Bài 2 Tìm x : a) -1/2.3x+3/4=-2,25
b)13-3.|x-2|=10
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9 tháng 9 2023
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
AH
Akai Haruma
Giáo viên
9 tháng 9 2023
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.
$2.3^{x+1}-4^{10}:4^7=(7^5:7^3).(3^3-5^2)=7^2.2=98$
$2.3^{x+1}-4^3=98$
$2.3^{x+1}=98+4^3=162$
$3^{x+1}=162:2=81=3^4$
$\Rightarrow x+1=4$
$\Rightarrow x=3$
PH
16 tháng 5 2022
a.-1,75-(-\(\dfrac{1}{9}\)-2\(\dfrac{1}{8}\))
-1,75-\(\dfrac{1}{9}+\dfrac{17}{8}\)
\(-\dfrac{7}{4}-\dfrac{1}{9}+\dfrac{17}{8}\)
\(\dfrac{-126}{72}-\dfrac{8}{72}+\dfrac{153}{72}\)
=\(\dfrac{19}{72}\)
PH
16 tháng 5 2022
b.\(\dfrac{-1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\dfrac{21}{8}+\dfrac{1}{3}\)
\(\dfrac{-2}{24}-\dfrac{63}{24}+\dfrac{64}{24}\)
=\(\dfrac{-1}{24}\)
LT
1
T
22 tháng 12 2023
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) \(2x^5-4x^3+2x=0\)
\(\Rightarrow2x\left(x^4-2x^2+1\right)=0\)
\(\Rightarrow2x\left[\left(x^2\right)^2-2\cdot x^2\cdot1+1^2\right]=0\)
\(\Rightarrow2x\left(x^2-1\right)^2=0\)
\(\Rightarrow2x\left(x-1\right)^2\left(x+1\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
\(\text{#}Toru\)
5 tháng 10 2020
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 :>
20 tháng 4 2018
-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
20 tháng 4 2018
\(-\frac{1}{2}.3x+\frac{3}{4}=-2,25\)
\(\Rightarrow\frac{1}{2}.(-3).x+\frac{1}{2}.\frac{3}{2}=\frac{9}{4}\)
\(\Rightarrow\frac{1}{2}.\left(-3x+\frac{3}{2}\right)=\frac{9}{4}\)
\(\Rightarrow-3x+\frac{3}{2}=\frac{9}{4}:\frac{1}{2}\)
\(\Rightarrow-3x+\frac{3}{2}=\frac{9}{2}\)
\(\Rightarrow-3x=3\)
\(\Rightarrow x=-1\)
H9
HT.Phong (9A5)
CTVHS
10 tháng 6 2023
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: \(X-2019\dfrac{2}{13}=3\dfrac{7}{26}+4\dfrac{7}{52}\)
\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{85}{26}+\dfrac{215}{52}\)
\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{385}{52}\)
\(\Rightarrow X=\dfrac{105381}{52}\)
a) \(\frac{-1}{2}.3x+\frac{3}{4}=-2,25\)
\(\frac{-1}{2}.3x+\frac{3}{4}=\frac{9}{4}\)
\(\frac{-1}{2}.3x=\frac{9}{4}-\frac{3}{4}\)
\(\frac{-1}{2}.3x=\frac{6}{4}=\frac{3}{2}\)
\(3x=\frac{3}{2}:\frac{-1}{2}\)
\(3x=-3\)
\(x=\left(-3\right):3\)
\(x=-1\)
b) \(13-3\left|x-2\right|=10\)
\(3\left|x-2\right|=13-10\)
\(3\left|x-2\right|=3\)
\(\left|x-2\right|=3:3\)
\(\left|x-2\right|=1\)
\(\Rightarrow\hept{\begin{cases}\left|x-2\right|=1\\\left|x-2\right|=-1\end{cases}\Rightarrow\hept{\begin{cases}x=3\\x=1\end{cases}}}\)