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\(\dfrac{37\cdot5^4}{25^2}=\dfrac{37\cdot5^4}{5^4}=37\\ \dfrac{2^4\cdot2^6\cdot3^8\cdot9^2}{4^4\cdot3^{11}}=\dfrac{2^{10}\cdot3^8\cdot3^4}{2^8\cdot3^{11}}=2^2\cdot3=12\\ \dfrac{3\cdot9^4\cdot9^3}{3^2\cdot9}=\dfrac{3\cdot3^8\cdot3^6}{3^2\cdot3^2}=3^{11}\\ \dfrac{125\cdot5\cdot64-25^3\cdot10\cdot4}{5^7\cdot8}=\dfrac{5^3\cdot5\cdot2^6-5^6\cdot2\cdot5\cdot2^2}{5^7\cdot2^3}=\dfrac{5^4\cdot2^3\left(2^3-5^3\right)}{5^7\cdot2^3}=\dfrac{8-125}{5^3}=\dfrac{-117}{125}\)
= 25 ( x² + 10x + 25) - 9 ( x² + 14x + 49)
= 25x² + 250x + 625 - 9x² - 126x - 441
= 16x² + 124x 148
c: Ta có: \(25\left(x+5\right)^2-9\left(x+7\right)^2\)
\(=\left(5x+25-3x-21\right)\left(5x+25+3x+21\right)\)
\(=4\cdot\left(x+2\right)\cdot\left(4x+23\right)\)
a) = (x+3).(x-3)^2-(x-3)(x+3)^2
=(x^2-9)(x-3)-(x^2-9)(x+3)
=(x^2-9)(x-3-x-3)
=-6(x^2-9)
các câu còn lại tương tự
\(a,\left(x+3\right)\left(x^2-3x+9\right)-\left(x-3\right)\left(x^2+3x+9\right)\)
\(=x^3+3-\left(x^3-3\right)\)
\(=x^3+3-x^3+3\)
\(=6\)
\(b,\left(x-5\right)\left(x^2+5x+25\right)-\left(x+5\right)\left(x^2-5x+25\right)\)
\(=x^3-5^3-x^3-5^3\)
\(=-125-125\)
\(=-250\)
`a) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3`
`<=>x(x^2-25)-(x^3-8)=3`
`<=>x^3-25x-x^3+8=3`
`<=>-25x=-5`
`<=>x=1/5`
`b) (x – 3)^3 – (x – 3)(x^2 + 3x + 9) + 9(x + 1)^2 = 15`
`<=>x^3-9x^2+27x-27-(x^3-27)+9(x^2+2x+1)=15`
`<=>-9x^2+27x+9x^2+18x+9=15`
`<=>45x+9=15`
`<=>45x=6`
`<=>x=6/45=2/15`
`c) (x+5)(x^2 –5x +25) – (x – 7) = x^3`
`<=>x^3-125-x+7=x^3`
`<=>x^3-x-118=x^3`
`<=>-x-118=0`
`<=>-x=118<=>x=-118`
`d) (x+2)(x^2 – 2x + 4) – x(x^2 + 2) = 4 `
`<=>x^3+8-x^3-2x=4`
`<=>8-2x=4`
`<=>2x=4<=>x=2`
g) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\)
\(\Rightarrow-2\left(2x-5\right)=0\Rightarrow x=\dfrac{5}{2}\)
i) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-2x\right)=0\Rightarrow x\left(x+3\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)
a) 4x^2 - 25 - ( 2x - 5) .( 2x + 7) = 0
<=>4x2-25-(4x2+14x-10x-35)=0
<=>4x2-25-4x2-14x+10x+35= 0
<=>-4x+10= 0
<=>x= 5/2
b) x^3 + 27 + ( x+3). ( x -9) = 0
<=>x3+33+(x+3)(x-9)=0
<=>(x+3)(x2-3x+9)+(x+3)(x-9)=0
<=>(x+3)(x2-3x+9+x-9) =0
<=>(x+3)(x2-2x)=0
<=>(x+3)(x-2)x= 0
<=>x=-3 hoặc x=2 hoặc x=2