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16 tháng 9 2023

\(Bài.1:\\ a,104^2-16=104^2-4^2=\left(104+4\right)\left(104-4\right)=108.100=10800\\ b,9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\\ =\left(9.2\right)^8-\left(18^8-1\right)=18^8-18^8+1=1\\ c,999^3+3.999^2+3.999+1\\ =999^3+3.999^2.1+3.999.1^2+1^3=\left(999+1\right)^3=1000^3=1000000000\\ d,42^3-6.42^2+12.42-8\\ =42^3-3.42^2.2+3.42.2^2-2^3\\ =\left(42-2\right)^3=40^3=64000\)

16 tháng 9 2023

Bài 1

a) 104² - 16

= 104² - 4²

= (104 - 4)(104 + 4)

= 100.108

= 10800

b) 9⁸.2⁸ - (18⁴ - 1)(18⁴ + 1)

= 18⁸ - (18⁸ - 1)

= 18⁸ - 18⁸ + 1

= 1

c) 999³ + 3.999² + 3.999 + 1

= (999 + 1)³

= 1000³

= 1000000000

d) 42³ - 6.42² + 12.42 - 8

= (42 - 2)³

= 40³

= 64000

6 tháng 9 2018

nốt ý b:

\(\left(x-1\right)^3+1+3x\left(x-4\right)=0\)

\(\Leftrightarrow x^3-3x^2+3x-1+1+3x^2-12x=0\)

\(\Leftrightarrow x^3-9x=0\Leftrightarrow x\left(x^2-9\right)=0\)

\(\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

Vậy ..............

6 tháng 9 2018

\(a,x\left(x-2012\right)-2013x+2012.2013=0\)

\(=x\left(x-2012\right)+2013\left(-x+2012\right)=0\)

\(\Rightarrow x\left(x-2012\right)-2013\left(x-2012\right)=0\)

\(\Rightarrow\left(x-2013\right)\left(x-2012\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2013=0\\x-2012=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2013\\x=2012\end{matrix}\right.\)

Vậy...

16 tháng 10 2021

\(1,\\ a,\Leftrightarrow x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow\left[{}\begin{matrix}x-4=4\\x-4=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=0\end{matrix}\right.\\ c,\Leftrightarrow2x+1=-2\Leftrightarrow x=-\dfrac{3}{2}\\ 2,\\ a,=1\\ b,=\left(\dfrac{13}{4}\right)^2=\dfrac{169}{16}\\ c,=\left(-\dfrac{7}{4}\right)^2=\dfrac{49}{16}\\ d,=\left(\dfrac{3}{7}\right)^{20}:\left(\dfrac{3}{7}\right)^{12}=\left(\dfrac{3}{7}\right)^8=...\\ e,=\left(3\cdot5\cdot\dfrac{2}{3}\right)^2=10^2=100\)

10 tháng 12 2021

\(a,\Leftrightarrow2x^2+10x-2x^2=12\Leftrightarrow x=\dfrac{12}{10}=\dfrac{6}{5}\\ b,\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(9-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\\ d,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ e,\Leftrightarrow4x^2-4x+1-4x^2+196=0\\ \Leftrightarrow-4x=-197\Leftrightarrow x=\dfrac{197}{4}\)

\(f,\Leftrightarrow x^2+8x+16-x^2+1=16\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\\ g,Sửa:\left(3x+1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x+1-x-1\right)\left(3x+1+x+1\right)=0\\ \Leftrightarrow2x\left(4x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\ h,\Leftrightarrow x^2+8x-x-8=0\\ \Leftrightarrow\left(x+8\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\\ i,\Leftrightarrow2x^2-13x+15=0\\ \Leftrightarrow2x^2+2x-15x-15=0\\ \Leftrightarrow\left(x+1\right)\left(2x-15\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{15}{2}\end{matrix}\right.\)

Bài 3: 

b: \(\Leftrightarrow x^2\left(x+1\right)^2=0\)

hay \(x\in\left\{0;-1\right\}\)

c: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=0\)

=>x-1=0

hay x=1

d: \(\Leftrightarrow6x^2-3x-4x+2=0\)

\(\Leftrightarrow\left(2x-1\right)\left(3x-2\right)=0\)

hay \(x\in\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}\)

2 tháng 11 2021

Bài 1:

a) \(\Rightarrow3x^2+3x-2x^2-4x+x+1=0\)

\(\Rightarrow x^2=-1\left(VLý\right)\Rightarrow S=\varnothing\)

b) \(\Rightarrow\left(x-2020\right)\left(2x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2}\end{matrix}\right.\)

c) \(\Rightarrow\left(x-10\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)

d) \(\Rightarrow\left(x+4\right)^2=0\Rightarrow x=-4\)

e) \(\Rightarrow\left(x+6\right)\left(x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)

f) \(\Rightarrow\left(5x-4\right)\left(5x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)

Bài 2:

a) \(\Rightarrow3x\left(x^2-4\right)=0\Rightarrow3x\left(x-2\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

b) \(\Rightarrow x\left(x-2\right)+5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

4 tháng 1 2022

giúp mình với

4 tháng 1 2022

\(a,\Leftrightarrow\left(4-5x\right)\left(4+5x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+1-2\right)\left(x+1+2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(3x+1-2x\right)\left(3x+1+2x\right)=0\\ \Leftrightarrow\left(x+1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{1}{5}\end{matrix}\right.\\ d,Sửa:\left(4x+1\right)^2-\left(x-2\right)^2=0\\ \Leftrightarrow\left(4x+1-x+2\right)\left(4x+1+x-2\right)=0\\ \Leftrightarrow\left(3x+3\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{5}\end{matrix}\right.\\ e,\Leftrightarrow\left(2x+1-x-3\right)\left(2x+1+x+3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)

a: Ta có: \(\left(x^2+2\right)\left(x-4\right)-\left(x+2\right)^3=-16\)

\(\Leftrightarrow x^3-4x^2+2x-8-x^3-6x^2-12x-8=-16\)

\(\Leftrightarrow-10x^2-10x=0\)

\(\Leftrightarrow-10x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

c: Ta có: \(x^3+3x^2+3x+28=0\)

\(\Leftrightarrow\left(x+1\right)^3=-27\)

\(\Leftrightarrow x+1=-3\)

hay x=-4

Chọn D