9^2x+1=27^3
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1.A =( x-3)( x+3) + 15 - x2
A=X2-3X+3X+15-X3
A=15-X
2.B=(X -1) (X2+X+1) - X (X2+2) + 2X
B=X3+ X2+ X - X2 - X - 1 - X3 - 2X + 2X
B= -1
3.C=(2X - 1 ) (4X2 + 2X + 1) - X ( 8 X 2 + 1 ) + X
C=8X3 - 4X2 +4X2 - 2X +2 X - 1 - 8X22 - X + X
C=8X3 - 1 - 8X22
MK CHỈ LM ĐC TỚI ĐÓ THUI SAI CHỖ NÀO ĐỪNG TRÁCH VÌ MK YẾU PHẦN NÀY

Bài giải
\(4x\left(2x^2-1\right)+27=\left(4x^2+6x+9\right)\left(2x+3\right)\)
\(8x^3-4x+27=8x^3+12x^2+18x+12x^2+18x+27\)
\(8x^3-4x+27=8x^3+24x^2+36x+27\)
\(8x^3-4x+27-8x^3-36x-27=24x^2\)
\(-40x=24x^2\)
\(\frac{3}{5}x^2=x\)
\(\frac{3}{5}x^2-x=0\)
\(x\left(\frac{3}{5}x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\\frac{3}{5}x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\\frac{3}{5}x=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{5}{3}\end{matrix}\right.\)
\(\Rightarrow\text{ }x\in\left\{0\text{ ; }\frac{5}{3}\right\}\)

`@` `\text {Ans}`
`\downarrow`
`3)`
`(2x + 1) \div 9 - 104 \div 13 = 27^2 \div 3^5`
`\Rightarrow (2x + 1) \div 9 - 8 = (3^3)^2 \div 3^5`
`\Rightarrow (2x + 1) \div 9 - 8 = 3^6 \div 3^5`
`\Rightarrow (2x + 1) \div 9 - 8 = 3`
`\Rightarrow (2x + 1) \div 9 = 3 + 8`
`\Rightarrow (2x + 1) \div 9 = 11`
`\Rightarrow 2x + 1 = 11 . 9`
`\Rightarrow 2x + 1 = 99`
`\Rightarrow 2x = 99 - 1`
`\Rightarrow 2x = 98`
`\Rightarrow x = 98 \div 2`
`\Rightarrow x = 49`
Vậy, `x = 49.`
3) \(\left(2x+1\right):9-104:13=27^2:3^5\)
\(\left(2x+1\right):9-8=729:243\)
\(\left(2x+1\right):9-8=3\)
\(\left(2x+1\right):9=8+3\)
\(\left(2x+1\right):9=11\)
( 2x + 1) = 11.9
2x+1 = 99
2x = 99-1
2x = 98
x = 98 : 2
x = 49

\(a,\Rightarrow x=3\)
\(b,\Rightarrow2x-1=2\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\dfrac{3}{2}\)
\(c,\Rightarrow x-2=4\)
\(\Rightarrow x=6\)
\(d,\Rightarrow2x-3=3\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)

\(\left(3-x\right)^3=-\dfrac{27}{64}\)
\(\left(3-x\right)^3=\left(\dfrac{-3}{4}\right)^3\)
\(=>3-x=\dfrac{-3}{4}\)
\(x=3-\dfrac{-3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\)
\(x=\dfrac{15}{4}\)
________
\(\left(x-5\right)^3=\dfrac{1}{-27}\)
\(\left(x-5\right)^3=\left(\dfrac{-1}{3}\right)^3\)
\(=>x-5=\dfrac{-1}{3}\)
\(x=\dfrac{-1}{3}+5=\dfrac{-1}{3}+\dfrac{15}{3}\)
\(x=\dfrac{14}{3}\)
_____________
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8}\)
\(\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{3}{2}\right)^3\)
\(=>x-\dfrac{1}{2}=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}+\dfrac{1}{2}\)
\(x=2\)
________
\(\left(2x-1\right)^2=\dfrac{1}{4}\)
\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\) hoặc \(\left(2x-1\right)^2=\left(\dfrac{-1}{2}\right)^2\)
\(=>2x-1=\dfrac{1}{2}\) \(2x-1=\dfrac{-1}{2}\)
\(2x=\dfrac{1}{2}+1=\dfrac{1}{2}+\dfrac{2}{2}\) \(2x=\dfrac{-1}{2}+1=\dfrac{-1}{2}+\dfrac{2}{2}\)
\(2x=\dfrac{3}{2}\) \(2x=\dfrac{1}{2}\)
\(x=\dfrac{3}{2}:2=\dfrac{3}{2}.\dfrac{1}{2}\) \(x=\dfrac{1}{2}:2=\dfrac{1}{2}.\dfrac{1}{2}\)
\(x=\dfrac{3}{4}\) \(x=\dfrac{1}{4}\)
____________
\(\left(2-3x\right)^2=\dfrac{9}{4}\)
\(\left(2-3x\right)^2=\left(\dfrac{3}{2}\right)^2\) hoặc \(\left(2-3x\right)^2=\left(\dfrac{-3}{2}\right)^2\)
\(=>2-3x=\dfrac{3}{2}\) \(2-3x=\dfrac{-3}{2}\)
\(3x=2-\dfrac{3}{2}=\dfrac{4}{2}-\dfrac{3}{2}\) \(3x=2-\dfrac{-3}{2}=\dfrac{4}{2}+\dfrac{3}{2}\)
\(3x=\dfrac{1}{2}\) \(3x=\dfrac{7}{2}\)
\(x=\dfrac{1}{2}.\dfrac{1}{3}\) \(x=\dfrac{7}{2}.\dfrac{1}{3}\)
\(x=\dfrac{1}{6}\) \(x=\dfrac{7}{6}\)
______________
\(\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\) -> Kiểm tra đề câu này
(3-x)3=(-\(\dfrac{3}{4}\))3
3-x=-\(\dfrac{3}{4}\)
x=3-(-\(\dfrac{3}{4}\))
x=\(\dfrac{15}{4}\)

Ta có: \(4x\left(2x^2-1\right)+27=\left(4x^2+6x+9\right)\left(2x+3\right)\)
\(\Leftrightarrow8x^3-4x+27=8x^3+12x^2+12x^2+18x+18x+27\)
\(\Leftrightarrow8x^3-4x+27-8x^3-24x^2-36x-27=0\)
\(\Leftrightarrow-24x^2-40x=0\)
\(\Leftrightarrow-8x\left(3x+5\right)=0\)
mà -8≠0
nên \(\left[{}\begin{matrix}x=0\\3x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{-5}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\frac{-5}{3}\right\}\)


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
\(9^{2x+1}=27^3\\ =>\left(3^2\right)^{2x+1}=\left(3^3\right)^3\\ =>3^{2\left(2x+1\right)}=3^{3\cdot3}\\ =>3^{4x+2}=3^9\\ =>4x+2=9\\ =>4x=9-2\\ =>4x=7\\ =>x=\dfrac{7}{4}\)
Vậy: ...
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