Tim X thuoc Z biet :
(x-3).(x+2)<0
tim x thuoc Z biet x^3-x^2+x-1=0
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: ...
(x-1)^2 =(x-3)^4=\(\left\{{}\begin{matrix}1+1\\2+2\\3+3\\4+4\end{matrix}\right.=2+4+6+8=\sqrt[]{251234=\Sigma\dfrac{2}{2}22\dfrac{2}{2}}\max\limits_{212}=\dfrac{21}{23}2123=\sum\limits1^{ }_{ }\text{(x-1)^2 =x=}\sum1\)
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}
tim x,y thuoc z biet
-24/-6 = x/3 = 4/y^2 = z^3/-2
Tim x biet
(x-2)(x+3)<0 voi x thuoc z
tim x, y thuoc Z biet : x.y - 2.x - 3.y = 5
xy - 2x - 3y = 5
<=> xy - 2x - 3y + 6 = 11
<=> x(y - 2) - 3(y - 2) = 11
<=> (x - 3)(y - 2) = 11
x - 3 | 1 | -1 | 11 | -11 |
y - 2 | 11 | -11 | 1 | -1 |
x | 4 | 2 | 14 | -8 |
y | 13 | -9 | 3 | 1 |
Vậy...
tim x,y thuoc Z biet -3/6=x/-2=-18/y=3/24
ta có : \(\dfrac{-3}{6}=\dfrac{x}{-2}=\dfrac{-18}{y}=\dfrac{3}{24}\)
\(\Rightarrow\dfrac{-3}{6}=\dfrac{3}{24}\) (vô lí)
\(\Rightarrow\) đề sai
tim x thuoc z biet 1+2+3+...+x=210
vì sao lại bằng 20
Nguyễn Trần Thành Đạt không biết làm copy Dark Killer
tim x, y thuoc Z biet 3/x/ phan 2 +5/y/ phan 3=8
tim x thuoc z biet x+5:x-2
x + 5 chia hết cho x - 2
=> x + 5 = x - 2 + 7
ta có : x - 2 chia hết cho x - 2 nên để x + 5 chia hết cho x - 2 thì 7 phải chia hết cho x - 2
=> x - 2 \(\in\)Ư ( 7 ) = { 1 ; 7 ; -1 ; -7 }
Lập bảng ta có :
x - 2 | 1 | 7 | -1 | -7 |
x | 3 | 9 | 1 | 5 |
Vậy x = { 3 ; 9 ; 1 ; 5 }