tính giá trị :\(Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
Tinh gia tri bieu thuc :
\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right).0\)
\(=0\)
\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)=\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right).0=0\)
tính nhanh\(\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)x\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right)\)
Thực hiện phép tính :
a, \(\frac{5922.6001-69}{5932+6001.5931}\)
b, \(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}+\frac{1234}{99999}\right).\left(\frac{1}{2}_{ }-\frac{1}{3}-\frac{1}{6}\right)\)
Tính:
\(\left(\frac{99^9}{11^9}-\frac{99^{99}}{11^{99}}-\frac{99^{999}}{11^{999}}\right).\left(\frac{1}{5}-\frac{1}{7}-\frac{2}{35}\right)\)
\(\left(\frac{99^9}{11^9}-\frac{99^{99}}{11^{99}}-\frac{99^{999}}{11^{999}}\right)\left(\frac{1}{5}-\frac{1}{7}-\frac{2}{35}\right)\)
\(=\left(\frac{99^9}{11^9}-\frac{99^{99}}{11^{99}}-\frac{99^{999}}{11^{999}}\right)\left(\frac{7}{35}-\frac{5}{35}-\frac{2}{35}\right)\)
\(=\left(\frac{99^9}{11^9}-\frac{99^{99}}{11^{99}}-\frac{99^{999}}{11^{999}}\right).0\)
\(=0\)
bài dễ thế không ai làm được hay thật
Cho 3 số a,b,c\(\ne\)0 và \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\)
Tính giá trị của biểu thức \(\left(a^9+b^9\right)\left(b^{99}+c^{99}\right)\left(c^{999}+a^{999}\right)\)
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\Rightarrow\frac{1}{a+b+c}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)
\(\Leftrightarrow\frac{a+b}{ab}=\frac{-a-b}{c\left(a+b+c\right)}\Rightarrow c\left(a+b\right)\left(a+b+c\right)=ab\left(-a-b\right)\)
\(\Rightarrow\left(a+b\right)\left(ca+cb+c^2\right)+ab\left(a+b\right)=0\Rightarrow\left(a+b\right)\left(ca+cb+c^2+ab\right)=0\)
\(\Rightarrow\left(a+b\right)\left(c+a\right)\left(b+c\right)=0\)
=> Trong 3 số a,b,c có 2 số đối nhau.Giả sử a = -b thì a9 + b9 = 0.
Tương tự giả sử b = -c hay a = -c thì b99 + c99 = 0 hay c999 + a999 = 0
Vậy biểu thức cần tính bằng 0.
\(A=\left(\frac{9}{1999}+\frac{99}{999}+\frac{999}{9999}\right)\cdot\left(\frac{1}{5}-\frac{1}{4}+\frac{1}{20}\right)\)
A=(9/1999+99/999+999/9999).(1/5-1/4+1/20)
A=(9/1999+99/999+999/9999).(-1/20+1/20)
A=(9/1999+99/999+999/9999).0
A=0
Vì mọi số nhân vs 0 thì đều = 0 kể cả phân số
mk nhanh nhất ủng hộ nha
\(A=\left(\frac{9}{1999}+\frac{99}{999}+\frac{999}{9999}\right)\cdot0\)
A=0
\(\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right)...\left(\frac{1}{999}+1\right)\)\(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right).\left(\frac{1}{4}-1\right)...\left(\frac{1}{999}-1\right)\)
Tính:
\(\left(\frac{1000}{1}+\frac{999}{2}+\frac{998}{3}+\frac{997}{4}+...+\frac{2}{999}+\frac{1}{1000}\right)\)\(:\)\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{1000}\right)\)
Tính
a)\(\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right)...\left(\frac{1}{999}+1\right);\)
b)\(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right).\left(\frac{1}{4}-1\right)...\left(\frac{1}{1000}-1\right)\)
c)\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{99}{10^2}\)
lam on ai biet thi chi trong toi nay tui se cho ma ngay mai la phai nop rui
( 1/2 + 1 ) . ( 1/3 + 1 ) . ( 1/4 + 1 ) . ... . ( 1/999 + 1 )
= 3/2 . 4/3 . 5/4 . ... . 1000/999
= 3 . 4 . 5 . ... . 1000 / 2 . 3 . 4 . ... . 999
= 500