a, 25/12

b,  25/72

a) = 3/4 + 4/3 = 25/12

b) = 3/8 - 16/9 = -101/72 * hình như hơi sai sai:"> *

Ta có :

\(\frac{x-3}{97}+\frac{x-27}{73}+\frac{x-67}{33}+\frac{x-73}{27}=4\)

\(\Leftrightarrow\left(\frac{x-3}{97}-1\right)+\left(\frac{x-27}{73}-1\right)+\left(\frac{x-67}{33}-1\right)+\left(\frac{x-73}{27}-1\right)=0\)

\(\Leftrightarrow\frac{x-100}{97}+\frac{x-100}{73}+\frac{x-100}{33}+\frac{x-100}{27}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\right)=0\)

Vì \(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}>0\) Nên \(x-100=0\)

\(\Leftrightarrow x=100\)

Vậy \(x=100\)

\(\Leftrightarrow\frac{x-3}{87}+\frac{x-27}{79}+\frac{x-67}{33}+\frac{x-73}{27}-4=0\)

\(\Leftrightarrow\left(\frac{x-3}{97}-1\right)+\left(\frac{x-27}{73}-1\right)+\left(\frac{x-67}{33}-1\right)+\left(\frac{x-73}{27}-1\right)=0\)

\(\Leftrightarrow\left(\frac{x-3-97}{97}\right)+\left(\frac{x-27-73}{73}\right)+\left(\frac{x-67-33}{33}\right)+\left(\frac{x-73-27}{27}\right)=0\)

\(\Leftrightarrow\frac{x-100}{97}+\frac{x-100}{73}+\frac{x-100}{33}+\frac{x-100}{27}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\right)=0\)

Vì \(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\ne0\)

\(\Rightarrow x-100=0\Leftrightarrow x=100\)

\(\frac{x-3}{97}+\frac{x-27}{73}+\frac{x-67}{33}+\frac{x-73}{27}=4\)    

\(\frac{x-3}{97}-1+\frac{x-27}{73}-1+\frac{x-67}{33}-1+\frac{x-73}{27}-1=4-1-1-1-1\)     

\(\frac{x-100}{97}+\frac{x-100}{73}+\frac{x-100}{33}+\frac{x-100}{27}=0\)    

\(\left(x-100\right)\left(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\right)=0\)   

\(x-100=0\) 

\(x=100\)

ĐKXĐ: \(-2\le x\le3\)

\(\dfrac{\sqrt{-x^2+x+6}}{2x+5}-\dfrac{\sqrt{-x^2+x+6}}{x-4}\ge0\)

\(\Leftrightarrow\sqrt{-x^2+x+6}\left(\dfrac{1}{2x+5}-\dfrac{1}{x-4}\right)\ge0\)

\(\Leftrightarrow\dfrac{\left(-x-9\right)\sqrt{x^2+x+6}}{\left(2x+5\right)\left(x-4\right)}\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x^2+x+6=0\\\dfrac{-x-9}{\left(2x+5\right)\left(x-4\right)}\ge0\end{matrix}\right.\) \(\Leftrightarrow-2\le x\le3\)

Hoặc có thể biện luận như sau:

Ta có: \(\left\{{}\begin{matrix}2x+5>0;\forall x\in\left[-2;3\right]\\x-4< 0;\forall x\in\left[-2;3\right]\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{-x^2+x+6}}{2x+5}\ge0\\\dfrac{\sqrt{-x^2+x+6}}{x-4}\le0\end{matrix}\right.\) ; \(\forall x\in\left[-2;3\right]\)

Do đó nghiệm của BPT là \(-2\le x\le3\)

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

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

\(\Rightarrow3x^2+3x=0\Rightarrow3x\left(x+1\right)=0\)

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

\(=x^3+3x^2+3x+1-x^3+1-2=0\\ \Leftrightarrow3x^2+3x=0\\ \Leftrightarrow3x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Đề trước đó: 

(x-7)(x+1)-(x-3)^2=(3x-5)(3x+5)-(3x+1)^2+(x-2)^2-x

<=>x^2+x-7x-7-x^2+6x-9=9x^2-25-9x^2-6x-1+x^2-4x+4-x

<=>x^2-11x-6=0

<=>x^2-2x. 11/2 + 121/4-145/4=0

<=>(x-11/2)^2=145/4

<=>|x-11/2|=căn(145)/2

<=>x=[11+-căn(145)]/2

3/25 x ( 15/7 - 2/7 ) + 3/7 x 1/25 

= 3/25 x 13/7 + 3/7 x 1/25 

= (3 x 13/7 + 3/7 ) x 1/25 

= 42/7 x 1/25 

= 6 x 1/25 

= 6/25

\(\dfrac{3}{25}\times\dfrac{15}{7}+\dfrac{3}{7}\times\dfrac{1}{25}-\dfrac{2}{7}\times\dfrac{3}{25}\)

\(=\dfrac{3}{25}\times\left(\dfrac{15}{7}-\dfrac{2}{7}\right)+\dfrac{3}{7}\times\dfrac{1}{25}\)

\(=\dfrac{3}{25}\times\dfrac{13}{7}+\dfrac{3}{7}\times\dfrac{1}{25}\)

\(=\dfrac{3\times13}{25\times7}+\dfrac{3\times1}{7\times25}\)

\(=\dfrac{39}{175}+\dfrac{3}{175}\)

\(=\dfrac{39+3}{175}\)

\(=\dfrac{42}{175}\)

\(=\dfrac{6}{25}\)

A = \(\dfrac{3}{25}\) \(\times\) \(\dfrac{15}{7}\) + \(\dfrac{3}{25}\) \(\times\) \(\dfrac{1}{7}\) - \(\dfrac{2}{7}\times\) \(\dfrac{3}{25}\)

A = \(\dfrac{3}{25}\) \(\times\)( \(\dfrac{15}{7}\) + \(\dfrac{1}{7}-\dfrac{2}{7}\))

A = \(\dfrac{3}{25}\) \(\times\) \(\dfrac{14}{7}\)

A = \(\dfrac{6}{25}\)

\(\dfrac{1+x}{1-x}+3=\dfrac{x-3}{x-1}\)

\(ĐK:x\ne1\)

\(\Leftrightarrow\dfrac{1+x}{1-x}+3=\dfrac{3-x}{1-x}\)

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

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

\(\Leftrightarrow1+x+3-3x=3-x\)

\(\Leftrightarrow-x=-1\)

\(\Leftrightarrow x=1\left(ktm\right)\)

Vậy pt vô nghiệm

\(\dfrac{1+x}{1-x}+3=\dfrac{x-3}{x-1}\) đề như thế này phải ko?

\(\dfrac{1+x}{1-x}+3=\dfrac{x-3}{x-1}\)

\(ĐK:x\ne1\)

\(\Leftrightarrow\dfrac{1+x}{1-x}+3=\dfrac{3-x}{1-x}\)

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

\(\Leftrightarrow1+x+3\left(1-x\right)=3-x\)

\(\Leftrightarrow1+x+3-3x=3-x\)

\(\Leftrightarrow x=2\)

Vậy \(S=\left\{2\right\}\)

\(-\dfrac{4}{3}\cdot x=\dfrac{2}{3}:\dfrac{7}{12}:\dfrac{4}{18}\)

\(\Rightarrow-\dfrac{4}{3}\cdot x=\dfrac{36}{7}\)

\(\Rightarrow x=\dfrac{\dfrac{36}{7}}{-\dfrac{4}{3}}=-\dfrac{27}{7}\)

-\(\dfrac{4}{3}\).\(x\) = \(\dfrac{2}{3}\): \(\dfrac{7}{12}\):\(\dfrac{4}{18}\)

-\(\dfrac{4}{3}.x\) = \(\dfrac{2}{3}\times\)\(\dfrac{12}{7}\)\(\times\)\(\dfrac{18}{4}\)

-\(\dfrac{4}{3}.\)\(x\)= \(\dfrac{36}{7}\)

     \(x\) = \(\dfrac{36}{7}\):(-\(\dfrac{4}{3}\))

    \(x\) = - \(\dfrac{27}{7}\)

−34​⋅x=32​:127​:184​

⇒−43⋅�=367⇒−34​⋅x=736​

⇒�=367−43=−277⇒x=−34​736​​=−727​
 

đáp án :1216800
Học tốt nhé!!!