If a^3+b^3=637 and a+b=13 , find the value of (a-b)^2
If and
, find the value of
.
Answer:
\(a^3+b^3=637\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)=637\Rightarrow a^2-ab+b^2=\frac{637}{13}=49\)\(\left(a+b\right)=13\Rightarrow\left(a+b\right)^2=13^2=169\Leftrightarrow a^2+2ab+b^2=169\)
Ta có: \(a^2-ab+b^2=49\left(1\right)\)
\(a^2+2ab+b^2=169\left(2\right)\)
Lấy (2) trừ 1 ta được 3ab=120=>ab=40
ab=40=>-ab=-40=>a2+b2=49+40=89
\(\left(a-b\right)^2=a^2-2ab+b^2=a^2+b^2-2ab=89-2.40=89-80=9\)Nhập kết quả: 9
give a%b=(a-3)b+b-a
find the value of b if 7%b=13
Have a%b = (a-3)b +b-a
But 7%b=13 so (7-3)b +b-7=13=> 5b=20 => b=4
The average of A ,B and C is 70 . The value of A is 2 more than B . The value of B is in more than C. Find value A , B and C
Find the value of |b-a-1|-|a-b-3| for a<b and ab<0
giúp với,mik cần gấp!
Ta có : \(\hept{\begin{cases}a< b\\ab< 0\end{cases}}\)
\(\Rightarrow\)a là số âm còn b là số dương
\(\Rightarrow\hept{\begin{cases}\left|b-a-1\right|=b-a-1\\\left|a-b-3\right|=3+b-a\end{cases}}\)
\(pt\Leftrightarrow\left(b-a-1\right)-\left(3+b-a\right)\)
\(=b-a-1-3-b+a\)
\(=\left(b-b\right)+\left(a-a\right)+\left(-1-3\right)\)
\(=0+0-4\)
\(=-4\)
Vậy ...
if the denominator of a fraction is 2 units more than it's numerator and this fraction equals 2/3 then this fraction is a/b . the value of a is?
bn ơi hình như bài này là toán tiếng anh nhỉ
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