a. 3x² - 2x - 1 = 0

<=> 3x² - 3x + x - 1 = 0

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

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

<=> \(\left[{}\begin{matrix}3x+1=0\\x-1=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=1\end{matrix}\right.\)

b. \(\dfrac{x+1}{3}+\dfrac{2x+3}{5}=\dfrac{3}{4}\)

<=> \(\dfrac{20\left(x+1\right)}{60}+\dfrac{12\left(2x+3\right)}{60}=\dfrac{45}{60}\)

<=> 20x + 20 + 24x + 36 = 45

<=> 44x = -11

<=> x = \(-\dfrac{1}{4}\)

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

    \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

b) Pt\(\Rightarrow\)\(5\cdot4\left(x+1\right)+3\cdot4\cdot\left(2x+3\right)=3\cdot3\cdot5\)

       \(\Leftrightarrow44x=-11\Rightarrow x=-\dfrac{1}{4}\)

\(a,\Leftrightarrow x^3-4x^2+4x=0\\ \Leftrightarrow x\left(x^2-4x+4\right)=0\\ \Leftrightarrow x\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ b,\Leftrightarrow4\left(x-1\right)=3x+6\left(2x-3\right)\\ \Leftrightarrow4x-4=3x+12x-18\\ \Leftrightarrow11x=14\Leftrightarrow x=\dfrac{14}{11}\)

a/ \(x^3-4x^2=-4x\)

\(\Leftrightarrow x^3-4x^2+4x=0\)

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

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

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

b/ \(\dfrac{x-1}{3}=\dfrac{x}{4}+\dfrac{2x-3}{2}\)

\(\Leftrightarrow8\left(x-1\right)=6x+12\left(2x-3\right)\)

\(\Leftrightarrow8x-8=6x+24x-36\)

\(\Leftrightarrow8x-8=30x-36\)

\(\Leftrightarrow8x-30x=8-36\)

\(\Leftrightarrow-22x=-28\)

\(\Leftrightarrow x=\dfrac{14}{11}\)

1/ \(=\lim\limits_{x\rightarrow1}\dfrac{\left(2x+7-9\right)\left(2+\sqrt{x+3}\right)}{\left(4-x-3\right)\left(\sqrt{2x+7}+3\right)}=\lim\limits_{x\rightarrow1}\dfrac{2\left(x-1\right)\left(2+\sqrt{x+3}\right)}{\left(x-1\right)\left(-\sqrt{2x+7}-3\right)}=\dfrac{2.4}{-6}=-\dfrac{4}{3}\)

2/ \(=\lim\limits_{x\rightarrow1^-}\dfrac{2.1-3}{1-1}=-\infty\)

3/ \(=\lim\limits_{x\rightarrow2^+}\dfrac{3-x}{x-2}=+\infty\)

4/ \(=\lim\limits_{x\rightarrow\pm\infty}\dfrac{-\dfrac{8x^3}{x^2}+\dfrac{9x^2}{x^2}+\dfrac{x}{x^2}-\dfrac{1}{x^2}}{\dfrac{5x^2}{x^2}+\dfrac{1}{x^2}}=\lim\limits_{x\rightarrow\pm\infty}\dfrac{-8x}{5}=\pm\infty\)

5/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{\dfrac{x^2}{x^2}}+\dfrac{2x}{x}-\dfrac{1}{x}}{\dfrac{2x}{x}+\dfrac{7}{x}}=\dfrac{1}{2}\)

cam on a

b) Vì \(\left|x+\dfrac{1}{1.3}\right| \ge0;\left|x+\dfrac{1}{3.5}\right|\ge0;...;\left|x+\dfrac{1}{97.99}\right|\ge0\)

\(\Rightarrow50x\ge0\Rightarrow x\ge0\)

Khi đó: \(\left|x+\dfrac{1}{1.3}\right|=x+\dfrac{1}{1.3};\left|x+\dfrac{1}{3.5}\right|=x+\dfrac{1}{3.5};...;\left|x+\dfrac{1}{97.99}\right|=x+\dfrac{1}{97.99}\left(1\right)\)

Thay (1) vào đề bài:

\(x+\dfrac{1}{1.3}+x+\dfrac{1}{3.5}+...+x+\dfrac{1}{97.99}=50x\)

\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\right)=50x\)

\(\Rightarrow49x+\left[\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\right]=50x\)

\(\Rightarrow49x+\dfrac{16}{99}=50x\)

\(\Rightarrow x=\dfrac{16}{99}\)

Vậy \(x=\dfrac{16}{99}.\)

thank bn nhìu nhìu

1.

\(y'=12x+\dfrac{4}{x^2}\)

2.

\(y'=\dfrac{3}{\left(-x+1\right)^2}\)

3.

\(y'=\dfrac{2x-3}{2\sqrt{x^2-3x+4}}\)

4.

\(y=\dfrac{x^3+3x^2-x-3}{x-4}\)

\(y'=\dfrac{\left(3x^2+6x-1\right)\left(x-4\right)-\left(x^3+3x^2-x-3\right)}{\left(x-4\right)^2}=\dfrac{2x^3-9x^2-24x+7}{\left(x-4\right)^2}\)

5.

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

6.

\(y'=\sqrt{x^2-1}+\dfrac{x\left(x+1\right)}{\sqrt{x^2-1}}\)

1, ĐKXĐ: x\(\ge0\);x\(\ne1\)

Rút gọn P với \(x\ge0;x\ne1\)ta có

P=\(\dfrac{-\sqrt{x}+\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\dfrac{-\left(\sqrt{x}-0,5\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-0,5\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\)

\(=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\dfrac{-\sqrt{x}+0,5}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}-0,5\right)}{x-\sqrt{x}+1}\right)\)

=\(\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\dfrac{-x\sqrt{x}+x-\sqrt{x}+0,5x-0,5\sqrt{x}+0,5+x\sqrt{x}-x-0,5x+0,5\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\right)\)

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

=\(\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)

2, Thay x=7-4\(\sqrt{3}\)thỏa mãn đk vào P ta có:

P\(=\dfrac{7-4\sqrt{3}-\sqrt{7-4\sqrt{3}}+1}{\sqrt{7-4\sqrt{3}}}\)

=\(\dfrac{7-4\sqrt{3}-\sqrt{\left(\sqrt{3}-2\right)^2}+1}{\sqrt{\left(\sqrt{3}-2\right)^2}}\)

=\(\dfrac{7-4\sqrt{3}-2+\sqrt{3}+1}{2-\sqrt{3}}\)

\(=\dfrac{6-3\sqrt{3}}{2-\sqrt{3}}=12+6\sqrt{3}-6\sqrt{3}-9\)=3

**\(\dfrac{2x-5}{6}=\dfrac{x-7}{5}\Leftrightarrow\dfrac{6\left(2x-5\right)}{30}=\dfrac{5\left(x-7\right)}{30}\)

\(\Leftrightarrow6\left(2x-5\right)=5\left(x-7\right)\Leftrightarrow12x-30=5x-35\)

\(\Leftrightarrow12x-5x=-35+30\Leftrightarrow7x=-5\Leftrightarrow x=-\dfrac{5}{7}\)

**\(\dfrac{4x+1}{3}=\dfrac{2x-3}{4}\Leftrightarrow\dfrac{4\left(x+1\right)}{12}=\dfrac{3\left(2x-3\right)}{12}\)

\(\Leftrightarrow4\left(x+1\right)=3\left(2x-3\right)\Leftrightarrow4x+4=6x-9\)

\(\Leftrightarrow4x-6x=-9-4\Leftrightarrow-2x=-13\Leftrightarrow x=\dfrac{13}{2}\)

toán 8 ạ mình lộn mất TvT