Tìm x:10-2(4-3x)
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\(3x\left(\dfrac{4}{3}x+1\right)-4x\left(x-2\right)=10\)
\(\Leftrightarrow4x^2+3x-4x^2+8x=10\)
\(\Leftrightarrow x=\dfrac{10}{11}\)
a) (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
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
\(x^3-2x^2+x-2=0\\ \Leftrightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Leftrightarrow\left(x^2+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ Vậy:x=2\\ ---\\ 2x\left(3x-5\right)=10-6x\\ \Leftrightarrow6x^2-10x-10+6x=0\\ \Leftrightarrow6x^2-4x-10=0\\ \Leftrightarrow6x^2+6x-10x-10=0\\ \Leftrightarrow6x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(6x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}6x-10=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)
\(4-x=2\left(x-4\right)^2\\ \Leftrightarrow4-x=2\left(x^2-8x+16\right)\\ \Leftrightarrow2x^2-16x+32+x-4=0\\ \Leftrightarrow2x^2-15x+28=0\\ \Leftrightarrow2x^2-8x-7x+28=0\\ \Leftrightarrow2x\left(x-4\right)-7\left(x-4\right)=0\\ \Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\\ ---\\ 4-6x+x\left(3x-2\right)=0\\ \Leftrightarrow4-6x+3x^2-2x=0\\ \Leftrightarrow3x^2-8x+4=0\\ \Leftrightarrow3x^2-6x-2x+4=0\\ \Leftrightarrow3x\left(x-2\right)-2\left(x-2\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)
\(10-2\left(4-3x\right)=-4\)
\(\Leftrightarrow10-8+6x=-4\)
\(\Leftrightarrow2+6x=-4\)
\(\Leftrightarrow6x=-4-2\)
\(\Leftrightarrow6x=-6\)
\(\Leftrightarrow x=\frac{-6}{6}\)
\(\Leftrightarrow x=-1\)
\(\text{Vậy }x=-1\)
10 - 2(4 - 3x)= -4
=> 10 - 8 + 6x = -4
=> 6x = -4 + 8 -10
=> 6x = -6
=> x = -1.
Vậy x = -1
$ 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)\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6.\)
\(\Leftrightarrow x^2-4x+4-x^2+9-6=0\)
\(\Leftrightarrow-4x+7=0\)
\(\Leftrightarrow4x=7\Leftrightarrow x=1,75\)
\(b,4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10.\)
\(\Leftrightarrow4\left(x^2-6x+9\right)-4x^2+1-10=0\)
\(\Leftrightarrow-24x+27=0\)
\(\Leftrightarrow24x=27\Leftrightarrow x=1,125\)