Gia tri cua x +244,biet:
Xx1=1:x
Những câu hỏi liên quan
bai 1 so cac gia tri nguyen cua x de gia tri tuyet doi cua x + gia tri tuyet doi cua x-2 =0
bai 2 so cac gia tri nguyen cua x thoa man gia tri tuyet doi cua x-3 + gia tri tuyet doi cua 8-2x = 1 là
tim gia tri cua x biet 1/4*1/5*x=1/2
gia tri cua x la
Gia tri tuyet doi cua am x +1 - 2 nhan gia tri tuyet doi cua x-2 -3 nhan gia tri tuyet doi cua am x+3=4
1.gia tri nho nhat cua x^2-x+1
2. gia tri nho nhat cua x^2+10x+2041
Bài 1:
\(x^2-x+1=x^2-x+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x^2-x+\frac{1}{4}\right)+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu "=" khi \(x=\frac{1}{2}\)
Vậy \(Min=\frac{3}{4}\) khi \(x=\frac{1}{2}\)
Bài 2:
\(x^2+10x+2041=x^2+10x+25+2016\)
\(=\left(x^2+10x+25\right)+2016\)
\(=\left(x+5\right)^2+2016\ge2016\)
Dấu "=" khi \(x=-5\)
Vậy \(Min=2016\) khi \(x=-5\)
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cho x+y =1 . tinh gia tri cua bieu thuc A=x^3+y^3+3xy
chox-y=1. tinh gia tri cua bieu thuc B=x^3-y^3-3xy
cho x+y=1 . tinh gia tri cua bieu thuc C=x^3+y^3+3xy(x^2+y^2)+6x^2*y^2(x+y)
Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)
\(=\left(x+y\right)^3=1^3=1\)
Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)
\(=x^2+xy+y^2-3xy=x^2-2xy+y^2=\left(x-y\right)^2=1^2=1\)
Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)
\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)
\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)
\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)
\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)
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Cho bieu thuc \(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-4x-1}{x^2-1}\right).\frac{x+2016}{x}\)
a, Voi gia tri nguyen nao cua x thi bieu thuc A co gia tri nguyen
b,Voi gia tri nao cua x thi A co gia tri duong
tim tap xac dinh cua ham so
y=3nhan x binh phuong -x tren cho( gia tri tuyet doi cua x binh phuog -x) +(giá tri tuyet doi cua x-1)
y=(can x+2)+(can 3-2x) tren cho (gia tri tuyet doi cua x) -1
mot phan thuc co gia tri bangh 0 khi gia tri cua tu thuc bang 0 con gia tri cua mau thuc khac 0. tim cac gia tri cua x de gia tri cua phan thuc \(\dfrac{x^2-10x+25}{x^2-5}\)bang 0
Đặt \(A=\dfrac{x^2-10x+25}{x^2-5}\)
ĐK : \(x^2-5\ne0\\ \Leftrightarrow\left\{{}\begin{matrix}x\ne\sqrt{5}\\x\ne-\sqrt{5}\end{matrix}\right.\)
\(A=0\\ \Leftrightarrow\dfrac{x^2-10x+25}{x^2-5}=0\\ \Leftrightarrow x^2-10x+25=0\\ \Leftrightarrow\left(x-5\right)^2=0\\ \Leftrightarrow x=5\left(TM\right)\)
Vậy x =5 thì A =0
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cho bieu thuc p=(x+1)(x+√x)/√x-x-√x, voi x>0
a/ rut gon bieu thuc
b/ tim gia tri cua x de gia tri cua bieu thuc p bang 2