Tinh gia tri bieu thuc:
\(D=\left(\frac{3}{2}\right)^2+\left(\frac{1}{4}\right)^2-\left(\frac{1}{2}\right)^3\)
Cho 2 so thuc a, b thoa man dieu kien ab= 1, a+ b\(\ne\)0. Tinh gia tri bieu thuc :
P= \(\frac{1}{\left(a+b\right)^3}\left(\frac{1}{a^3}+\frac{1}{b^3}\right)+\frac{3}{\left(a+b\right)^4}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+\frac{6}{\left(a+b\right)^3}\left(\frac{1}{a}+\frac{1}{b}\right)\)
Tinh hop ly gia tri bieu thuc sau
\(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{2}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{8}\right)\left(1-\frac{1}{10}\right)\)
(1-1/3)x(1-1/5)x(1-1/7)x(1-1/9)x(1-1/2)x(1-1/4)x(1-1/6)x(1-1/8)x(1-1/10)
=2/3x4/5x6/7x8/9x1/2x3/4x5/6x7/8x9/10
=2x4x6x8x1x3x5x7x9 /3x5x7x9x2x4x6x8x10
=1/10
Tinh hop ly gia tri cac bieu thuc sau
c) \(6\frac{5}{12}:2\frac{3}{4}+11\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{5}\right)\)
d) \(\left(\frac{3}{5}+0,415-\frac{3}{200}\right).2\frac{2}{3}.0,25\)
cau 1: tinh gia tri cua x thoa man
\(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\sqrt{2}\right)\left(2\sqrt{2}-x\right)=-3\)
cau 2.tinh GTLN cua bieu thuc
\(2x-2x^2+13\)
cau 3. tinh gia tri cua bieu thuc
\(\frac{3^{\left(x+y\right)^2}}{3^{\left(x-y\right)^2}}\)voi xy=\(\frac{1}{2}\)
cau 4. tim GTLN cua
\(-3x^2-6x-4\)
cau 5. cho ham so : f(x)=\(\frac{1}{5x+9}\)
tinh gia tri cua \(f\left(\frac{40}{25}\right)\)
cau 6. cho hinh thang can ABCD . Day nho AB,goc D bang 64 do. tinh so do goc ngoai tai A
A=\(\frac{\left(1^3+2^3+3^3+...+10^3\right)\cdot\left(x^2+y^2\right)\cdot\left(x^3+y^3\right)\cdot\left(x^4+y^4\right)}{1^2+2^2+3^2+...+10^2}\)
Tinh gia tri bieu thuc
1) Tinh gia tri cua bieu thuc:
A=\(\frac{\left(1+2+...+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\right)\left(2,4.42-21.4,8\right)}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
B=\(\frac{4^6.9^5+6^9.120}{-8^4.3^{12}+6^{11}}\)
\(A=\frac{\left(1+2+...+100\right)\left(\frac{1}{2}^2-...-\frac{1}{5}\right)\left(2,4.42-21.4,8\right)}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)
=> \(A=\frac{\left(1+2+...+100\right)\left(\frac{1}{2}-...-\frac{1}{5}\right).0}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)= 0
Tinh gia tri bieu thuc :
\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right).0\)
\(=0\)
\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)=\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right).0=0\)
tinh nhanh gia tri cua bieu thuc voi x,y nhan bat ki gia tri nao:
\(\frac{1}{3}x^2y\left(3xy\right)^2y^4+\frac{1}{2}x\left(-2xy\right)^3y^4+x^4y^7+18\)
cho f(x)=\(y=\frac{x^3}{3x^2-3x+1}\)
tinh gia tri bieu thuc sauA=\(y\left(\frac{1}{112}\right)+y\left(\frac{2}{112}\right)+...+y\left(\frac{111}{112}\right)\)