812x .27x =95
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Ta có:
\(x=28\)
\(\Rightarrow x-1=27\left(1\right)\)
Thay (1) vào biểu thức ta được:
\(x^4-27x^3-27x^2-27x+122\)
\(=x^4-\left(x-1\right)x^3-\left(x-1\right)x^2-\left(x-1\right)x+122\)
\(=x^4-x^4+x^3-x^3+x^2-x^2+x+122\)
\(=x+122\)
\(=28+122\)
\(=150\)
\(1-27x^3\)
\(=1-\left(3x\right)^3\)
\(=\left(1-3x\right)\left(1+3x+9x^2\right)\)
\(---\)
\(x-3^3+27\)
\(=x-27+27=x\)
\(---\)
\(27x^3+27x^2+9x+1\)
\(=\left(3x\right)^3+3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2+1^3\)
\(=\left(3x+1\right)^3\)
\(---\)
\(\dfrac{x^6}{27}-\dfrac{x^4y}{3}+x^2y^2-y^3\) (sửa đề)
\(=\left(\dfrac{x^2}{3}\right)^3-3\cdot\left(\dfrac{x^2}{3}\right)^2\cdot y+3\cdot\dfrac{x^2}{3}\cdot y^2-y^3\)
\(=\left(\dfrac{x^2}{3}-y\right)^3\)
#Ayumu
\(27x^3-27x^2+18x-4\)
\(=27x^3-9x^2-18x^2+6x+12x-4\)
\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)
\(=\left(9x^2-6x+4\right)\left(3x-1\right)\)
Ta có : \(\left(27x^3+27x^2+9x\right)=26\)
\(\Leftrightarrow\left(27x^3+27x^2+9x+1\right)=27\)
\(\Leftrightarrow\left(3x+1\right)^3=27\) \(\Leftrightarrow3x+1=3\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\)
Vậy ....
2x^3 + 27x^2 + 9x = 26
<=> 2x^3 + 27x^2 + 9x - 26 = 0
<=> (3x - 2)(9x^2 + 15x + 13) = 0
vì 9x^2 + 15x + 13 >= 0 nên:
<=> 3x - 2 = 0
<=> 3x = 2
<=> x = 2/3
a) 9x4+16y6-24x2y3
=(3x2)2-2.3x2.4y3+(4y3)2
=(3x2-4y3)2
b) 16x2-24xy+9y2
=(4x)2-2.4x.3y+(3y)2
=(4x-3y)2
c) 36x2-(3x-2)2
=(36x-3x+2)(36x+3x-2)
=(33x+2)(39x-2)
d) 27x3+54x2y+36xy2+8y3
=(3x)3+3.(3x)2.2y+3.3x.(2y)2+(2y)3
=(3x+2y)3
e) y9-9x2y6+27x4y3-27x6
=(y3)3-3.(y3)2.3x2+3.y3.(3x2)2-(3x2)3
=(y3-3x2)3
f) 64x3+1
= (4x)3+13
=(4x+1)[(4x)2-4x.1+12]
=(4x+1)(16x2-4x+1)
e) 27x6-8x3 *sửa đề*
=(3x2)3-(2x)3
=(3x2-2x)[(3x)2+3x2.2x+(2x)2]
=(3x2-2x)(9x2+6x3+4x2)
~~~
ta có :
812x .27x=95
(34)2x.(33)x=(32)5
38x.33x=310
311x=310
->11x=10
->x=\(\frac{10}{11}\)
Vậy x=\(\frac{10}{11}\)
812x .27x =95
\(\Leftrightarrow\)38x . 33x=95
\(\Leftrightarrow\)311x= 310
\(\Leftrightarrow\)11x = 10
\(\Leftrightarrow\)x= 10/11