cho \(a^3-a^2+a=5;b^3-2b^2+b=-4\)
chứng minh a+b=1
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Bài 2:
\(a^2+b^2=\left(a+b\right)^2-2ab=5^2-2\cdot\left(-2\right)=9\)
\(\dfrac{1}{a^3}+\dfrac{1}{b^3}=\dfrac{a^3+b^3}{a^3b^3}=\dfrac{\left(a+b\right)^3-3ab\left(a+b\right)}{\left(ab\right)^3}\)
\(=\dfrac{5^3-3\cdot5\cdot\left(-2\right)}{\left(-2\right)^3}=\dfrac{125+30}{8}=\dfrac{155}{8}\)
\(a-b=-\sqrt{\left(a+b\right)^2-4ab}=-\sqrt{5^2-4\cdot\left(-2\right)}=-\sqrt{33}\)
\(\frac{x-2}{\left(a+3\right)\left(5-a\right)}=\frac{1}{2\left(a+3\right)}+\frac{1}{2\left(5-a\right)}\)
\(\Rightarrow\frac{2\left(x-2\right)}{2\left(a+3\right)\left(5-a\right)}=\frac{5-a+a+3}{2\left(a+3\right)\left(5-a\right)}\)
\(\Rightarrow\) 2x - 4 = 8
\(\Rightarrow\) 2x = 12
\(\Rightarrow\) x = 6
Ta có \(a-b=5\Rightarrow\left(a-b\right)^2=25\Rightarrow a^2+b^2=25+2ab=25+2\cdot2=29\) (Do ab=2)
\(B=3\left[\left(a^2+b^2\right)^2-2a^2b^2\right]+2\left[\left(a-b\right)\left(a^4+b^4+a^3b^2+a^2b^3\right)\right]\)
= \(3\left[29^2-2\cdot4\right]+2\left\{5\left[\left(a^2+b^2\right)^2-2a^2b^2+ab\left(a^2+b^2\right)\right]\right\}\)
= 3\(\cdot833+10\left[29^2-2\cdot4+2\cdot29\right]\) \(=2499+10\cdot891=11409\)
A = 1 + 3 + 5 + 7 +... + 990
SSH : (990 - 1 ) : 2 + 1 = 495,5
=> tổng : (1 + 990) . 495,5 : 2 = 245520,25 (để xem số cuối có sai k vậy?)
B = 25 + 83 - 23 * 83
= 25 + 512 - 23 * 512 = -11239
C = 600 : {450 : [450 - (4 * 53 - 23 * 52)]}
= 600 : {450 : [450 - (4 * 125 - 8 * 25)]}
= 600 : {450 : [450 - ( 500 - 200)]
= 600 : {450 : [450 - 300]}
= 600 : {450 : 150}
= 600 : 3 = 200
Bài 2 : a) A chia hết cho 2 => x \(\in\){0;2;4;6;8}
b) A chia hết cho 5 => x \(\in\){0;5 }
c) A chia hết cho 2 và 5 => x = 0
d) A chia hết cho 2 nhưng A ko chia hết cho 5 => x \(\in\){2;4;6;8}
Bài 3 tương tự