(3x-23):5 =2
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TM
11 tháng 8 2021
1/ \(2\left(x-5\right)=\left(-x-5\right)\)
\(\Leftrightarrow2x-10=-x-5\)
\(\Leftrightarrow3x=5\)
\(\Leftrightarrow x=\dfrac{5}{3}\)
Vậy: \(S=\left\{\dfrac{5}{3}\right\}\)
==========
2/ \(2\left(x+3\right)-3\left(x-1\right)=2\)
\(\Leftrightarrow2x+6-3x+3=2\)
\(\Leftrightarrow-x=-7\)
\(\Leftrightarrow x=7\)
Vậy: \(S=\left\{7\right\}\)
==========
3/ \(4\left(x-5\right)-\left(3x-1\right)=x-19\)
\(\Leftrightarrow4x-20-3x+1=x-19\)
\(\Leftrightarrow0x=0\)
Vậy: \(S=\left\{x|x\text{ ∈ }R\right\}\)
===========
4/ \(7-\left(x-2\right)=5\left(2-3x\right)\)
\(\Leftrightarrow7-x+2=10-15x\)
\(\Leftrightarrow14x=1\)
\(\Leftrightarrow x=\dfrac{1}{14}\)
Vậy: \(S=\left\{\dfrac{1}{14}\right\}\)
==========
5/ \(2x-\left(5-3x\right)=7x+1\)
\(\Leftrightarrow2x-5+3x=7x+1\)
\(\Leftrightarrow-2x=6\)
\(\Leftrightarrow x=-3\)
Vậy: \(S=\left\{-3\right\}\)
[---]
Chúc bạn học tốt.
NT
11 tháng 8 2021
1. \(2\left(x-5\right)=-x-5\)
\(\Leftrightarrow3x=5\)
\(\Leftrightarrow x=\dfrac{5}{3}\)
Vậy \(S=\left\{\dfrac{5}{3}\right\}\)
2. \(2\left(x+3\right)-3\left(x-1\right)=2\)
\(\Leftrightarrow2x+6-3x+3=2\)
\(\Leftrightarrow x=7\)
Vậy \(S=\left\{7\right\}\)
3. \(4\left(x-5\right)-\left(3x-1\right)=x-19\)
\(\Leftrightarrow4x-20-3x+1-x+19=0\)
\(\Leftrightarrow0x=0\)
Vậy \(S=\left\{x\in R\right\}\)
4. \(7-\left(x-2\right)=5\left(2-3x\right)\)
\(\Leftrightarrow7-x+2-10+15x=0\)
\(\Leftrightarrow14x-1=0\)
\(\Leftrightarrow x=\dfrac{1}{14}\)
Vậy \(S=\left\{\dfrac{1}{14}\right\}\)
4. \(2x-\left(5-3x\right)=7x+1\)
\(\Leftrightarrow2x-5+3x-7x-1=0\)
\(\Leftrightarrow-2x-6=0\)
\(\Leftrightarrow x=-3\)
Vậy \(S=\left\{-3\right\}\)
3 tháng 3 2022
\(a,x-5\left(x-2\right)=6x\\ \Leftrightarrow x-5x+10-6x=0\\ \Leftrightarrow-10x+10=0\\ \Leftrightarrow x=1\\ b,2^3+3x^2-32x=48\\ \Leftrightarrow3x^2-32x+8=48\\ \Leftrightarrow3x^2-32x-40=0\)
Nghiệm xấu lắm bn
\(c,\left(3x+1\right)\left(x-3\right)^2=\left(3x+1\right)\left(2x-5\right)^2\\ \Leftrightarrow c,\left(3x+1\right)\left[\left(2x-5\right)^2-\left(x-3\right)^2\right]\\ \Leftrightarrow\left(3x+1\right)\left(2x-5-x+3\right)\left(2x-5+x-3\right)=0\\ \Leftrightarrow\left(3x+1\right)\left(x-2\right)\left(3x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=2\\x=\dfrac{8}{3}\end{matrix}\right.\)
\(d,9x^2-1=\left(3x+1\right)\left(4x+1\right)\\ \Leftrightarrow\left(3x+1\right)\left(4x+1\right)-\left(3x-1\right)\left(3x+1\right)=0\\ \Leftrightarrow\left(3x+1\right)\left(4x+1-3x+1\right)=0\\ \Leftrightarrow\left(3x+1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-2\end{matrix}\right.\)
3 tháng 3 2022
\(b,2x^3+3x^2-32x-48=0\\ \Leftrightarrow\left(2x^3-8x^2\right)+\left(11x^2-44x\right)+\left(12x-48\right)=0\\ \Leftrightarrow2x^2\left(x-4\right)+11x\left(x-4\right)+12\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(2x^2+11x+12\right)=0\\ \Leftrightarrow\left(x-4\right)\left[\left(2x^2+8x\right)+\left(3x+12\right)\right]=0\\ \Leftrightarrow\left(x-4\right)\left[2x\left(x+4\right)+3\left(x+4\right)\right]=0\\ \Leftrightarrow\left(x-4\right)\left(2x+3\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{3}{2}\\x=-4\end{matrix}\right.\)
NT
2
28 tháng 4 2015
a) \(x=-\frac{1}{2}\)
b) \(x=-\frac{29}{15}\)
c) \(x=\frac{5}{6}\)
**** cho mình nhé
12 tháng 10 2021
b: \(\left(17x-25\right):8+65=9^5:9^3\)
\(\Leftrightarrow\left(17x-25\right):8=9^2-65=81-65=16\)
\(\Leftrightarrow17x-25=128\)
hay x=9
12 tháng 10 2021
b: (17x−25):8+65=95:93(17x−25):8+65=95:93
⇔(17x−25):8=92−65=81−65=16⇔(17x−25):8=92−65=81−65=16
⇔17x−25=128⇔17x−25=128
hay x=9
20 tháng 1 2023
Bài 22:
a: =>-12x+60+21-7x=5
=>-19x+81=5
=>-19x=-76
=>x=4
b: =>30x+60-6x+30-24x=100
=>90=100(loại)
TN
12 tháng 7 2017
\(C=\left(23-x\right)\left(3x+5\right)+13\)
\(=69x+115-3x^2-5x+13\)
\(=-3x^2+64x+128\)
\(=-3\left(x^2-\dfrac{64}{3}x+\dfrac{1024}{9}\right)+\dfrac{1408}{3}\)
\(=-3\left(x-\dfrac{32}{3}\right)^2+\dfrac{1408}{3}\le\dfrac{1408}{3}\)
Vậy \(Max_C=\dfrac{1408}{3}\)
Để \(C=\dfrac{1408}{3}\) thì \(x-\dfrac{32}{3}=0\Rightarrow x=\dfrac{32}{3}\)
d, \(D=\left(2-3x\right)\left(3x+5\right)-7\)
\(=6x+10-9x^2-15x-7\)
\(=-9x^2-9x+3\)
\(=-9\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{21}{4}\)
\(=-9\left(x-\dfrac{1}{2}\right)^2+\dfrac{21}{4}\le\dfrac{21}{4}\)
Vậy \(Max_D=\dfrac{21}{4}\) khi \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)
NA
1
MN
24 tháng 4 2016
a) \(\frac{2}{5}+\frac{3}{5}.\left(3x-3,7\right)=\frac{-53}{10}\)
=> \(\frac{3}{5}.\left(3x-3,7\right)=\frac{-53}{10}-\frac{2}{5}\)
=> \(\frac{3}{5}.\left(3x-3,7\right)=\frac{-57}{10}\)
=> \(\left(3x-3,7\right)=\frac{-19}{5}\)
=>\(3x=\frac{-19}{5}+\frac{37}{10}\)
=>\(3x=\frac{-1}{10}\)
=>\(x=\frac{-1}{30}\)
Ko biết đúng hay sai!!?