1.

a.

ĐKXĐ: \(x^2-1>0\Rightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)

\(log_2\left(x^2-1\right)=3\)

\(\Rightarrow x^2-1=8\)

\(\Leftrightarrow x^2=9\)

\(\Rightarrow x=\pm3\) (tm)

b.

ĐKXĐ: \(x>0\)

\(log_3x+log_{\sqrt{3}}x+log_{\dfrac{1}{3}}x=6\)

\(\Leftrightarrow log_3x+2log_3x-log_3x=6\)

\(\Leftrightarrow log_3x=3\)

\(\Rightarrow x=3^3=27\)

c. ĐKXĐ: \(x>0\)

\(log_{\sqrt{2}}^2x+3log_2x+log_{\dfrac{1}{2}}x=2\)

\(\Leftrightarrow\left(2log_2x\right)^2+3log_2x-log_2x=2\)

\(\Leftrightarrow4log_2^2x+2log_2x-2=0\)

\(\Rightarrow\left[{}\begin{matrix}log_2x=-1\\log_2x=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\sqrt{2}\end{matrix}\right.\)

d.

ĐKXĐ: \(x>0\)

\(log_{\dfrac{1}{2}}^24x+log_2\dfrac{x^2}{8}=8\)

\(\Leftrightarrow\left(-log_24x\right)^2+log_2x^2-log_28=8\)

\(\Leftrightarrow\left(log_2x+log_24\right)^2+2log_2x-3=8\)

\(\Leftrightarrow\left(log_2x+2\right)^2+2log_2x-11=0\)

\(\Leftrightarrow log_2^2x+6log_2x-7=0\)

\(\Rightarrow\left[{}\begin{matrix}log_2x=1\\log_2x=-7\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2^7}\end{matrix}\right.\)

a) \(\dfrac{2x}{3}\)+\(\dfrac{2x-1}{6}\)=4 - \(\dfrac{x}{3}\)

<=>\(\dfrac{2x}{3}\)+\(\dfrac{2x-1}{6}\) - 4+\(\dfrac{x}{3}\)=0

<=>\(\dfrac{2x.2+2x-1-4.6+x.2}{6}\)=0

=>4x-2x-24+2x=0

<=>4x-24=0

<=>4x=24

<=>x=6

Vậy x=6

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

<=>\(\dfrac{x-1}{2}\)+\(\dfrac{x-1}{4}\)-1+\(\dfrac{2\left(x-1\right)}{3}\)=0

<=>\(\dfrac{6.\left(x-1\right)+3\left(x-1\right)-1.12+4.2\left(x-1\right)}{12}\)=0

=>6x-6+3x-3-12+4x-4+2x-2=0

<=>15x-27=0

<=>15x=27

<=>x=\(\dfrac{9}{5}\)

Vậy x=\(\dfrac{9}{5}\)

d)\(\dfrac{7x}{8}\)-5(x-9)=20x+\(\dfrac{1,5}{6}\)

<=>\(\dfrac{7x}{8}\)-5(x-9)-20x-\(\dfrac{1,5}{6}\)=0

<=>\(\dfrac{3.7x-24.5\left(x-9\right)-20x.24-1,5.4}{24}\)=0

=>-478x+165=0

<=> - 478= - 165

<=> x=\(\dfrac{165}{487}\)

Vậy x =\(\dfrac{165}{487}\)

Lời giải.

c.

$x^3-3x^2+3x-1=0$

$\Leftrightarrow (x-1)^3=0$

$\Leftrightarrow x-1=0$

$\Leftrightarrow x=1$

Vậy pt có tập nghiệm $S=\left\{1\right\}$

d. ĐKXĐ: $x\neq \frac{-1}{3}; -3$

PT $\Leftrightarrow \frac{(3x-1)(x+3)+(x-3)(3x+1)}{(3x+1)(x+3)}=2$

$\Leftrightarrow \frac{6x^2-6}{3x^2+10x+3}=2$

$\Leftrightarrow 6x^2-6=2(3x^2+10x+3)$

$\Leftrightarrow 20x+12=0$

$\Leftrightarrow x=\frac{-3}{5}$ (tm)

Vậy tập nghiệm của pt là $S=\left\{\frac{-3}{5}\right\}$

Bài 2:

a. 

\(\left\{\begin{matrix} 2x-3y=11\\ 5x-4y=3\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 10x-15y=55\\ 10x-8y=6\end{matrix}\right.\)

\(\Rightarrow (10x-8y)-(10x-15y)=6-55\)

\(\Leftrightarrow 7y=-49\Leftrightarrow y=-7\)

\(x=\frac{3y+11}{2}=\frac{3.(-7)+11}{2}=-5\)

Vậy hpt có nghiệm $(x,y)=(-5,-7)$

b. Không đủ cơ sở để tìm $x,y$

c. 

\(\left\{\begin{matrix} 5x+3y=\lambda\\ -x+\lambda y=-8\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 5x+3y=\lambda\\ -5x+5\lambda y=-40\end{matrix}\right.\)

\(\Rightarrow (3+5\lambda)y=\lambda-40\)

Nếu $\lambda = \frac{-3}{5}$ thì $0.y=\frac{-203}{5}$ (vô lý) nên hpt vô nghiệm

Nếu $\lambda \neq \frac{-3}{5}$ thì:

$y=\frac{\lambda - 40}{3+5\lambda}$

$x=8+\lambda y=\frac{\lambda ^2+24}{5\lambda +3}$

Bài nào em ??

Các bài ở hình ảnh trên

Bài 1: 

a) Ta có: \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2020\cdot2021}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}\)

\(=1-\dfrac{1}{2021}\)

\(=\dfrac{2020}{2021}\)

b) Ta có: \(B=\dfrac{3}{3\cdot5}+\dfrac{3}{5\cdot7}+...+\dfrac{3}{47\cdot49}\)

\(=\dfrac{3}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{47\cdot49}\right)\)

\(=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{47}-\dfrac{1}{49}\right)\)

\(=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)

\(=\dfrac{3}{2}\cdot\dfrac{46}{147}\)

\(=\dfrac{46}{2}\cdot\dfrac{3}{147}\)

\(=\dfrac{23}{49}\)

Bài 2: 

Ta có: \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}\)

\(=\dfrac{99}{100}< 1\)