Tim x thuoc Z: (x+1)+(x+2)+(x+3)+...+(x+40)=1000
NT
Những câu hỏi liên quan
Tim x thuoc Z de A thuoc Z va tim gia tri do .
a/ A= x+3/x-2 .
b/ A= 1-2x/x+3 .
tim x thuoc Z
(10-x)+48=40-(2-x)
10 - x + 48 = 40 - 2 + x
=>58-40 +2 = x - (-x )
=>20 = 2x
=>x=10
Đúng 0
Bình luận (0)
1 )Tim x, y thuoc Z
x + y = x.y
2) Tim x thuoc Z
(x + 1)+(x+3)+(x+5)+...+(x+99)=0(x-3)+(x-2)+(x-1)+...+10+11=11-12(x-5)+7(3-x)=530(x+2)-6(x-5)-24x=100
x + y = x.y
=> xy - x - y = 0
=> (xy - x) - y + 1 = 1
=> x(y - 1) - (y - 1) = 1
=> (x - 1)(y - 1) = 1
=> x - 1 = y - 1 = 1 hoặc x - 1 = y - 1 = -1
=> x = y = 2 hoặc x = y = 0
Đúng 0
Bình luận (0)
tim x thuoc Z biet x^3-x^2+x-1=0
cho p=(x+3)/(x^2+9) .tim x thuoc z de p thuoc z
b2, Cho a,b,m thuoc Z , m>0, CMR neu a,b chia cho m co cung so du thi a- b chia hết cho m
b3, tim x thuoc Z, biet
a,|x-1|+|3-x|=2
b,|x-2|+|x-3|=1
b4, cho 40 số nguyên, trong đó bất kì 3 số nào cũng có tích là 1 số âm. CMR tích của 40 số là 1 số dương
5, tim x,y biet
(y+1)*(x*y-1)=3
tim x thuoc Z biet :
(x-1)^2 =(x-3)^4
HELP ME:0!!
\(\left(x-1\right)^2=\left(x-3\right)^4\)
\(\Leftrightarrow\left(x-1\right)^2-\left(x-3\right)^4=0\)
\(\Leftrightarrow\left(x-1\right)^2-\left[\left(x-3\right)^2\right]^2=0\)
\(\Leftrightarrow\left[\left(x-1\right)-\left(x-3\right)^2\right]\left[\left(x-1\right)+\left(x-3\right)^2\right]=0\)
\(\Leftrightarrow\left(x-1-x^2+6x-9\right)\left(x-1+x^2-6x+9\right)=0\)
\(\Leftrightarrow\left(-x^2+7x-10\right)\left(x^2-5x+8\right)=0\)
\(\Leftrightarrow-\left(x-5\right)\left(x-2\right)\left(x^2-5x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)
Vậy: ...
Đúng 2
Bình luận (0)
(x-1)^2 =(x-3)^4=\(\left\{{}\begin{matrix}1+1\\2+2\\3+3\\4+4\end{matrix}\right.=2+4+6+8=\sqrt[]{251234=\Sigma\dfrac{2}{2}22\dfrac{2}{2}}\max\limits_{212}=\dfrac{21}{23}2123=\sum\limits1^{ }_{ }\text{(x-1)^2 =x=}\sum1\)
Đúng 0
Bình luận (0)
Bổ sung cho @ Huỳnh Thanh Phong.
(- \(x^2\) + 7\(x\) - 10).(\(x^2\) - 5\(x\) + 8) = 0
(- \(x^2\) + 5\(x\) + 2\(x\) - 10).(\(x^2\) - \(\dfrac{5}{2}\)\(x\) - \(\dfrac{5}{2}\)\(x\) + \(\dfrac{25}{4}\) + \(\dfrac{7}{4}\)) = 0
[(- \(x^2\) + 5\(x\)) + (2\(x\) - 10)].[(\(x^2\) - \(\dfrac{5}{2}\)\(x\)) - (\(\dfrac{5}{2}\)\(x\) - \(\dfrac{25}{4}\)) + \(\dfrac{7}{4}\)] = 0
[ -\(x\)(\(x\) - 5) + 2.(\(x\) - 5)]. [\(x\)(\(x\) - \(\dfrac{5}{2}\)) - \(\dfrac{5}{2}\).(\(x\) - \(\dfrac{5}{2}\)) + \(\dfrac{7}{4}\)] = 0
(\(x\) - 5).(-\(x\) + 2).[(\(x-\dfrac{5}{2}\)).(\(x\) - \(\dfrac{5}{2}\)) + \(\dfrac{7}{4}\)] = 0
(\(x\) - 5).(-\(x\) + 2).[(\(x\) - \(\dfrac{5}{2}\))2 + \(\dfrac{7}{4}\)] = 0 (1)
Vì (\(x\) - \(\dfrac{5}{2}\))2 ≥ 0 ⇒ (\(x\) - \(\dfrac{5}{2}\))2 + \(\dfrac{7}{4}\) ≥ \(\dfrac{7}{4}\) (2)
Kết hợp (1) và (2) ta có:
\(\left[{}\begin{matrix}x-5=0\\-x+2=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)
Vậy \(x\in\) {2; 5}
Đúng 1
Bình luận (0)
Xem thêm câu trả lời
Tim x thuoc Z
(x-3)+(x-2)+(x-1)+...+10+11=11
Tim x thuoc q de 3/x-2 thuoc z
Để \(\frac{3}{x-2}\in Z\)thì x-2 thuộc Ư(3)={-3;-1;1;3}
Ta có bảng:
| x-2 | -3 | -1 | 1 | 3 |
| x | -1 | 1 | 3 | 5 |
Vậy x=-1;1;3;5 thì \(\frac{3}{x-2}\in Z\)
Đúng 0
Bình luận (0)