Tìm x biết
x2 + 3x -10
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\(x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
\(x^2-2x-xy+2y=\left(x^2-xy\right)-2\left(x-y\right)=x\left(x-y\right)-2\left(x-y\right)=\left(x-y\right)\left(x-2\right)\)
a ) 10 x X - 1 - 3 - 5 - 7 - ... - 19 = 2 + 4 + 6 + ... + 20
10 x X - 1 - 3 - 5 - 7 - ... - 19 = 110
10 x X - ( 1 + 3 + 5 + 7 + ... + 19 ) = 110
10 x X - 100 = 110
10 x X = 110 + 100
10 x X = 210
X = 210 : 10
X = 21
\(3x\left(\dfrac{4}{3}x+1\right)-4x\left(x-2\right)=10\)
\(\Leftrightarrow4x^2+3x-4x^2+8x=10\)
\(\Leftrightarrow x=\dfrac{10}{11}\)
a: Ta có: \(x\left(2x-3\right)-\left(2x-1\right)\left(x+5\right)=17\)
\(\Leftrightarrow2x^2-3x-2x^2-10x+x+5=17\)
\(\Leftrightarrow-12x=12\)
hay x=-1
Ta có: |x - 10| + 10 = x
=> |x - 10| = x - 10
=> x - 10 = x - 10
x - 10 = -(x - 10)
=> x - 10 = 0
=> x = 10 + 0
=> x = 10
$ 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`
a) (x+3)² + (4+x)(4-x) = 10
x² + 6x + 9 + 16- x² = 10
6x + 25 = 10
6x = -15
x = -15/6
b) 9(x+1)² - (3x-2)(3x+2) = 10
9x² + 18x + 9 - 9x² + 4 =10
18x + 13 = 10
18x = -3
x = -1/6
a) ( x + 3 )2 + ( 4 + x )( 4 - x ) = 10
⇔ x2 + 6x + 9 + 16 - x2 = 10
⇔ 6x + 25 = 10
⇔ 6x = -15
⇔ x = -15/6 = -5/2
b) 9( x + 1 )2 - ( 3x - 2 )( 3x + 2 ) = 10
⇔ 9( x2 + 2x + 1 ) - ( 9x2 - 4 ) = 10
⇔ 9x2 + 18x + 9 - 9x2 + 4 = 10
⇔ 18x + 13 = 10
⇔ 18x = -3
⇔ x = -3/18 = -1/6
x^2+3x-10=x^2-5x+2x-10=(x^2-5x)+(2x-10)=x*(x-5)+2*(x-5)=(x+2)*(x+5)