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a) Ta có: \(33\cdot55+33\cdot67+45\cdot33+67^2\)
\(=\left(33\cdot55+33\cdot45\right)+\left(33\cdot67+67^2\right)\)
\(=33\cdot\left(55+45\right)+67\left(33+67\right)\)
\(=33\cdot100+67\cdot100\)
\(=100\cdot\left(33+67\right)\)
\(=100\cdot100\)
\(=10000\)
c) Ta có: \(2016\cdot2018-2017^2\)
\(=\left(2017-1\right)\left(2017+1\right)-2017^2\)
\(=2017^2-1-2017^2\)
\(=-1\)
b)\(3x^3+6x^2-75x-150=0\Leftrightarrow3\left(x^3+2x^2-25x-50\right)=0\Leftrightarrow x^3+2x^2-25x-50=0\)
<=>\(x^2\left(x+2\right)-25\left(x+2\right)=0\Leftrightarrow\left(x^2-25\right)\left(x+2\right)=0\Leftrightarrow\left(x-5\right)\left(x+5\right)\left(x+2\right)=0\)
<=>x-5=0 hoặc x+5=0 hoặc x+2=0<=>x=5 hoặc x=-5 hoặc x=-2
c)\(2x^5-3x^4+6x^3-8x^2+3=0\Leftrightarrow2x^5+x^4-4x^4-2x^3+8x^3+4x^2-12x^2+3=0\)
<=>\(x^4\left(2x+1\right)-2x^3\left(2x+1\right)+4x^2\left(2x+1\right)-3\left(4x^2-1\right)=0\)
<=>\(x^4\left(2x+1\right)-2x^3\left(2x+1\right)+4x^2\left(2x+1\right)-3\left(2x-1\right)\left(2x+1\right)=0\)
<=>\(\left(2x+1\right)\left(x^4-2x^3+4x^2-6x+3\right)=0\)
<=>\(\left(2x+1\right)\left(x^4-2x^3+x^2+3x^2-6x+3\right)=0\)
<=>\(\left(2x+1\right)\left[x^2\left(x^2-2x+1\right)+3\left(x^2-2x+1\right)\right]=0\)
<=>\(\left(2x+1\right)\left(x^2+3\right)\left(x^2-2x+1\right)=0\Leftrightarrow\left(2x+1\right)\left(x^2+3\right)\left(x-1\right)^2=0\)
Vì \(x^2\ge0\Rightarrow x^2+3\ge3>0\Rightarrow\orbr{\begin{cases}2x+1=0\\\left(x-1\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)
a) 2x3 - x2 - 8x + 4 = 0
x2.(2x - 1) - 4.(2x - 1) = 0
(x2 - 4)(2x - 1) = 0
\(\Rightarrow\orbr{\begin{cases}x^2-4=0\\2x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=4\\x=\frac{1}{2}\end{cases}}\)
Với x2 = 4
=> x = 2 hoặc x = -2
=> x = {-2 ; 2 ; \(\frac{1}{2}\))
c) Ta có: \(33\cdot55+33\cdot67+45\cdot33+67\cdot67\)
\(=33\left(55+45\right)+67\left(33+67\right)\)
\(=33\cdot100+67\cdot100\)
\(=100\cdot100=10000\)
\(\dfrac{x-2}{102}+\dfrac{x-3}{103}=\dfrac{x-4}{104}+\dfrac{x-5}{105}\)
\(\Leftrightarrow\dfrac{x-2}{102}+1+\dfrac{x-3}{103}+1=\dfrac{x-4}{104}+1+\dfrac{x-5}{105}+1\)
\(\Leftrightarrow\dfrac{x+100}{102}+\dfrac{x+100}{103}-\dfrac{x+100}{104}-\dfrac{x+100}{105}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{102}+\dfrac{1}{103}-\dfrac{1}{104}-\dfrac{1}{105}\right)=0\)
\(\Leftrightarrow x+100=0\Leftrightarrow x=-100\)
Câu 1 :
a) \(x^3-5x^2-14x\)
\(=x^3-7x^2+2x^2-14x\)
\(=x^2\left(x-7\right)+2x\left(x-7\right)\)
\(=\left(x-7\right)\left(x^2+2x\right)\)
\(=x\left(x-7\right)\left(x+2\right)\)
b) \(a^4+a^2+1\)
\(=\left(a^2\right)^2+2a^2+1-a^2\)
\(=\left(a^2+1\right)-a^2\)
\(=\left(a^2-a+1\right)\left(a^2+a+1\right)\)
c) \(x^4+64\)
\(=\left(x^2\right)^2+2\cdot x^2\cdot8+8^2-2\cdot x^2\cdot8\)
\(=\left(x^2+8\right)^2-\left(4x\right)^2\)
\(=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)
Câu 2 :
a) \(\left(a-b\right)^2=a^2-2ab+b^2\)
Ta có : \(\left(a+b\right)^2=a^2+2ab+b^2\)
\(\Rightarrow a^2+b^2=\left(a+b\right)^2-2ab=7^2-2\cdot14=25\)
\(\Rightarrow\left(a-b\right)^2=25-2\cdot12=1\)
b) tương tự
a) 10000. b) 875000. c) -1.