Tìm x biết:
(x - 2/9)3 = (2/3)6
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Bài làm
( 3 + 6 + x ) . 9 = ( 2 - x ) . 2
27 + 54 + 9x = 4 - 2x
9x + 2x = 4 - 27 - 54
11x = -77
x = -7
Vậy x = -7
b) ( 6 - x ) . 3 = ( 9 + x ) . 3
18 - 3x = 27 + 3x
-3x - 3x = 27 - 18
-6x = 9
x = 9/-6
x = -3/2
Vậy x = -3/2
a,\(\text{(3+6+x) . 9 = (2-x).2}\)
\(27+54+9x=4-2x\)
\(9x+2x=4-54-27\)
\(11x=-77\)
\(\Rightarrow x=-7\)
b, \(\text{(6-x) . 3 = (9+x) . 3}\)
\(18-3x=27+3x\)
\(-3x-3x=27-18\)
\(-6x=9\)
\(\Rightarrow x=\frac{9}{-6}\)
học tốt
\(\Leftrightarrow x^2-36-x^2+12x-9=9\)
\(\Leftrightarrow12x=54\)
hay x=9/2
Ta có
( x – 6 ) ( x + 6 ) – ( x + 3 ) 2 = 9 ⇔ x 2 – 36 – ( x 2 + 6 x + 9 ) = 9 ⇔ x 2 – 36 – x 2 – 6 x – 9 – 9 = 0
ó - 6x – 54 = 0 ó 6x = -54 ó x = -9
Vậy x = -9
Đáp án cần chọn là: A
d. (x - 3)(x2 + 3x + 9) + x(x + 2)(2 - x) = 1
<=> x3 - 9 + (x2 + 2x)(2 - x) = 1
<=> x3 - 9 + 2x2 - x3 + 4x - 2x2 = 1
<=> 4x = 10
<=> x = \(\dfrac{10}{4}=\dfrac{5}{2}\)
d)(x - 3)(x^2 + 3x + 9) + x(x + 2)(2 - x) = 1
\(<=> x^3-27-x(x^2-4)=1\)
\(<=> x^3-27-x^3-4x=1<=>-4x=28<=> x=-7\)
=> ptrình có tập nghiệm S={-7}
e) (x + 1)^3 - (x - 1)^3 - 6(x - 1)^2 = -19
\(<=> x^3+3x^2+3x+1-(x^3-3x^2+3x-1)-6(x^2-2x+1)+19=0\)
\(<=>x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+19=0\)
\(<=>12x=15<=>x=12/15 \)
=> ptrình có tập nghiệm S={12/15}
x^3 -9x^2 +27x -27 -(x^3 -27) +6(x^2 +2x+1) +3x^2 =-33
x^3 -9x^2 +27x -27 -x^3 +27 +6x^2 + 12x+ 6 +3x^2 =-33
39x+6=-33
39x=-39
x=-1
Vậy x=-1
\(a,\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=0\\ \Rightarrow\left(x^3-27\right)+x\left(4-x^2\right)=0\\ \Rightarrow x^3-27+4x-x^3=0\\ \Rightarrow4x-27=0\\ \Rightarrow4x=27\\ \Rightarrow x=\dfrac{27}{4}\)
\(b,\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\\ \Rightarrow\left(x^3+3x^2+3x+1\right)-\left(x^3-3x^2+3x-1\right)-6\left(x^2-2x+1\right)=-10\\ \Rightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+10=0\)
\(\Rightarrow12x+6=0\\ \Rightarrow12x=-6\\ \Rightarrow x=-\dfrac{1}{2}\)
a)-7.(x+9)-3.(5-x)=2
-7.x-63-15+3.x=2
-4.x-78=2
-4.x=2+78=80
x=80:(-4)
x=-20
b)3.(x+2)-6.(x-5)=2.(5-2x)
3x+6-6x+30=10-4.x2
9x+36=6.x2
Tới đây bn tự làm nha vì mk bí ý tưởng rùi
\(a,\left(3x+x\right)\left(x^2-9\right)-\left(x-3\right)\left(x^2+3x+9\right)\)
\(=4x\left(x^2-9\right)-x^3+27\)
\(=4x^3-36x-x^3+27\)
\(=3x^3-36x+27\)
\(\left(x+6\right)^2-2x.\left(x+6\right)+\left(x-6\right).\left(x+6\right)\)
\(=\left(x+6\right).\left(x+6-2x+x-6\right)\)
\(=\left(x+6\right).0\)
\(=0\)
\(\left(x-\frac{2}{9}\right)^3=\left(\left(\frac{2}{3}\right)^2\right)^3\)
\(\left(x-\frac{2}{9}\right)^3=\left(\frac{4}{9}\right)^3\)
\(x-\frac{2}{9}=\frac{4}{9}\)
\(x=\frac{2}{3}\)
Bài làm:
Ta có: \(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)
\(\Leftrightarrow\left(x-\frac{2}{9}\right)^3=\left(\frac{4}{9}\right)^3\)
\(\Leftrightarrow x-\frac{2}{9}=\frac{4}{9}\)
\(\Rightarrow x=\frac{6}{9}=\frac{2}{3}\)