BT1: Tìm x\(\in\)N biết
1) \(\dfrac{2^{2x-3}}{4^{10}}=8^3.16^5\)
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M
21 tháng 9 2018
\(\dfrac{2^{2x-3}}{4^{10}}=8^3.16^5\)
=> \(2^{2x-3}:4^{10}=8^3.16^5\)
=> \(2^{2x-3}.\left(2^2\right)^{10}=\left(2^3\right)^3.\left(2^4\right)^5\)
=> \(2^{2x-3}:2^{20}=2^9.2^{20}\)
=> \(2^{2x-3}:2^{20}\) = 229
=> \(2^{2x-3-20}=2^{29}\)
\(\Leftrightarrow\) \(2x-3-20=29\)
=> \(2x-3=29+20\)=49
=> \(2x=49+3\) = 52
=> \(x=52:2\) => x=26
D
2
BN
12 tháng 9 2019
a, Tự chép đề bài :v
=> 22x-3 = ( 83. 165 ) : 410
22x-3 = ( 29. 220 ) : 220
22x-3 = 229 : 220
22x-3 = 29
=> 2x - 3 = 9
2x = 9 + 3
2x = 12
x = 6
Vậy....
b, 7. 2x = 29 + 5. 28
7. 2x = 1792
2x = 1792 : 7
2x = 256
2x = 28
=> x = 8
Vậy ....
M
12 tháng 9 2019
a \(\frac{2^{2x-3}}{4^{10}}=8^3.16^5\)
\(\Leftrightarrow\frac{2^{2x-3}}{4^{10}}=2^{29}\)
\(\Leftrightarrow2^{2x-3}=2^{29}.4^{10}\)
\(\Leftrightarrow2^{2x-3}=2^{49}\)
\(\Leftrightarrow2x-3=49\)
\(\Leftrightarrow x=26\)
b \(7.2^x=2^9+5.2^8\)
\(\Leftrightarrow7.2^x=2^8.(2+5)\)
\(\Leftrightarrow7.2^x=7.2^8\)
\(\Leftrightarrow x=7\)
26 tháng 5 2022
a: \(\dfrac{5^5}{5^x}=5^{18}\)
=>5-x=18
hay x=-13
b: \(\dfrac{2^{4-x}}{16^5}=32^6\)
\(\Leftrightarrow2^{4-x}=\left(2^5\right)^6\cdot\left(2^4\right)^5=2^{30+20}=2^{50}\)
=>4-x=50
hay x=-46
c: \(\dfrac{2^{2x-3}}{4^{10}}=8^3\cdot16^5\)
\(\Leftrightarrow2^{2x-3}=2^9\cdot2^{20}\cdot2^{20}=2^{49}\)
=>2x-3=49
=>2x=52
hay x=26
d: \(\dfrac{2^3}{2^x}=4^5\)
\(\Leftrightarrow2^{3-x}=2^{10}\)
=>3-x=10
hay x=-7
e: \(9\cdot5^x=6\cdot5^6+3\cdot5^6\)
\(\Leftrightarrow9\cdot5^x=9\cdot5^6\)
\(\Leftrightarrow5^x=5^6\)
hay x=6
f: \(7\cdot2^x=2^9+5\cdot2^8\)
\(\Leftrightarrow2^x\cdot7=2^8\cdot7\)
\(\Leftrightarrow2^x=2^8\)
hay x=8
29 tháng 7 2021
a) Ta có: \(\dfrac{4}{5}-3\left|x\right|=\dfrac{1}{5}\)
\(\Leftrightarrow3\left|x\right|=\dfrac{4}{5}-\dfrac{1}{5}=\dfrac{3}{5}\)
\(\Leftrightarrow\left|x\right|=\dfrac{1}{5}\)
hay \(x\in\left\{\dfrac{1}{5};-\dfrac{1}{5}\right\}\)
b) Ta có: \(4x-\dfrac{1}{2}x+\dfrac{3}{5}x=\dfrac{4}{5}\)
nên \(\dfrac{41}{10}x=\dfrac{4}{5}\)
hay \(x=\dfrac{8}{41}\)
c) Ta có: \(\left(2x-8\right)\left(10-5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-8=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=8\\5x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
d) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}\)
\(\Leftrightarrow\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}-\dfrac{3}{4}=\dfrac{14}{4}-\dfrac{3}{4}=\dfrac{11}{4}\)
\(\Leftrightarrow\left|2x-1\right|=11\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=11\\2x-1=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=12\\2x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\)
18 tháng 9 2021
1) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{11}{12}-\dfrac{2}{5}-x=\dfrac{2}{3}\)
\(\Leftrightarrow x=\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}\)
\(\Leftrightarrow x=-\dfrac{3}{20}\)
2) \(2x\left(x-\dfrac{1}{7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
3) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4x}=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{1}{4x}=-\dfrac{7}{20}\)
\(\Leftrightarrow4x=-\dfrac{20}{7}\)
\(\Leftrightarrow x=-\dfrac{5}{7}\)
VD
21 tháng 3 2022
\(a,\dfrac{x}{8}=\dfrac{7}{-2}\\ \Rightarrow x=-28\\ b,\dfrac{1-2x}{6}=\dfrac{-1}{2}\\ \Leftrightarrow2-4x=-6\\ \Leftrightarrow4x=8\\ \Leftrightarrow x=2\\ c,\dfrac{x+2}{3}=\dfrac{x+3}{4}\\ \Leftrightarrow4x+8=3x+9\\ \Leftrightarrow x=1\\ d,\dfrac{10}{2-x}=2\\ \Leftrightarrow4-2x=10\\ \Leftrightarrow2x=-6\\ \Leftrightarrow x=-3\)
H9
HT.Phong (9A5)
CTVHS
11 tháng 7 2023
a) \(\left(\dfrac{x^2}{2}+y^2\right)^2\)
\(=\left(\dfrac{1}{2}x^2+y^2\right)^2\)
\(=\left(\dfrac{1}{2}x^2\right)^2+2\cdot\dfrac{1}{2}x^2\cdot y^2+\left(y^2\right)^2\)
\(=\dfrac{1}{4}x^4+x^2y^2+y^4\)
b) \(\left(\dfrac{4}{5}x^2-\dfrac{2}{3}y\right)^2\)
\(=\left(\dfrac{4}{5}x^2\right)^2-2\cdot\dfrac{4}{5}x^2\cdot\dfrac{2}{3}y+\left(\dfrac{2}{3}y\right)^2\)
\(=\dfrac{16}{25}x^4-\dfrac{16}{15}x^2y+\dfrac{4}{9}y^2\)
c) \(\left(2x+\dfrac{1}{2}\right)\left(2x-\dfrac{1}{2}\right)\)
\(=\left(2x\right)^2-\left(\dfrac{1}{2}\right)^2\)
\(=4x^2-\dfrac{1}{4}\)
10 tháng 7 2023
a: (1/2x^2+y^2)^2
=(1/2x^2)^2+2*1/2x^2*y^2+y^4
=1/4x^4+x^2y^2+y^4
b: (4/5x^2-2/3y)^2
=(4/5x^2)^2-2*4/5x^2*2/3y+4/9y^2
=16/25x^4-16/15x^2y+4/9y^2
c: =(2x)^2-(1/2)^2
=4x^2-1/4
1 tháng 6 2022
1/2+1/3<x<=1+1/2+1/5
=>5/6<x<=1+7/10
=>5/6<x<17/10
mà x là số nguyên
nên x=1
\(\dfrac{2^{2x-3}}{4^{10}}=8^3.16^5\)
\(\Rightarrow2^{2x-3}=8^3.16^5.4^{10}\)
\(\Rightarrow2^{2x-3}=2^9.2^{20}.2^{20}\)
\(\Rightarrow2^{2x-3}=2^{49}\)
Vì \(2\ne\pm1;2\ne0\) nên \(2x-3=49\)
\(\Rightarrow2x=52\Rightarrow x=26\)
Vậy..........
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