Gia tri cua bieu thuc 1+2+3+4...+199+200 la
gia tri bieu thuc a=1-2+3-4+5-6+....+199-200 la
voi A = 2 gia tri cua bieu thuc A la
A =
tim gia tri cua bieu thuc a de bieu thuc A co gia tri lon nhat tim gia tri lon nhat do
1. cho x+y=7 va xy=8 gia tri cua bieu thuc x3+y3 = ?
2. gia tri lon nhat cua bieu thuc B= 1+3x-x2 la ?
1. cho x+y=7 va xy=8 gia tri cua bieu thuc x3+y3 = ?
2. gia tri lon nhat cua bieu thuc B= 1+3x-x2 la ?
gia tri cua bieu thuc ; m+4050⋮5 voi m=2945 la ............... gia tri cua bieu thuc;n-57*13 voin=9876 la...................
\(m+4050=2945+4050=6995\)
\(n-57\times13=9876-57\times13=9876-741=9135\)
cho bieu thuc A=[x+2/x^2-x+x-2/x^2+x].x^2-1/x^2+2
a) tim dieu kien cua x de gia tri cua bieu thuc A duoc xac dinh
b) tinh gia tri cua bieu thuc A voi x = -200
a) \(A=\left[\dfrac{x+2}{x^2-x}+\dfrac{x-2}{x^2+x}\right].\dfrac{x^2-1}{x^2-x}\)
\(A=\left[\dfrac{x+2}{x\left(x-1\right)}+\dfrac{x-2}{x\left(x+1\right)}\right].\dfrac{x^2-1}{x^2+2}\)
\(A=\left[\dfrac{\left(x+2\right)\left(x+1\right)+\left(x-2\right)\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\right].\dfrac{x^2-1}{x^2+2}\)
\(A=\left[\dfrac{x^2+2x+x+2+x^2-2x-x+2}{x\left(x-1\right)\left(x+1\right)}\right].\dfrac{x^2-1}{x^2+2}\)
\(A=\dfrac{2x^2+4}{x\left(x^2-1\right)}.\dfrac{x^2-1}{x^2+2}\)
\(A=\dfrac{2\left(x^2+2\right)\left(x^2-1\right)}{x\left(x^2-1\right)\left(x^2+2\right)}=\dfrac{2}{x}\)
b) Thay \(x=-200\) vào biểu thức \(A=\dfrac{2}{x}\) ta được :
\(A=\dfrac{2}{x}=\dfrac{2}{-200}=\dfrac{-2}{200}=\dfrac{-1}{100}\)
1.cho phan thuc P=\(\frac{5x-4y}{5x+4y}\)voi 25x2+16y2=50xy.khi do gia tri cua bieu thuc P=\(\frac{1+P^2}{1-P^2}\)la
2.gia tri lon nhat cua bieu thuc B=\(\frac{14}{\frac{x^2}{4}-3x+16}la\)
3.gia tri cua a de da thuc A=2x3+7x2+ax+3 chia het cho B=(x+1)2la a=?
\(25x^2+16y^2=50xy\)
\(\Leftrightarrow\) \(\left(5x+4y\right)^2-40xy=50xy\)
\(\Leftrightarrow\) \(\left(5x+4y\right)^2=90xy\)
Mặt khác, ta cũng có: \(25x^2+16y^2=50xy\)
\(\Leftrightarrow\) \(\left(5x-4y\right)^2=10xy\)
Do đó:
\(P^2=\frac{\left(5x-4y\right)^2}{\left(5x+4y\right)^2}=\frac{10xy}{90xy}=\frac{1}{9}\)
Vậy, \(P'=\frac{1+\frac{1}{9}}{1-\frac{1}{9}}=1\frac{1}{4}\)
1)
\(25x^2-40xy+16y^2=10xy\Leftrightarrow\left(5x-4y\right)^2=10xy\)
\(25x^2+40xy+16y^2=10xy\Leftrightarrow\left(5x+4y\right)^2=90xy\)
\(P^2=\frac{1}{9}\Leftrightarrow Q=\frac{1+P^2}{1-P^2}=\frac{1+\frac{1}{81}}{1-\frac{1}{81}}=\frac{82}{80}=\frac{41}{40}\)
Cho bieu thuc m x 3 + n x 4 + p x 2 + m + 2 x p . Voi m + n + p = 2009 thi gia tri cua bieu thuc la :
=m x 3 + m + n x 4 + p x 2 + p x 2
=m x 3 + m x 1 + n x 4 + p x 2 + p x 2
=m x (3 + 1) + n x 4 + p x (2 + 2)
=m x 4 + n x 4 + p x 4
=(m + n + p) x 4
=2009 x 4
=8036
=m + n + p)
1) Cho bieu thuc: \(B=\left(\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right):\frac{x+16}{\sqrt{x}+2}\left(x\ge0,x\ne16\right)\)
a) Cho bieu thuc A= \(\frac{\sqrt{x}+4}{\sqrt{x}+2}\) ; voi cac cua bieu thuc A va B da cho, hay tim cac gia tri cua x nguyen de gia tri cua bieu thuc B(A;-1) la so nguyen