Tìm x biết
\(3^{2x+2}+9^x=270\)
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3x + 4 + 3x + 2 = 270
35 + 33 = 270
= > x = 5 - 4 = 1
=> x = 3 - 2 = 1
=> x = 1
3x + 4 + 3x + 2 = 270
10.3x + 2 = 270
3x + 2 = 270 : 10
3x + 2 = 27
3x + 2 = 33
x + 2 = 3
x = 3 - 2
x = 1
[(2x - 4) + 9]3 = 36
(2x - 4 + 9)3 = 729
x = 2
a)\(280-\left(x-140\right):35=270\)
\(\Leftrightarrow\left(x-140\right):35=10\)
\(\Leftrightarrow x-140=350\)
\(\Leftrightarrow x=490\)
b) \(\left(190-2x\right):35-32=16\)
\(\Leftrightarrow\left(190-2x\right):35=48\)
\(\Leftrightarrow190-2x=1680\)
\(\Leftrightarrow2x=-1490\Leftrightarrow x=-745\)
c) \(720:\left[41-\left(2x-5\right)\right]=2^3.5\)
\(\Leftrightarrow720:\left[41-\left(2x-5\right)\right]=40\)
\(\Leftrightarrow41-\left(2x-5\right)=18\)
\(\Leftrightarrow2x-5=23\)
\(\Leftrightarrow2x=28\Leftrightarrow x=14\)
280-(x-140):35=270
(x-140):35=280-270
(x-140):35=10
(x-140)=10×35
(x-140)=350
x=350-140
x=210
vậy x=210
(192-2x):35-32=16
(192-2x):35=16+32
(192-2x):35=48
(192-2x)=48×35
(192-2x)=1680
2x=192-1680
2x=-1488
x=-1488:2
x=-744
vậy x=-744
\(3^{x+4}+3^{x+2}=270\)
\(3^{x+2+2}+3^{x+2}=270\)
\(3^{x+2}\cdot3^2+3^{x+2}\cdot1=270\)
\(3^{x+2}\left(3^2+1\right)=270\)
\(3^{x+2}\cdot10=270\)
\(3^{x+2}=270:10\)
\(3^{x+2}=27\)
\(3^{x+2}=3^3\)
\(x+2=3\)
\(x=1\)
3x + 3x+2 = 270
=> 3x.(1 + 32) = 270
=> 3x.(1 + 9) = 270
=> 3x.10 = 270
=> 3x = 270 : 10
=> 3x = 27
=> 3x = 33
=> x = 3
\(3^x+3^{x+2}=270\)
\(=>3^x\left(1+3^2\right)=270\)
\(=>3^x.10=270\)
\(=>3^x=27\)
\(=>x=3\)
\(3^{x+2}+3^x=270\)
=> \(3^x\cdot3^2+3^x\cdot1=270\)
=> \(3^x\left(3^2+1\right)=270\)
=> \(3^x\cdot10=270\)
=> \(3^x=270:10=27\)
=> \(x=3\)
3x+2 + 3x = 270
=> 3x x 32 + 3x = 270
3x x(32 + 1) =270
3x x 10 = 270
3x = 27
3x = 33
=> x = 3
\(3^{x+4}+3^{x+2}=270\Rightarrow3^{x+2}\left(3^2+1\right)\)\(=270\Rightarrow3^{x+2}\cdot10=270\)\(=>3^{x+2}=270:10=27=>x=1\)
a, 7\(x\) - \(x\) = 521 : 519 + 3.22.7
6\(x\) = 53 + 3.4.7
6\(x\) = 125 + 12.7
6\(x\) = 125 + 84
6\(x\) = 209
\(x\) = 209 : 6
\(x\) = \(\dfrac{209}{6}\)
b; 11\(x\) - 7\(x\) + 34 : 33 = 54 + 2\(x\)
4\(x\) + 3 = 625 + 2\(x\)
4\(x\) - 2\(x\) = 625 - 3
2\(x\) = 622
\(x\) = 622 : 2
\(x\) = 311
c; 75 - 5.(\(x-3\))3 = 700
5.(\(x\) - 3)3 = 700 - 75
5.(\(x\) - 3)3 = - 625
(\(x\) - 30)3 = - 625 : 5
(\(x\) - 30)3 = - 125
(\(x-3\))3 = (-5)3
\(x\) - 3 = - 5
\(x\) = - 5 + 3
\(x\) = -2
d, 3.(2\(x\) - 1)2 = 75
(2\(x\) - 1)2 = 75 : 3
(2\(x\) - 1)2 = 25
\(\left[{}\begin{matrix}2x-1=-5\\2x-1=5\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=-5+1\\2x=5+1\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=-4\\2x=6\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
`Answer:`
a. \(x^3+6x^2+12=19\)
\(\Leftrightarrow x^3+6x^2+12x-19=0\)
\(\Leftrightarrow x^3-x^2+7x^2-7x+19x-19=0\)
\(\Leftrightarrow x^2.\left(x-1\right)+7x\left(x-1\right)+19\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+7x+19\right)=0\)
Ta có \(x^2+7x+19=x^2+2x.3,5+12,25+6,75=\left(x+3,5\right)^2+6,75>0\)
\(\Rightarrow x-1=0\Leftrightarrow x=1\)
b. \(5\left(x+9\right)^2.\left(x-4\right)^3-10\left(x+9\right)^3.\left(x-4\right)^2=0\)
\(\Leftrightarrow5\left(x+9\right)^2.\left(x-4\right)^2.[x-4-2\left(x+9\right)]=0\)
\(\Leftrightarrow\left(x+9\right)^2.\left(x-4\right)^2.\left(x-4-2x-18\right)=0\)
\(\Leftrightarrow\left(x+9\right)^2.\left(x-4\right)^2.\left(-x-22\right)=0\)
\(\Leftrightarrow\left(x+9\right)^2=0\) hoặc \(\left(x-4\right)^2=0\) hoặc \(-x-22=0\)
\(\Leftrightarrow x+9=0\) hoặc \(x-4=0\) hoặc \(-x=22\)
\(\Leftrightarrow x=-9\) hoặc \(x=4\) hoặc \(x=-22\)
c. \(\left(2x+3\right)^2+\left(x-2\right)^2-2\left(2x+3\right)\left(x-2\right)\)
\(=\left(2x+3\right)^2-2\left(2x+3\right)\left(x-2\right)+\left(x-2\right)^2\)
\(=\left(2x+3-x+2\right)^2\)
\(=\left(x+5\right)^2\)
\(3^{2x+2}+9^x=270\)
\(\Leftrightarrow3^{2x+2}+3^{2x}=270\)
\(\Leftrightarrow3^{2x}\left(3^2+1\right)=270\)
\(\Leftrightarrow3^{2x}.10=270\)
\(\Leftrightarrow3^{2x}=27\)
\(\Leftrightarrow3^{2x}=3^3\)
\(\Leftrightarrow2x=3\)
\(\Leftrightarrow x=\frac{3}{2}\)
Vậy: \(x=\frac{3}{2}\)