tinh gia tri cua bieu thuc
(10^2+11^2+12^2):(13^2+14^2)
TD
Những câu hỏi liên quan
Tinh gia tri cua bieu thuc:
a.1/42+1//30+1/20+1/12+1/6+1/2
b.14/6+1/9+19/13+17/9+7/13+4/6
\(\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}=\frac{1}{1\times2}+\frac{1}{2\times3}+....+\frac{1}{6\times7}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-...-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)= 1 - 1/7 = 6/7
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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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tinh gia tri cua bieu thuc \
A= (100+98+96+....................+2)-97-95-......-1
B=1+2-3-4+5+6-7-8+9+10-11-12+............-299-300+301+302
Tinh gia tri bieu thuc
2-4-6+8+10-12-14+16+.......-2006+2008+2010-2012
Lionel Messicho 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 :
\(A=\dfrac{2}{x}=\dfrac{2}{-200}=\dfrac{-2}{200}=\dfrac{-1}{100}\)
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tinh gia tri bieu thuc
2^10*13+2^10*65
Tinh gia tri cua bieu thuc:
a, 4+11/8-5/6=
b, (1/2+4/7)-(3/7-3/10)=
\(4+\frac{11}{8}-\frac{5}{6}=\frac{96+33-20}{24}=\frac{109}{24}\)
\(\left(\frac{1}{2}+\frac{4}{7}\right)-\left(\frac{3}{7}-\frac{3}{10}\right)=\frac{1}{2}+\frac{4}{7}-\frac{3}{7}+\frac{3}{10}\)
\(\left(\frac{1}{2}+\frac{3}{10}\right)+\left(\frac{4}{7}-\frac{3}{7}\right)=\frac{4}{5}+\frac{1}{7}=\frac{28+5}{35}=\frac{33}{35}\)
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ban Chitanda Eru oi to bao nhung ma nhan voi bao nhieu ma ra duoc 96
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tinh gia tri cua cac bieu thuc sau roi so sanh cac ket qua tinh duoc
(-25)-(-12), 18+(14-42), (-18)-13+17
(-25)-(-12)= - 13
18+(14-42)=18+(-28) = -10
(-18)-13+17=(-31) +17= -14
\(\rightarrow\)-13 < -10 > -14\(\Rightarrow\) (-25)-(-12) < 18+(14-42) > (-18)-13+17
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\(\frac{\text{ (21^7+5^17).(3^14-5^12).(2^4-4^2)}}{\text{13^2+5^3.67^7}}\)tinh gia tri bieu thuc