Ta có :
\(2x^2-3x=0\)
\(\Leftrightarrow\)\(x\left(2x-3\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\2x-3=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\2x=3\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}\)
Vậy \(x\in\left\{0;\frac{3}{2}\right\}\)
tim x de :
1) -2 / 3x - 1 co gia tri duong
2) 3 - 1/2x / 2x^2 + 11 co gia tri am
3) 2/3 - 2x / -x^2 - 1 co gia tri khong am
Tìm x
A= 2x-8 co gia tri am
B=6-x co gia tri ko duong
C=(x-2)(2x+6)co gia tri am
D=3x2+9x co gia tri duong
E=x-2/x1 co gia tri am
F=2x-5/x-4 co gia tri duong
G=x+1/3-X co gia tri ko am
cho bieu thucA=\(\frac{8-x}{x-3}\)
tim gia tri cua x de A co gia tri bang gia tri bieu thuc M=x+1
tim cac gia tri cua x de bieu thuc sau nhan gia tri am
a)x mu 2 + 5x b)x mu 2 - 5/2x c)3(2x+3)(3x-5) d)x - 2/x-6 e) x mu 2 -1/x mu 2
cho 4 so ko am a.b.c.d thoa man a+b+c+d=1.goi S la tong cac gia tri truyen doi cua hieu tung cap so co duoc tu 4 so nay..S co the dat duoc gia tri lon nhat bang bao nhieu
tim x de p=x2 -x co gia tri bang 0
cho bieu thuc A = (x-1)^2 + (y-1)^2 . De co gia tri bang 0 thi x, y phai nhan gia tri la bao nhieu
tim 2 p/s co tu bang 9. biet rang gia tri moi p/s do lon hon -11/13 va nho hon -11/15
tim cac gia tri cua x de
a)1/x_1>2
b)F=x^2_1/x^2 co gia tri am
