a: =>3[(2x-1)^2-4]=49*125:175+196=231

=>(2x-1)^2-4=77

=>(2x-1)^2=81

=>2x-1=9 hoặc 2x-1=-9

=>x=5 hoặc x=-4

b: \(\Leftrightarrow2\cdot3^x\cdot3-4^3=7^2\cdot\left(27-25\right)\)

=>\(6\cdot3^x=49\cdot2+64=162\)

=>3^x=27

=>x=3

Lời giải:
a.

$3[(2x-1)^2-4]-14^2=7^2.5^3:175=35$

$3[(2x-1)^2-4]=35+14^2=231$

$(2x-1)^2-4=231:3=77$

$(2x-1)^2=77+4=81=9^2=(-9)^2$

$\Rightarrow 2x-1=9$ hoặc $2x-1=-9$

$\Rightarrow x=5$ hoặc $x=-4$

b.

$2.3^{x+1}-4^{10}:4^7=(7^5:7^3).(3^3-5^2)=7^2.2=98$

$2.3^{x+1}-4^3=98$

$2.3^{x+1}=98+4^3=162$

$3^{x+1}=162:2=81=3^4$

$\Rightarrow x+1=4$

$\Rightarrow x=3$

a.-1,75-(-\(\dfrac{1}{9}\)-2\(\dfrac{1}{8}\))
-1,75-\(\dfrac{1}{9}+\dfrac{17}{8}\)
\(-\dfrac{7}{4}-\dfrac{1}{9}+\dfrac{17}{8}\)
\(\dfrac{-126}{72}-\dfrac{8}{72}+\dfrac{153}{72}\)
=\(\dfrac{19}{72}\)

b.\(\dfrac{-1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\dfrac{21}{8}+\dfrac{1}{3}\)
\(\dfrac{-2}{24}-\dfrac{63}{24}+\dfrac{64}{24}\)
=\(\dfrac{-1}{24}\)

\(\dfrac{-5}{6}-\left(\dfrac{-3}{8}+\dfrac{1}{10}\right)\)
\(\dfrac{-5}{6}-\dfrac{-3}{8}-\dfrac{1}{10}\)
 

a) \(3\left(x-1\right)^2\cdot3x\left(x-5\right)=0\)

\(\Rightarrow9x\left(x-1\right)^2\left(x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=5\end{matrix}\right.\)

b) \(\left(x+3\right)^2-5x-15=0\)

\(\Rightarrow\left(x+3\right)^2-5\left(x+3\right)=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

c) \(2x^5-4x^3+2x=0\)

\(\Rightarrow2x\left(x^4-2x^2+1\right)=0\)

\(\Rightarrow2x\left[\left(x^2\right)^2-2\cdot x^2\cdot1+1^2\right]=0\)

\(\Rightarrow2x\left(x^2-1\right)^2=0\)

\(\Rightarrow2x\left(x-1\right)^2\left(x+1\right)^2=0\)

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

\(\text{#}Toru\)

Bài 1 :

a) \(\frac{12}{21}-\frac{3}{7}+\left(-\frac{2}{3}\right)=\frac{4}{7}-\frac{3}{7}+\left(-\frac{2}{3}\right)=\frac{1}{7}-\frac{2}{3}=-\frac{11}{21}\)

b) \(\left(-\frac{25}{13}\right)+\left(-\frac{9}{17}\right)+\frac{12}{13}+\left(-\frac{25}{17}\right)\)

\(=\left[\left(-\frac{25}{13}\right)+\frac{12}{13}\right]+\left[\left(-\frac{9}{17}\right)+\left(-\frac{25}{17}\right)\right]\)

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

c) \(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{9}\cdot\frac{3}{13}=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)=\frac{5}{9}\cdot1=\frac{5}{9}\)

Bài 2 :

a)  \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)

=> \(\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}=-\frac{29}{70}\)

=> \(x=\left(-\frac{29}{70}\right):\frac{2}{3}=\left(-\frac{29}{70}\right)\cdot\frac{3}{2}=-\frac{87}{140}\)

b) \(x:\frac{5}{2}-\frac{1}{2}=-\frac{2}{3}\)

=> \(x:\frac{5}{2}=-\frac{2}{3}+\frac{1}{2}=-\frac{1}{6}\)

=> \(x=\left(-\frac{1}{16}\right)\cdot\frac{5}{2}=-\frac{5}{32}\)

c) Bạn chỉ cần xét hai trường hợp âm và dương thôi :>

-1/2.3x+3/4=-2.25

-1/2.3x+3/4=9/4

-1/2.3x      =9/4-3/4

-1/2.3x      =3/2

       3x     =3/2:-1/2

       3x     =-3

        x     =-3:3

        x     =-1

vậy   x=-1

\(-\frac{1}{2}.3x+\frac{3}{4}=-2,25\)

\(\Rightarrow\frac{1}{2}.(-3).x+\frac{1}{2}.\frac{3}{2}=\frac{9}{4}\)

\(\Rightarrow\frac{1}{2}.\left(-3x+\frac{3}{2}\right)=\frac{9}{4}\)

\(\Rightarrow-3x+\frac{3}{2}=\frac{9}{4}:\frac{1}{2}\)

\(\Rightarrow-3x+\frac{3}{2}=\frac{9}{2}\)

\(\Rightarrow-3x=3\)

\(\Rightarrow x=-1\)

gúp minh với ạ 

Bài 13: 

a: =>20-x=15-8+13=20

hay x=0

Bài 1: Ta có: \(4\dfrac{3}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)

\(\dfrac{23}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)

\(\dfrac{138}{30}< X< \dfrac{200}{3}\)

\(\Rightarrow X\in\left\{\dfrac{160}{30};\dfrac{161}{30};\dfrac{162}{30};...;\dfrac{198}{30};\dfrac{199}{30}\right\}\)

Bài 2: \(X-2019\dfrac{2}{13}=3\dfrac{7}{26}+4\dfrac{7}{52}\)

\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{85}{26}+\dfrac{215}{52}\)

\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{385}{52}\)

\(\Rightarrow X=\dfrac{105381}{52}\)