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AT
19 tháng 3 2020
a, 4.(x-3)=72-13
=> 4(x - 3) = 49 - 1
=> 4(x - 3) = 48
=> x - 3 = 12
=> x = 15
b, 5x+x=150:2+3
=> 6x + = 75 + 3
=> 6x = 78
=> x = 13
19 tháng 3 2020
a) 4(x-3)=72-13
<=> 4(x-3)=49-1
<=> 4(x-3)=48
<=> x-3=12
<=> x=15
Vậy x=15
b) 5x+x=150:2+3
<=> 6x=75+3
<=> 6x=78
<=> x=13
Vậy x=13
6 tháng 4 2023
a: =>x/27+1=-2/3
=>x/27=-5/3
=>x=-45
b: \(\Leftrightarrow x-4=\dfrac{2}{5}:\dfrac{20}{21}=\dfrac{2}{5}\cdot\dfrac{21}{20}=\dfrac{42}{100}=\dfrac{21}{50}\)
=>x=221/50
c: \(\Leftrightarrow x+\dfrac{2}{3}=\dfrac{4}{60}=\dfrac{1}{15}\)
=>x=1/15-2/3=1/15-10/15=-9/15=-3/5
d: \(\Leftrightarrow x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{15}{14}\cdot\dfrac{21}{20}\)
=>\(x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{3}{2}\cdot\dfrac{3}{4}=\dfrac{1}{5}-\dfrac{9}{8}=\dfrac{-37}{40}\)
=>x=-37/24
e: =>-3/7x=84/45
=>x=-196/45
f: =>11/10x=-2/3
=>x=-20/33
S
26 tháng 4 2019
\(a)\left(3\frac{1}{2}-x\right).1\frac{1}{4}=\frac{15}{6}\)
\(\left(\frac{7}{2}-x\right).\frac{5}{4}=\frac{15}{6}\)
\(\frac{7}{2}-x=\frac{15}{6}:\frac{5}{4}\)
\(\frac{7}{2}-x=2\)
\(x=\frac{7}{2}-2\)
\(\Rightarrow x=\frac{3}{2}\)
N
1
20 tháng 1 2022
a: \(\Leftrightarrow\left|x\cdot\dfrac{7}{3}-\dfrac{3}{4}\right|=1+\dfrac{1}{3}+\dfrac{2}{3}=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{7}{3}-\dfrac{3}{4}=2\\x\cdot\dfrac{7}{3}-\dfrac{3}{4}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{33}{28}\\x=-\dfrac{15}{28}\end{matrix}\right.\)
b: \(\Leftrightarrow\left|x\cdot\dfrac{2}{3}-\dfrac{1}{3}\right|=\dfrac{6}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{2}{3}-\dfrac{1}{3}=-\dfrac{6}{5}\\x\cdot\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-13}{10}\\x=\dfrac{23}{10}\end{matrix}\right.\)
H9
HT.Phong (9A5)
CTVHS
9 tháng 7 2023
a) Đặt: \(A=1+2^2+2^3+...+2^{10}\)
\(\Rightarrow2A=2\left(1+2^2+2^3+...+2^9+2^{10}\right)\)
\(\Rightarrow2A=2+2^3+2^4+...+2^{10}+2^{11}\)
\(\Rightarrow2A-A=\left(2+2^3+2^4+...+2^{10}+2^{11}\right)-\left(1+2^2+2^3+...+2^{10}\right)\)
\(\Rightarrow A=\left(2^3-2^3\right)+\left(2^4-2^4\right)+...+\left(2-1\right)+\left(2^{11}-2^2\right)\)
\(\Rightarrow A=0+0+...+1+\left(2^{11}-2^2\right)\)
\(\Rightarrow A=1+2^{11}-2^2=1+2048-4=2045\)
Vậy: \(1+2^2+2^3+...+2^{10}=2045\)
b)
a] \(60-3\left(x-1\right)=2^3\cdot3\)
\(\Rightarrow60-3\left(x-1\right)=24\)
\(\Rightarrow3\left(x-1\right)=36\)
\(\Rightarrow x-1=12\)
\(\Rightarrow x=13\)
b] \(\left(3x-2\right)^3=2\cdot2^5\)
\(\Rightarrow\left(3x-2\right)^3=2^6\)
\(\Rightarrow\left(3x-2\right)^3=\left(2^2\right)^3\)
\(\Rightarrow3x-2=2^2\)
\(\Rightarrow3x=6\)
\(x=2\)
c] \(5^{x+1}-5^x=500\)
\(\Rightarrow5^x\left(5-1\right)=500\)
\(\Rightarrow5^x\cdot4=500\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
d] \(x^2=x^4\)
\(\Rightarrow x=x^2\)
\(\Rightarrow x-x^2=0\)
\(\Rightarrow x\left(1-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\1-x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(a,4\left(x-3\right)=7^2-1^3\)
\(4\left(x-3\right)=49-1\)
\(4\left(x-3\right)=48\)
\(\Rightarrow x-3=48:4\)
\(x-3=24\)
\(\Rightarrow x=27\)
\(5x+x=\frac{150}{2}-3\)
\(6x=75-3\)
\(6x=72\)
\(\Rightarrow x=12\)
a, 4( x - 3) = 72 - 13
<=> 4x - 12 = 49 - 1
<=> 4x = 49 - 1 +12
<=> x = 15
Vậy....
b, 5x + x = \(\frac{150}{2}\)+ 3
<=> 5x + x = 75 + 3
<=> 6x = 78
<=> x = 13
Vậy......