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Tìm x,biết:
a/
\(\Leftrightarrow\) x = 0 hoặc 1 + 5x = 0
1) x = 0
2) 1+ 5x = 0 \(\Leftrightarrow\) x = \(\frac{-1}{5}\)
Vậy: S = \(\left\{0;\frac{-1}{5}\right\}\)
b/
\(\Leftrightarrow\) (x+1) - (x+1)2 = 0
\(\Leftrightarrow\) ( x+ 1)(1-x-1) = 0
\(\Leftrightarrow\) (x+1).(-x) = 0
\(\Leftrightarrow\) x+1 = 0 hoặc x = 0
\(\Leftrightarrow\) x= -1 ; 0
Vậy: S=\(\left\{-1;0\right\}\)
c/
\(\Leftrightarrow\) x(x2 + 1) = 0
\(\Leftrightarrow\) x = 0 hoặc x2 + 1 = 0
Ta có : x2 + 1 \(\ge\) 0 vs mọi x
Vậy: S = \(\left\{0\right\}\)
d/0
\(\Leftrightarrow\) 5x(x-2) + (x - 2) = 0
\(\Leftrightarrow\) (x - 2)(5x+1) = 0
\(\Leftrightarrow\) x - 2 = 0 hoặc 5x+ 1 = 0
\(\Leftrightarrow\) x = 2 hoặc x = \(\frac{-1}{5}\)
Vậy: S = \(\left\{\frac{-1}{5};2\right\}\)
g/
x = 4 hoặc x = 2
Vậy: S= \(\left\{2;4\right\}\)
h/
\(\Leftrightarrow\) x = 0 hoặc x = 3
Vậy: S = \(\left\{0;3\right\}\)
Vậy: S= \(\left\{0;3\right\}\)
i/
4x(x+1)-8(x+1) = 0
\(\Leftrightarrow\) 4(x+1) (x - 2) = 0
\(\Leftrightarrow\) x+1 = 0 hoặc x - 2 = 0
\(\Leftrightarrow\) x= -1 hoặc x = 2
Vậy: S=\(\left\{-1;2\right\}\)
Cho 2 số a,b thỏa mãn \(a^3+b^3+3\left(a^2+b^2\right)+4\left(a+b\right)+4=0\)
Tính giá trị của biểu thức \(M=2018\left(a+b\right)^2\)
Đặt \(\left\{{}\begin{matrix}x-2010=a\\2009-x=b\end{matrix}\right.\)
Theo đề bài ta có:
\(\dfrac{\left(2009-x\right)^2+\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}{\left(2009-x\right)^2-\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}=\dfrac{19}{49}\)
\(\Leftrightarrow\dfrac{b^2+ab+a^2}{b^2-ab+a^2}=\dfrac{19}{49}\)
\(\Leftrightarrow19\left(b^2-ab+a^2\right)=49\left(b^2+ab+a^2\right)\)
\(\Leftrightarrow19b^2-19ab+19a^2-49b^2-49ab-49a^2=0\)
\(\Leftrightarrow-30a^2-68ab-30b^2=0\)
\(\Leftrightarrow-2\left(15a^2+34ab+15b^2\right)=0\)
\(\Leftrightarrow15a^2+34ab+15b^2=0\)
\(\Leftrightarrow15a^2+25ab+9ab+15b^2=0\)
\(\Leftrightarrow5a\left(3a+5b\right)+3b\left(3a+5b\right)=0\)
\(\Leftrightarrow\left(3a+5b\right)\left(5a+3b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3a+5b=0\\5a+3b=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3\left(x-2010\right)+5\left(2009-x\right)=0\\5\left(x-2010\right)+3\left(2009-x\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-6030+10045-5x=0\\5x-10050+6027-3x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x+4015=0\\2x-4023=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2x=-4015\\2x=4023\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-4015}{-2}=2007,5\\x=\dfrac{4023}{2}=2011,5\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=2007,5\\x=2011,5\end{matrix}\right.\)
Đặt a=(2009-x)2
b=(x-2010)2
Theo đề bài ta có
\(\dfrac{\text{a^2+ab+b^2}}{a^2-ab+b^2}=\dfrac{19}{49}\)
\(\text{49(a^2+ab+b^2)}=19\left(a^2-ab+b^2\right)\)
\(\text{30a^2+68ab+30b^2=0}\)
\(\text{15a^2+34ab+15b^2=0}\)
\(\text{15a^2+9ab+25ab+15b^2=0}\)
\(\text{3a(5a+3b)+5(3b+5a)=0}\)
\(\text{(5a+3b)(3a+5b)=0}\)
\(\left[{}\begin{matrix}3a+5b=0\\3b+5a=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}3\left(2009-x\right)=5\left(x-2010\right)\\5\left(2009-x\right)=3\left(x-2010\right)\end{matrix}\right.\)
\(-8x=-6030-10045\) hay \(8x=-10050-6027\)
\(x\simeq2009\),375 hay \(x\simeq2009,625\)
Đặt x-2009=a\(\Leftrightarrow\dfrac{\left(x-2009\right)^2-\left(x-2009\right)\left(x-2010\right)+\left(x-2010\right)^2}{\left(x-2009\right)^2+\left(x-2009\right)\left(x-2010\right)+\left(x-2010\right)^2}=\dfrac{19}{49}\)
\(\Leftrightarrow\dfrac{a^2-a\left(a-1\right)+\left(a-1\right)^2}{a^2+a\left(a-1\right)+\left(a-1\right)^2}=\dfrac{19}{49}\)
\(\Leftrightarrow\dfrac{a^2-a^2+a+a^2-2a+1}{a^2+a^2-a+a^2-2a+1}=\dfrac{19}{49}\)
=>\(\dfrac{a^2-a+1}{3a^2-3a+1}=\dfrac{19}{49}\)
=>49a^2-49a+49-57a^2+57a-19=0
=>-8a^2+8a+30=0
=>a=5/2 hoặc a=-3/2
=>x-2009=5/2 hoặc x-2009=-3/2
=>x=4023/2 hoặc x=4015/2