\(5X\left(X-2020\right)+X=2020\)

\(\Leftrightarrow5X^2-10100X+X=2020\)

\(\Leftrightarrow5X^2-10099X=2020\)

\(\Leftrightarrow5X^2-10099X-2020=0\)

\(\Leftrightarrow5X^2-10100X+x-2020=0\)

\(\Leftrightarrow5X\left(X-2020\right)+X-2020=0\)

\(\Leftrightarrow\left(X-2020\right)\left(5X+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=-\frac{1}{5}\end{cases}}\)

\(4\left(x-5\right)^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left[2\left(x-5\right)\right]^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left[2\left(x-5\right)-2x-1\right]\left[2\left(x-5\right)+2x+1\right]=0\)

\(\Leftrightarrow\left(2x-10-2x-1\right)\left(2x-10+2x+1\right)=0\)

\(\Leftrightarrow-11\left(4x-9\right)=0\)

\(\Leftrightarrow x=\frac{9}{4}\)

\(a,5x\left(x-2020\right)+x=2020\)

\(< =>5x\left(x-2020\right)+x-2020=0\)

\(< =>\left(5x+1\right)\left(x-2020\right)=0\)

\(< =>\orbr{\begin{cases}5x+1=0\\x-2020=0\end{cases}}\)

\(< =>\orbr{\begin{cases}5x=-1\\x=2020\end{cases}< =>\orbr{\begin{cases}x=-\frac{1}{5}\\x=2020\end{cases}}}\)

\(b,4\left(x-5\right)^2-\left(2x+1\right)^2=0\)

\(< =>4\left(x^2-20x+25\right)-\left(4x^2+4x+1\right)=0\)

\(< =>4x^2-80x+100-4x^2-4x-1=0\)

\(< =>-84x+99=0< =>84x=99< =>x=\frac{99}{84}\)

f, \(\frac{12x+1}{11x-4}+\frac{10x-4}{9}=\frac{20x+17}{18}\)

\(\Leftrightarrow\) \(\frac{18\left(12x+1\right)}{18\left(11x-4\right)}+\frac{2\left(10x-4\right)\left(11x-4\right)}{18\left(11x-4\right)}=\frac{\left(20x+17\right)\left(11x-4\right)}{18\left(11x-4\right)}\)

\(\Leftrightarrow\) 18(12x + 1) + 2(10x - 4)(11x - 4) = (20x + 17)(11x - 4)

\(\Leftrightarrow\) 216x + 18 + 220x² − 168x + 32 = 220x² + 107x − 68

\(\Leftrightarrow\) 216x + 18 + 220x² − 168x + 32 - 220x² - 107x + 68 = 0

\(\Leftrightarrow\) −59x + 118 = 0

\(\Leftrightarrow\) -59x = -118

\(\Leftrightarrow\) x = 2

Vậy S = {2}

Chúc bạn học tốt!

a) \(\frac{4x-3}{x-5}\)=\(\frac{29}{3}\) (ĐKXĐ:x≠5)

⇔\(\frac{3\left(4x-3\right)}{3\left(x-5\right)}\)=\(\frac{29\left(x-5\right)}{3\left(x-5\right)}\)

⇒12x-9=29x-145

⇔12x-9-29x+145=0

⇔-17x+136=0

⇔-17x=-136

⇔x=8

Vậy tập nghiệm của phương trình đã cho là:S={8}

d: =>4x+6=15x-12

=>4x-15x=-12-6=-18

=>-11x=-18

hay x=18/11

e: =>\(45x+27=12+24x\)

=>21x=-15

hay x=-5/7

f: =>35x-5=96-6x

=>41x=101

hay x=101/41

g: =>3(x-3)=90-5(1-2x)

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

=>3x-9=10x+85

=>-7x=94

hay x=-94/7

a,x⁴-10x²+9=0

=>(x-1)(x³+x²-9x-9)=0

=> (x-1)(x+1)(x-3)(x+3)=0

=>\(\orbr{\begin{cases}x-1=0\\x+1=0\end{cases}}\)hoặc\(\orbr{\begin{cases}x-3=0\\x+3=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=\pm1\\x=\pm3\end{cases}}\)

Vậy tập nghiệm cuả pt là S={\(\pm1,\pm3\)}

trả lời

h bn tính theo đenta là ra thôi mà

hok tốt

a/ ĐKXĐ: ...

\(\Leftrightarrow2\left(x^2-5x-6\right)+\sqrt{x^2-5x-6}-3=0\)

Đặt \(\sqrt{x^2-5x-6}=a\ge0\)

\(2a^2+a-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2-5x-6}=1\Leftrightarrow x^2-5x-7=0\)

b/ ĐKXĐ: ...

\(\Leftrightarrow5\sqrt{3x^2-4x-2}-2\left(3x^2-4x-2\right)+3=0\)

Đặt \(\sqrt{3x^2-4x-2}=a\ge0\)

\(-2a^2+5a+3=0\) \(\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{1}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{3x^2-4x-2}=3\Leftrightarrow3x^2-4x-11=0\)

c/ \(\Leftrightarrow x^2+2x-6+\sqrt{2x^2+4x+3}=0\)

Đặt \(\sqrt{2x^2+4x+3}=a>0\Rightarrow x^2+2x=\frac{a^2-3}{2}\)

\(\frac{a^2-3}{2}-6+a=0\Leftrightarrow a^2+2a-15=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-5\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2x^2+4x+3}=3\Leftrightarrow2x^2+4x-6=0\)

d/ ĐKXĐ: ...

Đặt \(\sqrt{\frac{3x-1}{x}}=a>0\)

\(2a=\frac{1}{a^2}+1\Leftrightarrow2a^3-a^2-1=0\)

\(\Leftrightarrow\left(a-1\right)\left(2a^2+a+1\right)=0\)

\(\Rightarrow a=1\Rightarrow\sqrt{\frac{3x-1}{x}}=1\Leftrightarrow3x-1=x\)

e/ĐKXĐ: ...

\(\Leftrightarrow2\sqrt{\frac{6x-1}{x}}=\frac{x}{6x-1}+1\)

Đặt \(\sqrt{\frac{6x-1}{x}}=a>0\)

\(2a=\frac{1}{a^2}+1\Leftrightarrow2a^3-a^2-1=0\Leftrightarrow\left(a-1\right)\left(2a^2+a+1\right)=0\)

\(\Rightarrow a=1\Rightarrow\sqrt{\frac{6x-1}{x}}=1\Rightarrow6x-1=x\)

f/ ĐKXĐ: ...

Đặt \(\sqrt{\frac{x}{2x-1}}=a>0\)

\(\frac{1}{a}+1+a=3a^2\)

\(\Leftrightarrow3a^3-a^2-a-1=0\)

\(\Leftrightarrow\left(a-1\right)\left(3a^2+2a+1\right)=0\)

\(\Leftrightarrow a=1\Rightarrow\sqrt{\frac{x}{2x-1}}=1\Rightarrow x=2x-1\)

a)\(\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4\)

=> 2x + 7 = 4 

     2x        = 4 - 7 

     2x        = -3

       x        = -3 : 2

       x         = -1,5

   Vậy x = -1,5

Bài 1:

a) Ta có: \(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)

\(=\frac{2x}{x\left(x+2y\right)}+\frac{y}{y\left(x-2y\right)}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{2}{x+2y}+\frac{y}{x-2y}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{2\left(x-2y\right)}{\left(x+2y\right)\left(x-2y\right)}+\frac{y\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{2x-4y+xy+2y^2+4}{\left(x-2y\right)\cdot\left(x+2y\right)}\)

b) Ta có: \(\frac{1}{x-y}+\frac{3xy}{y^3-x^3}+\frac{x-y}{x^2+xy+y^2}\)

\(=\frac{x^2+xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}-\frac{3xy}{\left(x-y\right)\left(x^2+xy+y^2\right)}+\frac{\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)

\(=\frac{x^2+xy+y^2-3xy+x^2-2xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)

\(=\frac{2x^2-4xy+2y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)

\(=\frac{2\left(x^2-2xy+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)

\(=\frac{2\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)

\(=\frac{2x-2y}{x^2+xy+y^2}\)

c) Ta có: \(\frac{xy}{2x-y}-\frac{x^2-1}{y-2x}\)

\(=\frac{xy}{2x-y}+\frac{x^2-1}{2x-y}\)

\(=\frac{x^2+xy-1}{2x-y}\)

d) Ta có: \(\frac{2\left(x+y\right)\left(x-y\right)}{x}-\frac{-2y^2}{x}\)

\(=\frac{2\left(x^2-y^2\right)+2y^2}{x}\)

\(=\frac{2x^2-2y^2+2y^2}{x}\)

\(=\frac{2x^2}{x}=2x\)

Bài 2:

a) Ta có: \(\frac{4x+1}{2}-\frac{3x+2}{3}\)

\(=\frac{3\left(4x+1\right)}{6}-\frac{2\left(3x+2\right)}{6}\)

\(=\frac{12x+3-6x-4}{6}\)

\(=\frac{6x-1}{6}\)

b) Ta có: \(\frac{x+3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}\)

\(=\frac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}-\frac{x^2}{x\left(x-3\right)}+\frac{9}{x\left(x-3\right)}\)

\(=\frac{x^2-9-x^2+9}{x\left(x-3\right)}=\frac{0}{x\left(x-3\right)}=0\)

c) Ta có: \(\frac{x+3}{x^2+1}-\frac{1}{x^2+2}\)

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

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

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

e) Ta có: \(\frac{3}{2x^2+2x}+\frac{2x-1}{x^2-1}-\frac{2}{x}\)

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

\(=\frac{3\left(x-1\right)}{2x\left(x+1\right)\left(x-1\right)}+\frac{2x\left(2x-1\right)}{2x\left(x+1\right)\left(x-1\right)}-\frac{2\cdot2\cdot\left(x+1\right)\left(x-1\right)}{2x\left(x+1\right)\left(x-1\right)}\)

\(=\frac{3x-3+4x^2-2x-4\left(x^2-1\right)}{2x\left(x+1\right)\left(x-1\right)}\)

\(=\frac{4x^2+x-3-4x^2+4}{2x\left(x+1\right)\left(x-1\right)}\)

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

d) Ta có: \(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10x+8}{9x^2-4}\)

\(=\frac{3x+2}{\left(3x-2\right)\left(3x+2\right)}-\frac{4\left(3x-2\right)}{\left(3x+2\right)\left(3x-2\right)}-\frac{-10x+8}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\frac{3x+2-12x+8+10x-8}{\left(3x-2\right)\left(3x+2\right)}\)

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

f) Ta có: \(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)

\(=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x-y\right)}\)

\(=\frac{3x\cdot2\cdot\left(x-y\right)}{10\left(x+y\right)\left(x-y\right)}-\frac{x\cdot\left(x+y\right)}{10\left(x-y\right)\left(x+y\right)}\)

\(=\frac{6x^2-6xy-x^2-xy}{10\left(x-y\right)\left(x+y\right)}\)

\(=\frac{5x^2-7xy}{10\left(x-y\right)\left(x+y\right)}\)