1 . Số ab , biết a > b và a + b là hai số nguyên tố
2. Gia tri cua x thuoc N thoa man : 5^2 -3^2 =x^2
Những câu hỏi liên quan
1.gia tri a de da thuc P(x)=3x3-8x2+6x-a chia het cho Q(x)=3x2-5x+1 la a =?
2.khi a+b=1 .gia tri nho nhat cua bieu thuc a3+b3+ab la?
3.tong cac gia tri cua x thoa man x5=x4+x3+x2+x+2 la?
1.gia tri x<0 thoa man (2x-3)2=(x+5)2
2.tap hop cac gia tri cua x thoa man x4-2x3+10x2-20x=0 la S(....)
3.phan tich da thuc xy-12+3x-4 ta duoc (x+a)(y+b) khi do a+b=?
tim cac gia tri x y thoa man
a,(x^2-1)(x^2-16)<0 va x thuoc Z
B,1/x-y/8=1/16 va x y thuoc N
1. so x0 thoa man frac{5}{x}frac{x+1}{22}2. GTLN cua x thoa man 5^{left(x-2right)left(x+3right)}3. Gia tri x0 thoa man /x^2+50//-2x^2-1/4. Gia tri nguyen cua x thoa man left(3x-4right)^5^{left(3x-4right)^7}la x5. So gia tri cua x thoa man /x-2014/+/x-2015/06. Gia tri cua x de bieu thuc Afrac{x-3+5}{3-x+2}dat gia tri lon nhat la x
Đọc tiếp
1. so x>0 thoa man \(\frac{5}{x}=\frac{x+1}{22}\)
2. GTLN cua x thoa man \(5^{\left(x-2\right)\left(x+3\right)}\)
3. Gia tri x<0 thoa man /\(x^2+50\)/=/\(-2x^2\)-1/
4. Gia tri nguyen cua x thoa man \(\left(3x-4\right)^5\)=\(^{\left(3x-4\right)^7}\)la x=
5. So gia tri cua x thoa man /x-2014/+/x-2015/=0
6. Gia tri cua x de bieu thuc A=\(\frac{x-3+5}{3-x+2}\)dat gia tri lon nhat la x=
Tim cac gia tri x, y thoa man:
a, ( x2 - 1 ) ( x2 - 16 ) < 0 ca x thuoc z
b, 1/x - y/8 = 1/16 va x, y thuoc N
c, | x+1 | + | x+2 | + | x+3 | = x
1. tich cac gia tri x thoa man x2-\(\frac{1}{10}\)x-\(\frac{1}{5}\)+0
2.gia tri cua x de bieu thuc A=6x2+7x-8 dat gia tri nho nhat
3.tích cac gia tri cua x thoa man (2x-1)2=(2-x)2
1/tong cac gia tri nguyen cua x thoa man /x-0,2/<13
2/ti so gia A va B biet 2/7 cua A gap 4/9 lan 3/10 cua B
Cho a,b thuoc N thoa man ucln(a,b)=1
tinh gia tri lon nhat cua UCLN(a+10b;b+10a
tim gia tri lon nhat cua A=2018-/x-7/-/y+2/
tim gia tri nho nhat cua B /x-500/+/x-300/
tim n thuoc Z,biet: a,3.n+2 chia het cho n-1; b, n^2 +5 chia het cho n+1
\(A=2018-\left|x-7\right|-\left|y+2\right|\)
Ta có: \(\hept{\begin{cases}\left|x-7\right|\ge0\forall x\\\left|y+2\right|\ge0\forall y\end{cases}}\Rightarrow2018-\left|x-7\right|-\left|y+2\right|\le2018\)
\(A=2018\Leftrightarrow\hept{\begin{cases}\left|x-7\right|=0\\\left|y+2\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=7\\y=-2\end{cases}}}\)
Vậy \(A_{m\text{ax}}=2018\Leftrightarrow\hept{\begin{cases}x=7\\y=-2\end{cases}}\)
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