Tìm x biết ( x – 6 ) ( x + 6 ) – ( x + 3 ) 2 = 9
A. x = -9
B. x = 9
C. x = 1
D. x = -6
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a) 6 . x - 15 = 15
6 . x = 15 + 15
6 . x = 30
x = 30 : 6
x = 5
b) 2 . x + 3 = 9
2 . x = 9 - 3
2 . x = 6
x = 6 : 2
x = 3
c) 17 - 6 : x = 15
6 : x = 17 - 15
6 : x = 2
x = 6 : 2
x = 3
d) x : 3 + 12 = 14
x : 3 = 14 - 12
x : 3 = 2
x = 2 . 3
x = 6
a) 6 . x -15 = 15
⇒ 6 .x = 15 + (-15)
⇒ 6 .x = 0
⇔ x = 0
Vậy x = 0.
a) \(\dfrac{x}{12}=\dfrac{6}{9}\)
\(\Rightarrow x=\dfrac{12\cdot6}{9}=8\)
b) \(\dfrac{9}{15}=x+\dfrac{11}{35}\)
\(\Rightarrow x=\dfrac{9}{15}-\dfrac{11}{35}\)
\(\Rightarrow x=\dfrac{2}{7}\)
c) \(x+\dfrac{6}{5}=\dfrac{x}{2}\)
\(\Rightarrow2\left(x+\dfrac{6}{5}\right)=x\)
\(\Rightarrow2x+\dfrac{12}{5}=x\)
\(\Rightarrow2x-x=-\dfrac{12}{5}\)
\(\Rightarrow x=-\dfrac{12}{5}\)
\(a,\dfrac{x}{12}=\dfrac{6}{9}\\ \Leftrightarrow x=12\cdot\dfrac{6}{9}\\ \Leftrightarrow x=8\\ b,\dfrac{9}{15}=x+\dfrac{11}{35}\\ \Leftrightarrow x=\dfrac{9}{15}-\dfrac{11}{35}\\ \Leftrightarrow x=\dfrac{2}{7}\\ c,x+\dfrac{6}{5}=\dfrac{x}{2}\\ \Leftrightarrow\dfrac{x}{2}=-\dfrac{6}{5}\\ \Leftrightarrow x=-\dfrac{12}{5}\)
\(a,\left(3x+x\right)\left(x^2-9\right)-\left(x-3\right)\left(x^2+3x+9\right)\)
\(=4x\left(x^2-9\right)-x^3+27\)
\(=4x^3-36x-x^3+27\)
\(=3x^3-36x+27\)
\(\left(x+6\right)^2-2x.\left(x+6\right)+\left(x-6\right).\left(x+6\right)\)
\(=\left(x+6\right).\left(x+6-2x+x-6\right)\)
\(=\left(x+6\right).0\)
\(=0\)
\(a,2+x=6\)
\(x=6-2\)
\(x=4\)
\(b,\left(2-2\right).9+x=6:\left(1+2+3\right)\)
\(x=6:6\)
\(x=1\)
a, 2 + x = 6
x = 6 - 2
x = 4
b) [(2 - 2) . 9 + x = [6 : (1 + 2 + 3 )]
[ 0 . 9 + x ] = [ 6 : 6 ]
[ 0 + x ] = 1
x = 1 - 0
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)\(x^2-4^2+6x-x^2=0\)
\(16+6x=0\)
\(x=\frac{8}{3}\)
b)x=3
a,\(\left(x-15\right):50+22=24\)
\(< =>\frac{\left(x-15\right)}{50}=2< =>x-15=100\)
\(< =>x=100+15=115\)
b,\(42-\left(2x+32\right)+12:2=6\)
\(< =>42-2x-32=0\)
\(< =>10-2x=0< =>x=\frac{10}{2}=5\)
Làm nốt :
c) \(134-2\left\{156-6\cdot\left[54-2\cdot\left(9+6\right)\right]\right\}\cdot x=86\)
=> 134 - 2{156 - 6 . [54 - 2 . 15]} . x = 86
=> 134 - 2{156 - 6 . [54 - 30]} . x = 86
=> 134 - 2{156 - 6. 24} . x = 86
=> 134 - 2{156 - 144} . x = 86
=> 134 - 2.12 . x = 86
=> 134 - 24 . x = 86
=> 24.x = 48
=> x = 2
Bài 2 : a) 120 : [21 - (4x - 4)] = 23.3
=> 120 : [21 - (4x - 4)] = 8.3
=> 120 : [21 - (4x - 4)] = 24
=> 21 - (4x - 4) = 5
=> 4x - 4 = 16
=> 4x = 20
=> x = 5
b) 3.[205 - (x - 9)] - 486 = 0
=> 3.[205 - (x - 9)] = 486
=> 205 - (x - 9) = 162
=> x - 9 = 205 - 162 = 43
=> x = 43 + 9 = 52
c) 204 - 2{200 - 5.[64 - 2.(11 + 6)]} . x = 4
=> 204 - 2{200 - 5.[64 - 2.17]} . x = 4
=> 204 - 2{200 - 5 .[64 - 34]}.x = 4
=> 204 - 2{200 - 5.30} . x = 4
=> 204 - 2{200 - 150}.x = 4
=> 204 - 2.50 . x = 4
=> 2.50.x = 200
=> 100.x = 200
=> x = 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