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TH
25 tháng 10 2021
\(a,\Rightarrow x=3\)
\(b,\Rightarrow2x-1=2\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\dfrac{3}{2}\)
\(c,\Rightarrow x-2=4\)
\(\Rightarrow x=6\)
\(d,\Rightarrow2x-3=3\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
TG
12 tháng 4 2020
a/ \(2x+\frac{1}{7}=\frac{1}{3}\)
=> \(2x=\frac{1}{3}-\frac{1}{7}=\frac{7}{21}-\frac{3}{21}\)
=> \(2x=\frac{4}{21}\)
=> \(x=\frac{4}{21}:2=\frac{4}{21}.\frac{1}{2}=\frac{2}{21}\)
b/ \(3\left(x-\frac{1}{2}\right)=\frac{4}{9}\)
=> \(x-\frac{1}{2}=\frac{4}{9}:3=\frac{4}{9}.\frac{1}{3}\)
=> \(x-\frac{1}{2}=\frac{4}{27}\)
=> \(x=\frac{4}{27}+\frac{1}{2}=\frac{8}{54}+\frac{27}{54}=\frac{35}{54}\)
c/ \(\left(x-5\right)^2+4=68\)
=> \(\left(x-5\right)^2=68-4=64\)
=> \(\left[{}\begin{matrix}x-5=8\\x-5=-8\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=8+5=13\\x=-8+5=-3\end{matrix}\right.\)
d/ \(\left(\left|x\right|-\frac{1}{2}\right)\left(2x+\frac{3}{2}\right)=0\)
=> \(\left[{}\begin{matrix}\left|x\right|-\frac{1}{2}=0\\2x+\frac{3}{2}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left|x\right|=0+\frac{1}{2}=\frac{1}{2}\\2x=0-\frac{3}{2}=-\frac{3}{2}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\\x=-\frac{3}{2}:2=-\frac{3}{2}.\frac{1}{2}=-\frac{3}{4}\end{matrix}\right.\)
e) \(5x+2=3x+8\)
=> \(5x-3x=8-2=6\)
=> \(2x=6\)
=> \(x=6:2=3\)
f/ \(26-\left(5-2x\right)=27\)
=> \(5-2x=26-27=-1\)
=> \(2x=5-\left(-1\right)=5+1=6\)
=> \(x=6:2=3\)
g/ \(\left(4x-8\right)-\left(2x-6\right)=4\)
=> \(4x-8-2x+6=4\)
=> \(\left(4x-2x\right)+\left(-8+6\right)=4\)
=> \(2x+-2=4\)
=> \(2x=4+2=6\)
=> \(x=6:2=3\)
h/ \(\left(x+3\right)^3:3-1=-10\)
=> \(\left(x+3\right)^3:3=-10+1=-9\)
=> \(\left(x+3\right)^3=-9.3=-27\)
=> \(x+3=-3\)
=> \(x=-3-3=-6\)
H9
HT.Phong (9A5)
CTVHS
18 tháng 7 2023
\(A=5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9\)
\(A=5\cdot\left(2^2\right)^{15}\cdot\left(3^2\right)^9-2^2\cdot3^{20}\cdot\left(2^3\right)^9\)
\(A=5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}\)
\(A=5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}\)
\(A=2^{29}\cdot3^{18}\cdot\left(5\cdot2^1\cdot1-1\cdot3^2\right)\)
\(A=2^{29}\cdot3^{18}\cdot\left(5-9\right)\)
\(A=-2^2\cdot2^{29}\cdot3^{18}\)
\(A=-2^{31}\cdot3^{18}\)
_______________
\(B=5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6\)
\(B=5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot\left(3^3\right)^6\)
\(B=5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}\)
\(B=2^{28}\cdot3^{18}\cdot\left(5\cdot1\cdot3-7\cdot2\cdot1\right)\)
\(B=2^{28}\cdot3^{18}\cdot\left(15-14\right)\)
\(B=2^{28}\cdot3^{18}\)
Ta có: \(A:B\)
\(=\left(-2^{31}\cdot3^{18}\right):\left(2^{28}\cdot3^{18}\right)\)
\(=\left(-2^{31}:2^{28}\right)\cdot\left(3^{18}:3^{18}\right)\)
\(=-2^3\cdot1\)
\(=-8\)
KM
16 tháng 8 2019
a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(=>2x+\frac{3}{5}=\frac{3}{5}\)
\(2x=\frac{3}{5}-\frac{3}{5}\)
\(2x=0\)
\(x=0:2\)
\(x=0\)
KM
16 tháng 8 2019
b) \(\left(3x-1\right).\left(-\frac{1}{2x}+5\right)=0\)
=> \(\left(3x-1\right)=0\)hoặc \(\left(-\frac{1}{2x}+5\right)=0\)hoặc \(\left(3x-1\right)\)và\(\left(-\frac{1}{2x}+5\right)\)cùng bằng 0.
\(\orbr{\begin{cases}3x-1=0\\-\frac{1}{2x}+5=0\end{cases}}=>\orbr{\begin{cases}3x=1\\-\frac{1}{2x}=-5\end{cases}}=>\orbr{\begin{cases}x\in\varnothing\\2x=\frac{1}{5}\end{cases}}=>x=\frac{1}{5}:2=>x=\frac{1}{10}\)
3T
2
31 tháng 10 2021
\(a,\Rightarrow x=3\\ b,\Rightarrow2x-1=2\Rightarrow x=\dfrac{3}{2}\\ c,\Rightarrow\left[{}\begin{matrix}x-2=4\\x-2=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\\ d,\Rightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\\ e,\Rightarrow2x+5=3^2=9\Rightarrow x=2\)
14 tháng 10 2021
\(1,\\ a,\Leftrightarrow4^{5-x}=4^2\Leftrightarrow5-x=2\Leftrightarrow x=3\\ b,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x+1=3\Leftrightarrow x=2\\ 2,\\ a,3^{100}=\left(3^2\right)^{50}=9^{50}\\ b,2^{98}=\left(2^2\right)^{49}=4^{49}< 9^{49}\\ c,5^{30}=5^{29}\cdot5< 6\cdot5^{29}\\ d,3^{30}=\left(3^3\right)^{10}=27^{10}>8^{10}\\ 4,\\ a,\Leftrightarrow5\left(x-10\right)=10\\ \Leftrightarrow x-10=2\Leftrightarrow x=12\\ b,\Leftrightarrow3\left(70-x\right)+5=92\\ \Leftrightarrow3\left(70-x\right)=87\\ \Leftrightarrow70-x=29\\ \Leftrightarrow x=41\\ c,\Leftrightarrow16+x-5=315-230=85\\ \Leftrightarrow x=74\\ d,\Leftrightarrow2^x-5+74=707:\left(16-9\right)=707:7=101\\ \Leftrightarrow2^x=32=2^5\\ \Leftrightarrow x=5\)
S
3 tháng 5 2020
a) (4x – 2)(x + 5) = 0
⇔ \(\left[{}\begin{matrix}4x-2=0\\x+5=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}4x=2\\x=-5\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=\frac{1}{2}\\x=-5\end{matrix}\right.\)
Vậy ...
b) 2x – 9 = -8 – 9
⇔ 2x = - 8
⇔ x = - 4
Vậy x = - 4
c) 3.| x -1 | - 27 = 0
⇔ 3 . | x - 1| = 27
⇔ | x - 1| = 9
⇔ \(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=10\\x=-8\end{matrix}\right.\)
Vậy x ∈ { 10 ; - 8}
d) 5.(3x + 8) –7.(2x + 3) = 16
⇔ 15x + 40 - 14x - 21 = 16
⇔ x + 19 = 16
⇔ x = - 3
Vậy ..
NB
3 tháng 5 2020
a)\(\left(4x-2\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}4x-2=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-5\end{matrix}\right.\)
Vậy...
b)\(2x-9=-8-9\\ \Leftrightarrow2x-9=-17\\ \Leftrightarrow2x=-8\\ \Leftrightarrow x=-4\)
Vậy...
c)\(2\left|x-1\right|-27=0\\ \Leftrightarrow\left|x-1\right|=\frac{27}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x-1=\frac{27}{2}\\x-1=-\frac{27}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{29}{2}\\x=-\frac{25}{2}\end{matrix}\right.\)
Vậy...
a. 2x - 9 = -8 -9
=> 2x - 9 = -17
=> 2x = - 17 + 9
=> 2x = -8
=> x = -4
Vậy x = -4
b. 3 . | x - 1 | - 27 = 0
=> 3 . | x - 1 | = 27
=> | x - 1| = 9
=> x -1 =9 hoặc x-1 = -9
Với x - 1 =9
=> x = 10
Với x - 1 = -9
=> x = -8
Vậy .....
a)2x-9=-8-9
2x-9=-17
2x=-17+9
2x=-8
x=-8:2
x=-4
vậy x=-4
b)3.|x-1|-27=0
3.|x-1|=0+27
3.|x-1|=27
|x-1|=27:3
|x-1|=9
* x-1=9 * x-1=-9
x=9+1 x=-9+1
x=10 x=-8
vậy x=10 hoặc x=-8