giúp me được ko

a) \(2x^2+20x+52=0\Rightarrow x^2+10x+26=0\Rightarrow\left(x+5\right)^2+1=0\)

\(\Rightarrow\) vô nghiệm

b) ĐK: \(x\ne1;-1\)

\(\dfrac{2x-19}{5x^2-5}-\dfrac{17}{x-1}=\dfrac{8}{1-x}\Rightarrow\dfrac{2x-19}{5\left(x-1\right)\left(x+1\right)}-\dfrac{17}{x-1}+\dfrac{8}{x-1}=0\)

\(\Rightarrow\dfrac{2x-19}{5\left(x-1\right)\left(x+1\right)}-\dfrac{9}{x-1}=0\Rightarrow\dfrac{2x-19-45\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}=0\)

\(\Rightarrow-43x-64=0\Rightarrow x=-\dfrac{64}{43}\)

a)  Ta có: \(\Delta'=100-104=-4< 0\)

Vậy phương trình vô nghiệm.

b) ĐKXĐ: \(x\ne1;x\ne-1\)

\(\Leftrightarrow\dfrac{2x-19}{5\left(x^2-1\right)}=\dfrac{17}{x-1}-\dfrac{8}{x-1}\)

\(\Leftrightarrow\dfrac{2x-19}{5\left(x-1\right)\left(x+1\right)}=\dfrac{9}{x-1}\)

\(\Leftrightarrow\dfrac{2x-19}{5\left(x-1\right)\left(x+1\right)}=\dfrac{45\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow2x-19=45x+45\)

\(\Leftrightarrow43x=-64\)

\(\Leftrightarrow x=-\dfrac{64}{43}\)(TM)

Vậy phương trình có nghiệm là: \(x=-\dfrac{64}{43}\)

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

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

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

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

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

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

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

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

a) Ta có: \(3a=2b\Leftrightarrow\frac{a}{2}=\frac{b}{3}\Leftrightarrow\frac{a}{10}=\frac{b}{15}\) (1)

Và \(4b=5c\Leftrightarrow\frac{b}{5}=\frac{c}{4}\Leftrightarrow\frac{b}{15}=\frac{c}{12}\) (2)

Từ (1) và (2) => \(\frac{a}{10}=\frac{b}{15}=\frac{c}{12}\)

Áp dụng t/c dãy tỉ số bằng nhau: \(\frac{a}{10}=\frac{b}{15}=\frac{c}{12}=\frac{-a-b+c}{-10-15+12}=\frac{-52}{-13}=4\)

\(\Rightarrow\hept{\begin{cases}a=40\\b=60\\c=48\end{cases}}\)

a) \(\hept{\begin{cases}3a=2b\\4b=5c\end{cases}}\Rightarrow\hept{\begin{cases}\frac{a}{2}=\frac{b}{3}\\\frac{b}{5}=\frac{c}{4}\end{cases}\Rightarrow}\hept{\begin{cases}\frac{a}{10}=\frac{b}{15}\\\frac{b}{15}=\frac{c}{12}\end{cases}\Rightarrow}\frac{a}{10}=\frac{b}{15}=\frac{c}{12}\)

-a - b + c = -52 => -( a + b - c ) = -52

                         => a + b - c = 52

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{a}{10}=\frac{b}{15}=\frac{c}{12}=\frac{a+b-c}{10+15-12}=\frac{52}{13}=4\)

\(\Rightarrow\hept{\begin{cases}a=40\\b=60\\c=48\end{cases}}\)

b) \(C=\frac{2x^2-5x+3}{2x-1}\)( ĐKXĐ : \(x\ne\frac{1}{2}\))

\(\left|x\right|=\frac{3}{2}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{3}{2}\end{cases}}\)

Với x = 3/2 ( tmđk )

=> C = \(\frac{2\cdot\left(\frac{3}{2}\right)^2-5\cdot\frac{3}{2}+3}{2\cdot\frac{3}{2}-1}=\frac{0}{2}=0\)

Với x = -3/2 ( tmđk )

=> C = \(\frac{2\cdot\left(-\frac{3}{2}\right)^2-5\cdot\left(-\frac{3}{2}\right)+3}{2\cdot\left(-\frac{3}{2}\right)-1}=\frac{15}{-4}=-\frac{15}{4}\)

b) Ta có: \(\left|x\right|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{3}{2}\end{cases}}\)

+ Nếu: \(x=\frac{3}{2}\Rightarrow C=0\)

+ Nếu: \(x=-\frac{3}{2}\Rightarrow C=-\frac{15}{4}\)

\(a,\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\) (x khác -3; khác 0)

\(=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x.\left(x+3\right)}=\frac{3x}{2x.\left(x+3\right)}-\frac{x-6}{2x.\left(x+3\right)}=\frac{3x-x+6}{2x.\left(x+3\right)}=\frac{2x+6}{x.\left(2x+6\right)}=\frac{1}{x}\)

\(b,\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}\) (x khác 0 , khác 1/2 khác -1/2 )

\(=\left(\frac{\left(2x+1\right)^2}{\left(2x-1\right)\left(2x+1\right)}-\frac{\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)

\(=\left(\frac{4x^2+4x+1}{\left(2x-1\right)\left(2x+1\right)}-\frac{4x^2-4x+1}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)

\(=\frac{8x}{\left(2x-1\right)\left(2x+1\right)}.\frac{5.\left(2x-1\right)}{4x}=\frac{10}{2x+1}\)

\(c,\frac{x^2+x}{5x^2-10x+5}:\frac{3x+3}{5x-5}\) (x khác 1 ; khác -1)

\(=\frac{x.\left(x+1\right)}{5.\left(x^2-2x+1\right)}.\frac{5x-5}{3x+3}=\frac{x.\left(x+1\right)}{5.\left(x-1\right)^2}.\frac{5\left(x-1\right)}{3.\left(x+1\right)}=\frac{x}{3.\left(x-1\right)}=\frac{x}{3x-3}\)

\(\frac{2x-1}{2}\)-1=\(\frac{x^2+x-3}{x-1}\)-\(\frac{5x-2}{2-2x}\)

\(\frac{2x-1}{2}\)-1=\(\frac{x^2+x-3}{x-1}\)-\(\frac{5x-2}{2\left(1-x\right)}\)

\(\frac{\left(2x-1\right)\left(x-1\right)}{2\left(x-1\right)}\)-\(\frac{2\left(x-1\right)}{2\left(x-1\right)}\)=\(\frac{2\left(x^2+x-3\right)}{2\left(x-1\right)}\)+\(\frac{5x-2}{2\left(x-1\right)}\)

2x²-x-2x+1-2x+2=2x²+2x-6+5x-2

2x²-x-2x+1-2x+2-2x²-2x+6-5x+2=0

2x²-2x²-x-2x-2x-2x-5x+1+2+6+2=0

11-12x=0

12x=11

x=\(\frac{11}{12}\)

   x²-4x+5=0

=>(x-2)²+1=0

=>(x-2)² =-1

=> pt vô nghiệm

   (x²+5x)(x³+3x²-18x)=0

=>\(\int^{x^2+5x=0}_{x^3+3x^2-18x=0}=>\int^{\int^{x=0}_{x=-5}}_{x=3;x=0;x=-6}\)