+5}{2x-3}\cdot\dfrac{1}{3x-5}\)

\(=\dfrac{-1}{2x-3}\)

c) Để A>0 thì 2x-3<0

hay \(x< \dfrac{3}{2}\)

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x\ne1\end{matrix}\right.\)

Tham khảo:Cho biểu thức P= \((\frac{2x}{2x^2-5x+3}-\frac{5}{2x-3}):(3+\frac{2}{1-x})\) a) Rút gọn P b) Tính P với |3x-2|+1=5 c)... - Hoc24

a) ĐKXĐ: $x\notin \left\{1;\frac{3}{2}\right\}$

Ta có: $A=\left(\frac{2x}{2{}^{}-5x+3}-\frac{5}{2x-3}\right):\left(3+\frac{2}{1-x}\right)$

$=\left(\frac{2x}{\left(x-1\right)\left(2x-3\right)}-\frac{5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}\right):(\frac{3\left(x-1\right)\left(2x-3\right)}{\left(x-1\right)(2x-}$

a, \(A=\left(\dfrac{2x}{2x^2-5x+3}-\dfrac{5}{2x-3}\right):\left(3+\dfrac{2}{1-x}\right)\)ĐK : \(x\ne1;\dfrac{3}{2}\)

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

\(=\dfrac{3x-5}{\left(2x-3\right)\left(5-x\right)}\)

b, \(\left|3x-2\right|+1=5\Leftrightarrow\left[{}\begin{matrix}3x-2=4\\3x-2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Với x = 2 \(A=\dfrac{6-5}{1.3}=\dfrac{1}{3}\)

Với x = -2/3 \(A=\dfrac{3\left(-\dfrac{2}{3}\right)-5}{\left(-\dfrac{2}{3}-3\right)\left(5+\dfrac{2}{3}\right)}=\dfrac{-2-5}{-\dfrac{11}{3}.\dfrac{17}{3}}=-\dfrac{7}{-\dfrac{187}{9}}=\dfrac{63}{187}\)

a, đk : x khác -2 ; 2 

\(\left(x+2\right)^2-8x=0\Leftrightarrow x^2-4x+4=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)(ktm) 

pt vô nghiệm 

b, đk : x khác -1 ; 1 

\(x\left(x+1\right)-5x+3=0\Leftrightarrow x^2-4x+3=0\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\Leftrightarrow x=1\left(ktm\right);x=3\left(tm\right)\)

\(\lim\limits_{x\rightarrow+\infty}\dfrac{\left(x+1\right)\sqrt{2x+1}}{\sqrt{5x^3+x+2}}=\lim\limits_{x\rightarrow+\infty}\dfrac{\left(1+\dfrac{1}{x}\right)\sqrt{2+\dfrac{1}{x}}}{\sqrt{5+\dfrac{1}{x^2}+\dfrac{2}{x^3}}}=\sqrt{\dfrac{2}{5}}\)

Bạn coi lại, \(x\rightarrow-\infty\) hay \(+\infty\) nhỉ? (Dù a; b không đổi, vẫn là 2 và 5 nhưng \(x\rightarrow+\infty\) thì kết quả phải dương, ko có dấu trừ đằng trước)

Bài 1: 

a: Ta có: \(x^2-2\sqrt{5}x+5=0\)

\(\Leftrightarrow x-\sqrt{5}=0\)

hay \(x=\sqrt{5}\)

b: Ta có: \(\sqrt{x+3}=1\)

\(\Leftrightarrow x+3=1\)

hay x=-2

`P <= 1` là `P` ở đâu cậu nhỉ cộng `A` với `B` lại với nhau à?