\(log_3\sqrt{3}=log_33^{\dfrac{1}{2}}=\dfrac{1}{2}\)

\(lne^3=log_ee^3=3\)

\(log_{27}3=log_{3^3}3=\dfrac{1}{3}\)

\(\log_{\sqrt{3}}3=log_{3^{\dfrac{1}{2}}}3=1:\dfrac{1}{2}=2\)

\(\log_{0,125}2=log_{2^{-3}}2=\dfrac{1}{-3}\)

\(\log_{\sqrt[3]{49}}7=\log_{7^{\dfrac{2}{3}}}7=1:\dfrac{2}{3}=\dfrac{3}{2}\)

\(\log_{\dfrac{1}{125}}5=\log_{5^{-3}}5=-\dfrac{1}{3}\)

\(\log_84=log_{2^3}2^2=\dfrac{1}{3}\cdot2=\dfrac{2}{3}\)

\(\log_{25}\left(\dfrac{1}{5}\right)=\log_{5^2}5^{-1}=\dfrac{1}{2}\cdot\left(-1\right)=-\dfrac{1}{2}\)

\(\log_{\dfrac{1}{5}}\sqrt{5}=\log_{5^{-1}}5^{\dfrac{1}{2}}=\dfrac{1}{-1}\cdot\dfrac{1}{2}=-\dfrac{1}{2}\)

\(log_{\dfrac{1}{7}}\sqrt[5]{49}=\log_{7^{-1}}7^{\dfrac{2}{5}}=\dfrac{1}{-1}\cdot\dfrac{2}{5}=-\dfrac{2}{5}\)

\(\log_4\left(\dfrac{1}{\sqrt{2}}\right)=\log_{2^2}\left(\sqrt{2}\right)^{-1}\)

\(=\log_{2^{-2}}\left(\sqrt{2}\right)^{-\dfrac{1}{2}}=\dfrac{1}{-2}\cdot\dfrac{-1}{2}=\dfrac{1}{4}\)

\(\log_{27}3\sqrt{3}=\log_{3^3}3^{\dfrac{3}{2}}=\dfrac{1}{3}\cdot\dfrac{3}{2}=\dfrac{1}{2}\)

Đáp án C.

Bài 1:

ĐKXĐ: x≠1

Ta có: \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)

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

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

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

\(\Leftrightarrow3x^2-3x=0\)

\(\Leftrightarrow3x\left(x-1\right)=0\)

Vì 3≠0

nên \(\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\)

Vậy: x=0

Bài 2:

ĐKXĐ: x≠2; x≠3; \(x\ne\frac{1}{2}\)

Ta có: \(\frac{x+4}{2x^2-5x+2}+\frac{x+1}{2x^2-7x+3}=\frac{2x+5}{2x^2-7x+3}\)

\(\Leftrightarrow\frac{x+4}{\left(x-2\right)\left(2x-1\right)}+\frac{x+1-\left(2x+5\right)}{\left(x-3\right)\left(2x-1\right)}=0\)

\(\Leftrightarrow\frac{x+4}{\left(x-2\right)\left(2x-1\right)}+\frac{x+1-2x-5}{\left(x-3\right)\left(2x-1\right)}=0\)

\(\Leftrightarrow\frac{\left(x+4\right)\left(x-3\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\frac{\left(-x-4\right)\left(x-2\right)}{\left(x-3\right)\left(2x-1\right)\left(x-2\right)}=0\)
\(\Leftrightarrow x^2+x-12-x^2-2x+8=0\)

\(\Leftrightarrow-x-4=0\)

\(\Leftrightarrow-x=4\)

hay x=-4(tm)

Vậy: x=-4

Bài 3:

ĐKXĐ: x≠1; x≠-1

Ta có: \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=3x\left(1-\frac{x-1}{x+1}\right)\)

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

\(\Leftrightarrow\frac{x+1}{x-1}-\frac{x-1}{x+1}-3x+\frac{3x\left(x-1\right)}{x+1}=0\)

\(\Leftrightarrow\frac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{3x\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{3x\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\)

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

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

\(\Leftrightarrow-6x^2+10x=0\)

\(\Leftrightarrow2x\left(-3x+5\right)=0\)

Vì 2≠0

nên \(\left[{}\begin{matrix}x=0\\-3x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-3x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{5}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{0;\frac{5}{3}\right\}\)

Bài 4:

ĐKXĐ: x≠1; x≠-3

Ta có: \(\frac{2x}{x-1}+\frac{4}{x^2+2x-3}=\frac{2x-5}{x+3}\)

\(\Leftrightarrow\frac{2x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}+\frac{4}{\left(x-1\right)\left(x+3\right)}-\frac{\left(2x-5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}=0\)

\(\Leftrightarrow2x^2+6x+4-\left(2x^2-7x+5\right)=0\)

\(\Leftrightarrow2x^2+6x+4-2x^2+7x-5=0\)

\(\Leftrightarrow13x-1=0\)

\(\Leftrightarrow13x=1\)

hay \(x=\frac{1}{13}\)(tm)

Vậy: \(x=\frac{1}{13}\)

Bài 5:

ĐKXĐ: x≠1; x≠-2

Ta có: \(\frac{1}{x-1}-\frac{7}{x+2}=\frac{3}{x^2+x-2}\)

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

\(\Leftrightarrow x+2-7\left(x-1\right)-3=0\)

\(\Leftrightarrow x+2-7x+7-3=0\)

\(\Leftrightarrow-6x+6=0\)

\(\Leftrightarrow-6\left(x-1\right)=0\)

Vì -6≠0

nên x-1=0

hay x=1(ktm)

Vậy: x∈∅

Bài 6:

ĐKXĐ: x≠4; x≠2

Ta có: \(\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{6x-8-x^2}\)

\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}-\frac{2}{6x-8-x^2}=0\)

\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}-\frac{2}{-\left(x^2-6x+8\right)}=0\)

\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}+\frac{2}{\left(x-4\right)\left(x-2\right)}=0\)

\(\Leftrightarrow\frac{\left(x+3\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}+\frac{\left(x-1\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\frac{2}{\left(x-4\right)\left(x-2\right)}=0\)

\(\Leftrightarrow x^2+x-6+x^2-5x+4+2=0\)

\(\Leftrightarrow2x^2-4x=0\)

\(\Leftrightarrow2x\left(x-2\right)=0\)

Vì 2≠0

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

Vậy: x=0

Bài 7:

ĐKXĐ: x≠1; x≠-2; x≠-1

Ta có: \(\frac{1}{x-1}-\frac{7}{x+2}=\frac{3}{1-x^2}\)

\(\Leftrightarrow\frac{1}{x-1}-\frac{7}{x+2}+\frac{3}{x^2-1}=0\)

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

\(\Leftrightarrow x^2+3x+2-7\left(x^2-1\right)+3x+6=0\)

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

\(\Leftrightarrow-6x^2+13x+8=0\)
\(\Leftrightarrow-6x^2+16x-3x+8=0\)

\(\Leftrightarrow2x\left(-3x+8\right)+\left(-3x+8\right)=0\)

\(\Leftrightarrow\left(-3x+8\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-3x+8=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=-8\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{3}\\x=\frac{-1}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{8}{3};\frac{-1}{2}\right\}\)

\( 1)\dfrac{1}{{x - 1}} + \dfrac{{2{x^2} - 5}}{{{x^3} - 1}} = \dfrac{4}{{{x^2} + x + 1}}\\ DK:x \ne 1\\ \Leftrightarrow \dfrac{{{x^2} + x + 1 + 2{x^2} - 5}}{{{x^3} - 1}} = \dfrac{{4\left( {x - 1} \right)}}{{{x^3} - 1}}\\ \Leftrightarrow {x^2} + x + 1 + 2{x^2} - 5 = 4x - 4\\ \Leftrightarrow 3{x^2} - 3x = 0\\ \Leftrightarrow 3x\left( {x - 1} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0\left( {tm} \right)\\ x = 1\left( {ktm} \right) \end{array} \right.\\ 2)\dfrac{{x + 4}}{{2{x^2} - 5x + 2}} + \dfrac{{x + 1}}{{2{x^2} - 7x + 3}} = \dfrac{{2x + 5}}{{2{x^2} - 7x + 3}}\\ + DK:x \ne \dfrac{1}{2};x \ne 2;x \ne 3\\ \Leftrightarrow \dfrac{{x + 4}}{{\left( {2x - 1} \right)\left( {x - 2} \right)}} + \dfrac{{x + 1}}{{\left( {x - 3} \right)\left( {2x - 1} \right)}} = \dfrac{{2x + 5}}{{\left( {x - 3} \right)\left( {2x - 1} \right)}}\\ \Leftrightarrow \left( {x + 4} \right)\left( {x - 3} \right) + \left( {x + 1} \right)\left( {x - 2} \right) = \left( {2x + 5} \right)\left( {x - 2} \right)\\ \Leftrightarrow {x^2} + x - 12 + {x^2} - x - 2 = 2{x^2} + x - 10\\ \Leftrightarrow x = - 4\left( {tm} \right)\\ 3)\dfrac{{x + 1}}{{x - 1}} - \dfrac{{x - 1}}{{x + 1}} = 3x\left( {1 - \dfrac{{x - 1}}{{x + 1}}} \right)\\ DK:x \ne \pm 1\\ \Leftrightarrow {\left( {x + 1} \right)^2} - {\left( {x - 1} \right)^2} = 3x\left( {x - 1} \right)\left( {x + 1 - x + 1} \right)\\ \Leftrightarrow {x^2} + 2x + 1 - {x^2} + 2x - 1 = 6x\left( {x - 1} \right)\\ \Leftrightarrow 4x = 6{x^2} - 6x\\ \Leftrightarrow 2x\left( {3x - 5} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = \dfrac{5}{3} \end{array} \right.\left( {tm} \right) \)

Còn lại tương tự mà làm nhé!

a:=>x^2-1-x=2x-1

=>x^2-x-1=2x-1

=>x^2-3x=0

=>x=0(loại) hoặc x=3(nhận)

b:=>x+2=0 hoặc 5-3x=0

=>x=-2 hoặc x=5/3

c:=>20(1-2x)+6x=9(x-5)-24

=>20-40x+6x=9x-45-24

=>-34x+20=9x-69

=>-43x=-89

=>x=89/43

d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3

=>2x^2+4x-19=-2x+7

=>2x^2+6x-26=0

=>x^2+3x-13=0

=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)

e: =>(2x-3)(2x-3-x-1)=0

=>(2x-3)(x-4)=0

=>x=4 hoặc x=3/2

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

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

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

\(\Rightarrow x^2-3x-6=0\)

.....

\(\frac{x+1}{x-2}+\frac{2x-1}{x-1}=\frac{2}{x^2-3x+2}+\frac{11}{2}\)

\(\Rightarrow\frac{2\left(x+1\right)\left(x-1\right)}{2\left(x-2\right)\left(x-1\right)}+\frac{2\left(2x-1\right)\left(x-2\right)}{2\left(x-1\right)\left(x-2\right)}\)\(=\frac{4}{2\left(x-1\right)\left(x-2\right)}+\frac{22\left(x-1\right)\left(x-2\right)}{2\left(x-1\right)\left(x-2\right)}\)

\(\Rightarrow2x^2-2+4x^2-10x+4=4+22x^2-66x+44\)

.....

\(3,\)\(-2x^2+4x+3\)

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

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

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

Đa thức này lớn nhất =5 khi và chỉ khi \(\left(x-1\right)^2\)nhỏ nhất 

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)