\(\left(x+2\right)+\left(x+12\right)+\left(x+42\right)+\left(x+47\right)=655\)
\(x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+2009\right)=2009\cdot2010\)
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GG
2 tháng 4 2019
(x+2)+(x+12)(x+42)+(x+47)=655
(x+x+x+x) + (2+12+42+47) =655
4x + 103 =655
4x =552
x =138
x+(x+1)+(x+2)+(x+3)+...+(x+2009)=2009⋅2010
(x+x+x+...+x+x)+(1+2+3+...+2009)=2009.2010 (có 2010 số hạng x)
2010x +(2009+1).2009:2 =2009.2010
2010x +2010.2009:2 =2009.2010
2010x =(2009.2010)-(2010.2009:2)
2010x =(2009.2010)(1-\(\frac{1}{2}\))
2010x =2009.2010.\(\frac{1}{2}\)
x =2009.\(\frac{1}{2}\)
x =\(\frac{2009}{2}\)
12 tháng 3 2017
đặt 2009-x=a,x-2010=b
suy ra a^2+ab+b^2/a^2-ab+b^2=19/49
suy ra 49(a^2+ab+b^2)=19(a^2-ab+b^2)
49a^2+49ab+49b^2=19a^2-19ab+19b^2
30a^2+68ab+30b^2=0
30a^2+50ab+18ab+30b^2=0
10a(3a+5b)+6b(3a+5b)=0
(3a+5b)(10a+6b)=0
suy ra 3a+5b=0 hoặc 10a+6b=0
thế vào lại rồi tìm x
14 tháng 7 2017
1. \(\left(2x-1\right)^3+\left(x+2\right)^3=\left(3x+1\right)^3\)
\(\Rightarrow8x^3-12x^2+6x-1+x^3+6x^2+12x+8=27x^3+27x^2+9x+1\)
\(\Rightarrow-18x^3-33x^2+9x+6=0\)\(\Rightarrow\left(x+2\right)\left(-18x^2+3x+3\right)=0\)
\(\Rightarrow\left(x+2\right)\left(2x-1\right)\left(-9x-3\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{1}{2};x=-\frac{1}{3}\end{cases}}\)
Vậy \(x=-2;x=\frac{1}{2};x=-\frac{1}{3}\)
2. \(\frac{x-1988}{15}+\frac{x-1969}{17}+\frac{x-1946}{19}+\frac{x-1919}{21}=10\)
\(\Rightarrow\left(\frac{x-1988}{15}-1\right)+\left(\frac{x-1969}{17}-2\right)+\left(\frac{x-1946}{19}-3\right)+\left(\frac{x-1919}{21}-4\right)=0\)
\(\Rightarrow\frac{x-2003}{15}+\frac{x-2003}{17}+\frac{x-2003}{19}+\frac{x-2003}{21}=0\)
\(\Rightarrow x-2003=0\)do \(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\ne0\)
Vậy \(x=2003\)
3. Đặt \(\hept{\begin{cases}2009-x=a\\x-2010=b\end{cases}}\)
\(\Rightarrow\frac{a^2+ab+b^2}{a^2-ab+b^2}=\frac{19}{49}\Rightarrow49a^2+49ab+49b^2=19a^2-19ab+19b^2\)
\(\Rightarrow30a^2+68ab+30b^2=0\Rightarrow\left(5a+3b\right)\left(3a+5b\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5a=-3b\\3a=-5b\end{cases}}\)
Với \(5a=-3b\Rightarrow5\left(2009-x\right)=-3\left(x-2010\right)\)
\(\Rightarrow-2x=-4015\Rightarrow x=\frac{4015}{2}\)
Với \(3a=-5b\Rightarrow3\left(2009-x\right)=-5\left(x-2010\right)\)
\(\Rightarrow2x=4023\Rightarrow x=\frac{4023}{2}\)
Vậy \(x=\frac{4023}{2}\)hoặc \(x=\frac{4015}{2}\)
8 tháng 1 2018
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 \(\left[{}\begin{matrix}x=2007,5\\x=2011,5\end{matrix}\right.\)
MS
28 tháng 12 2017
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 \(x\simeq2009,625\)
21 tháng 12 2022
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
tìm X, pleas
\(\left(x+2\right)+\left(x+12\right)+\left(x+42\right)+\left(x+47\right)=655\)
\(x+2+x+12+x+42+x+72=655\)
\(4x+\left(2+12+42+47\right)=655\)
\(4x+103=655\)
\(4x=655-103\)
\(4x=552\)
\(x=138\)