tìm x biết 3x + 1 = 9x
Bài 2: Tìm x, biết:
a/ 12x(x – 5) – 3x(4x - 10) = 120
b/ 9x(x + 4) – 5x(3x + 2) = 112 - 2x(3x + 1)
c/ 3x(1 – x) - 5x(3x + 7) = 154 + 9x(5 – 2x)
$ a/ 12x(x – 5) – 3x(4x - 10) = 120$
`<=>12x^2-60x-12x^2+30x=120`
`<=>-30x=120`
`<=>x=-4`
Vậy `x=-4`
$b/ 9x(x + 4) – 5x(3x + 2) = 112 - 2x(3x + 1)$
`<=>9x^2+36x-15x^2-10x=112-6x^2-2x`
`<=>-6x^2+26x=112-6x^2-2x`
`<=>28x=112`
`<=>x=4`
Vậy `x=4`
$c/ 3x(1 – x) - 5x(3x + 7) = 154 + 9x(5 – 2x)$
`<=>3x-3x^2-15x^2-35x=154+45x-18x^2`
`<=>-32x-18x^2=154+45x-18x^2`
`<=>77x=-154`
`<=>x=-2`
Vậy `x=-2`
tìm x biết :
(1-3x2) - (x-2)(9x+1) = (3x-4)(3x+4)-9(x+3)2
(1-3x2)-(x-2)(9x+1)=(3x-4)(3x+4)-9(x+3)2
⇒1-3x2-(9x2+x-18x-2)=9x2-16-9(x2+6x+9)
⇒1-3x2-(9x2-17x-2)= -56x-97
⇒1-3x2-9x2+17x+2=-56x-97
⇒3-12x2+17x=-56x-97
⇒3-12x2+17x+56x+97=0
⇒-12x2+73x+100=0
⇒-(12x2-73x-100)=0
Tìm x biết 3 ( 2 x - 1 ) ( 3 x - 1 ) - ( 2 x - 3 ) ( 9 x - 1 ) = 0
Tìm x biết 3( 2x - 1 )( 3x - 1 ) - ( 2x - 3 )( 9x - 1 ) = 0
Tìm x,biết;
a) (3x+1)^2 -9x (x-1) =46
b)5x (x-3) = 7 (3-x)
a)
(3x+1)^2 -9x (x-1) =46
9x2+6x+1-9x2+9x = 46
15x+1=46
15x=45
x=3
b)
5x (x-3) = 7 (3-x)
5x ( x - 3 ) - 7 ( 3 - x ) = 0
5x ( x - 3 ) + 7 ( x - 3 ) = 0
(5x + 7 ) ( x - 3 ) = 0
5x+7=0 hoặc x-3 = 0
5x=-7 hoặc x=3
x= -7 / 5 hoặc x=3
Tìm x biết:9x+5 chia hết 3x-1
tìm x , biết :
7/3x-4/9x=5/6x-1/3x+7/3
Tìm x biết :
a/ ( x +1 )^2 - 3 ( x + 1 ) = 0
b/ 2 ( 3x - 2 )^2 = 9x^2 - 4
a: Ta có: \(\left(x+1\right)^2-3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
b: Ta có: \(2\left(3x-2\right)^2=9x^2-4\)
\(\Leftrightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(6x-4-3x-2\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(3x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)
Tìm x, biết:
2.|3x+1|=1/3.|9x+3|+5