Tim x: (x+1) + (x+3) + (x+5) +...+ (x+41) = 1272
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Bài 1:
a.
$=(x^3+2^3)-(x^3-2)=2^3+2=10$
b.
$=(x^2+10x+25)-4x(4x^2+12x+9)-(2x-1)(x^2-9)$
$=x^2+10x+25-16x^3-48x^2-36x-(2x^3-18x-x^2+9)$
$=-18x^3-46x^2-8x+16$
2.
a.
$301^2=(300+1)^2=300^2+2.300+1=90000+600+1$
$=90601$
b.
$198^2=(200-2)^2=4(100-1)^2=4(100^2-2.100+1)$
$=4(10000-200+1)=4.9801=39204$
c.
$93.107=(100-7)(100+7)=100^2-7^2$
$=10000-49=9951$
d.
$127^2+146.127+73^2$
$=127^2+2.73.127+73^2$
$=(127+73)^2=200^2=40000$
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Tìm x thuoc z:
1) \(26-\left|x+9\right|=-13\)
\(\Leftrightarrow\left|x+9\right|=26-\left(-13\right)\)
\(\Leftrightarrow\left|x+9\right|=39\)
\(\Leftrightarrow\left[{}\begin{matrix}x+9=39\\x+9=-39\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=39-9=30\\x=-39-9=-48\end{matrix}\right.\)
Vậy: \(x\in\left\{30;-48\right\}\)
2) \(\left|x+7\right|-13=25\)
\(\Leftrightarrow\left|x+7\right|=25+13=38\)
\(\Leftrightarrow x+7\in\left\{38;-38\right\}\)
\(\Leftrightarrow x\in\left\{31;-45\right\}\)
Vậy:.................
tim x biet
\(1)123-3.\left(x+4\right)=23\)
\(\Leftrightarrow3\left(x+4\right)=123-23\)
\(\Leftrightarrow3\left(x+4\right)=100\)
\(\Leftrightarrow x+4=\frac{100}{3}\)
\(\Leftrightarrow x=\frac{100}{3}-4=\frac{100-12}{3}=\frac{88}{3}\)
Vậy:................
2) Tương tự
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1, \(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\) ( Trừ mỗi vế cho 2 ta được phương trình như này nhé ! )
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
Do \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\) nên \(x-2010=0\Leftrightarrow x=2010\)
2, \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
\(\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{55-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)+\left(\dfrac{51-x}{49}+1\right)=0\)
\(\Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\) \(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\) Do \(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\ne0\) nên \(100-x=0\Leftrightarrow x=100\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Khó quá thì nên hoc24 .vn
Nếu đúng thì ấnĐúng 2 không những thees sau khi ấn sẽ may mắn cả năm
(x+1) + (x+3) + (x+5) +...+ (x+41) = 1272
21x +(1+3+5+....+41)= 1272
21x + 441= 1272
21x= 1272 -441
21x= 831
x= 831 : 21
x= 39,57