a: \(P=\dfrac{x+\sqrt{x}-x-2}{\sqrt{x}+1}:\dfrac{x-\sqrt{x}+\sqrt{x}}{x-1}\)

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

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

b: Để P<0 thì \(\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)< 0\)

=>1<x<4

\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)

=> \(\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\frac{x+349}{5}=0+1+1+1+1\)

=> \(\frac{x+2+327}{327}+\frac{x+3+326}{326}+\frac{x+4+325}{325}+\frac{x+5+324}{324}+\frac{x+329+20}{5}=4\)

=> \(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}+\frac{20}{5}=4\)

=> \(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)+4=4\)

=> \(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

Ta có : \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\ne0\)

=> \(x+329=0\)

=> \(x=-329\)

a) \(\frac{3x+2}{-4x+5}=-\frac{4}{3}\left(ĐKXĐ:x\ne\frac{5}{4}\right)\)
\(\Rightarrow3\left(3x+2\right)=-4\left(-4x+5\right)\)
\(\Leftrightarrow9x+6=16x-20\)
\(\Leftrightarrow7x=26\)
\(\Leftrightarrow x=\frac{26}{7}\)
b) \(\frac{2\left|x\right|+5}{-4x+3}=-\frac{5}{4}\)(Thôi bài sau tự tìm đkxđ nhá)
\(\Rightarrow8\left|x\right|+20=20x-15\)
\(\Leftrightarrow8\left|x\right|-20x+35\)\(\left(1\right)\)
TH1: Nếu \(x\ge0\)thì \(\left(1\right)\Leftrightarrow8x-20x+35=0\Leftrightarrow x=\frac{35}{12}\left(tm\right)\)
TH2: Nếu \(x< 0\)thì \(\left(1\right)\Leftrightarrow-8x-20x+35=0\Leftrightarrow x=\frac{35}{28}\left(ktm\right)\)
Vậy x=35/12
c)\(\frac{2x+1}{5}=\frac{3}{2x-1}\)
\(\Rightarrow4x^2-1=15\)
\(\Leftrightarrow4x^2=16\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
d)\(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)=\left(2x+1\right)\left(0,5x+2\right)\)
\(\Leftrightarrow x^2+4x+3=x^2+4,5x+2\)
\(\Leftrightarrow0,5x=1\)
\(\Leftrightarrow x=2\)
e) \(\frac{\left|6x+1\right|}{4}=\frac{2}{4}\)
\(\Leftrightarrow\left|6x+1\right|=2\)
\(\Leftrightarrow\orbr{\begin{cases}6x+1=2\\6x+1=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{6}\\x=-\frac{1}{2}\end{cases}}}\)
g)\(\frac{\left|3x-5\right|}{3}=\frac{\left|x\right|}{2}\)
\(\Leftrightarrow\frac{\left|3x-5\right|}{\left|x\right|}=\frac{3}{4}\)
\(\Leftrightarrow\left|\frac{3x-5}{x}\right|=\frac{3}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3x-5}{x}=\frac{3}{4}\\\frac{3x-5}{x}=-\frac{3}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{20}{9}\\x=\frac{4}{3}\end{cases}}}\)
Mỏi tay quá, xin tý cho sảng khoái nào!!
\(\)

Hôm nay tặng bn 3 k , ngày mai mk tặng tiếp 3 k nx nhé ^^ Thanks bn nhiều lắm !

mt cũng tính đc tìm x mk thử rùi

BN ƠI ĐÂY LÀ BÀI TÌM X MÀ ?? BN CÓ BỊ SAO KO THẾ ???

nHƯNG MK CẦN LỜI GIẢI NX MÀ

6+2x+2=2x-1-12

vo nghiem

\(\Rightarrow\frac{x+1}{3}=\frac{2x-1}{6}-3\)

\(\Rightarrow\frac{x+1}{3}=\frac{2x-19}{6}\)

=> 2(x + 1) = 2x - 19

2x + 2 = 2x - 19

2x - 2x = 2 + 19

0 = 21

Vậy không tồn tại x 

vô nghiệm . mình chắc 100 %

\(=18x\left(\frac{19}{21}+\frac{8}{9}\right)\)

\(=18x\frac{113}{63}\)

\(=\frac{226}{7}\)

266/7

18 x ( \(\frac{19}{21}\)+ \(\frac{8}{9}\)) = 18 x \(\frac{35}{21}\)= 30

giúp mình với

\(\frac{4\frac{1}{2}.5\frac{2}{3}}{6\frac{3}{4}}=\frac{\frac{9}{2}.\frac{17}{3}}{\frac{27}{4}}=\frac{\frac{153}{6}}{\frac{27}{4}}=\frac{153}{6}.\frac{4}{27}=\frac{34}{9}\)

\(\frac{4\frac{1}{2}\times5\frac{2}{3}}{6\frac{3}{4}}\)

\(\frac{\frac{4\times2+1}{2}+\frac{5\times3+2}{3}}{\frac{6\times4+3}{4}}\)

\(=\frac{\frac{9}{2}+\frac{17}{3}}{\frac{27}{4}}\)

\(=\frac{\frac{27}{6}+\frac{34}{6}}{\frac{27}{4}}\)

\(=\frac{61}{6}\div\frac{27}{4}=\frac{61}{6}\times\frac{4}{27}=\frac{244}{162}=\frac{122}{81}\)