Bài làm

2+4+...+2016+2018/1019090 = -3x² - 4x

Ta có: số số hạng tử của phân số 2+4+...+2016+2018/1019090 là:( 2018 - 2 ) : 2 + 1 = 1009 ( số hạng)

Tổng của tử đó là: ( 2018 + 2 ) . 1009 : 2 = 1019090

=> Ta được: 1019090/1019090 = -3x² - 4x

<=> -3x² - 4x = 1

<=> -3x² - 4x - 1 = 0

<=> -3x² - 3x - x - 1 = 0

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

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

<=> x + 1 = 0 hoặc -3x - 1 = 0

<=> x = -1 hoặc x = 1/-3

Vậy nghiệm phương trình là: S = { -1; -1/3 }

\(\left(3x-5\right)\left(-2x-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-5=0\\-2x-7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=5\\-2x=7\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{-7}{2}\end{cases}}}\)

\(9x^2-1=\left(1+3x\right)\left(2x-3\right)\)

\(\Leftrightarrow9x^2-1=2x-3+6x^2-9x\)

\(\Leftrightarrow9x^2-1=-7x-3+6x^2\)

\(\Leftrightarrow9x^2-1+7x+3-6x^2=0\)

\(\Leftrightarrow3x^2+2+7x=0\)

\(\Leftrightarrow3x^2+6x+x+2=0\)

\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\3x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{3}\end{cases}}\)

\(\frac{2+4+...+2016+2018}{1019090}=-3x^2-4x.\)

\(\Leftrightarrow\frac{1009\left(1009+1\right)}{1019090}=-3x^2-4x\)

\(\Leftrightarrow\frac{1009.1010}{1009.1010}=-3x^2-4x\)

\(\Leftrightarrow3x^2+4x+1=0\)

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

\(\Leftrightarrow\left(x+1\right)\left(3x+1\right)=0\)

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

Vậy....

\(ĐKXĐ:x\ne-1;x\ne\frac{2}{3}\)

\(pt\Leftrightarrow\frac{7x-2\left(x+1\right)+\left(3x-2\right)}{\left(3x-2\right)\left(x+1\right)}=1\)

\(\Leftrightarrow7x-2\left(x+1\right)+\left(3x-2\right)=\left(3x-2\right)\left(x+1\right)\)

\(\Leftrightarrow8x-4=3x^2-2x+3x-2\)

\(\Leftrightarrow3x^2-7x+2=0\)

\(\Delta=7^2-4.3.2=25,\sqrt{\Delta}=5\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{7+5}{6}=2\\x=\frac{7-5}{6}=\frac{1}{3}\end{cases}}\)

Tự cho đkxđ nha!!!

<=> \(\frac{x+1-x}{x+1}=\frac{7x}{\left(3x-2\right)\left(x+1\right)}-\frac{2}{3x-2}\)

<=> \(\frac{3x-2}{\left(3x-2\right)\left(x+1\right)}=\frac{7x}{\left(3x-2\right)\left(x+1\right)}-\frac{2\left(x+1\right)}{\left(3x-2\right)\left(x+1\right)}\)

<=> \(\frac{7x-2x-2-3x+2}{\left(3x-2\right)\left(x+1\right)}=0\)

<=> \(\frac{2x}{\left(3x-2\right)\left(x+1\right)}=0\)

=> 2x = 0

<=> x = 0 (TM)

Vậy ...

\(1-\frac{x}{x+1}=\frac{7x}{\left(3x-2\right)\left(x+1\right)}+\frac{2}{2-3x}\)

\(\left(x+1\right)\left(3x-2\right)\left(2-3x\right)-x\left(3x-2\right)\left(2-3x\right)=7x\left(2-3x\right)+2\left(x+1\right)\left(3x-2\right)\)

\(-9x^2+12x-4=16x-15x^2-4\)

\(-9x^2+12x=16x-15x^2\)

\(9x^2-12x+16x-15x^2=0\)

\(-6x^2+4x=0\)

\(-2x\left(3x-2\right)=0\)

\(Th1:-2x=0\Leftrightarrow x=0\)

\(Th2:3x-2=0\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\)

x²+4x-5=0

<=> x²-5x+x-5=0

<=> x(x-5)+(x-5)=0

<=> (x-5)(x+1)=0

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

lollolol là j

\(x^2-4x-5=0\)

\(\Rightarrow x^2-x+5x-5=0\)

\(\Rightarrow x\left(x-1\right)+5\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x+5\right)=0\Rightarrow\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}=>\orbr{\begin{cases}x=1\\x=-5\end{cases}}}\)

Học tốt

Trả lời:

\(\frac{2}{\sqrt{5}+\sqrt{3}}-\sqrt{\frac{2}{4-\sqrt{15}}}+6\sqrt{\frac{1}{3}}\)

\(=\frac{2.\left(\sqrt{5}-\sqrt{3}\right)}{5-3}-\sqrt{\frac{2\times2}{2\times\left(4-\sqrt{15}\right)}}+6\times\frac{1}{\sqrt{3}}\)

\(=\frac{2.\left(\sqrt{5}-\sqrt{3}\right)}{2}-\sqrt{\frac{4}{8-2\sqrt{15}}}+6\times\frac{\sqrt{3}}{3}\)

\(=\sqrt{5}-\sqrt{3}-\sqrt{\frac{4}{5-2\sqrt{15}+3}}+2\sqrt{3}\)

\(=\sqrt{5}-\sqrt{3}-\sqrt{\frac{4}{\left(\sqrt{5}-\sqrt{3}\right)^2}}+2\sqrt{3}\)

\(=\sqrt{5}-\sqrt{3}-\frac{2}{\sqrt{5}-\sqrt{3}}+2\sqrt{3}\)

\(=\sqrt{5}+\sqrt{3}-\frac{2}{\sqrt{5}-\sqrt{3}}\)

\(=\frac{\left(\sqrt{5}-\sqrt{3}\right).\left(\sqrt{5}+3\right)-2}{\sqrt{5}-\sqrt{3}}\)

\(=\frac{5-3-2}{\sqrt{5}-\sqrt{3}}\)

\(=0\)

Học tốt 

-2.6314...

vô nghiệm

\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2015}\)

\(A=1-\frac{1}{2016}+1-\frac{1}{2017}+1-\frac{1}{2018}+1+\frac{3}{2015}\)

\(A=4-\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{3}{2015}\right)\)

Xét : 

\(\frac{1}{2016}< \frac{1}{2015}\)\(;\)\(\frac{1}{2017}< \frac{1}{2015}\)\(;\)\(\frac{1}{2018}< \frac{1}{2015}\)

\(\Rightarrow\)\(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}< \frac{1}{2015}+\frac{1}{2015}+\frac{1}{2015}\)

\(\Leftrightarrow\)\(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{3}{2015}< 0\)

Suy ra : \(A=4-\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{3}{2015}\right)>4-0=4\) ( đpcm ) 

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