Cho \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\) .CMR:
\(\dfrac{1}{a^{1995}}+\dfrac{1}{b^{1995}}+\dfrac{1}{c^{1995}}=\dfrac{1}{a^{1995}+b^{1995}+c^{1995}}\)
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bài 1:
Cho \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\) .Chứng minh rằng:
\(\dfrac{1}{a^{1995}}+\dfrac{1}{b^{1995}}+\dfrac{1}{c^{1995}}=\dfrac{1}{a^{1995}+b^{1995}+c^{1995}}\)
30 tháng 4 2021 lúc 20:17
rút gọn phân số:
\(a.\dfrac{-315}{540}\)
\(b.\dfrac{25.13}{26.35}\)
\(c.\dfrac{3.13-13.18}{15.40-80}\)
\(d.\dfrac{-1997.1996+1}{\text{(}-1995\text{)}.\left(-1997\right)+1996}\)
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19 tháng 7 2023 lúc 13:20
so sánh các số hữu tỉ sau bằng cách hợp lí:
a) -0,2; \(\dfrac{1}{1000}\)
b) \(\dfrac{\text{13}}{\text{-35}};\dfrac{\text{-11}}{\text{28}}\)
c) \(\dfrac{2022}{-2021};\dfrac{-1995}{1996}\)
d) \(\dfrac{\text{-18}}{\text{13}};\dfrac{\text{181818}}{\text{131313}}\)
Cô làm rồi em nhé:
https://olm.vn/cau-hoi/giup-em-voiii.8161766187032
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13 tháng 7 2023 lúc 18:41
1.thực hiện phép tính
a.3990-(463 x72 -39 x 39):15 b.1995-321x3-6020:35
c.\(3-\dfrac{7}{11}\) d.\(\dfrac{1}{4}:\dfrac{1}{6}+5\)
a: =3990-(463*72-39^2):15
=3990-2121=1869
b: =1995-963-172=860
c: =33/11-7/11=26/11
d: =3/2+5=13/2
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`@` `\text {Answer}`
`\downarrow`
`a,`
`3990 - (463 \times 72 - 39 \times 39) \div 15`
`= 3990 - (33336 - 1521) \div 15`
`= 3990 - 31815 \div 15`
`= 3990 - 2121`
`= 1869`
`b,`
`1995 - 321 \times 3 - 6020 \div 35`
`= 1995 - 963 - 172`
`= 1032 - 172`
`= 860`
`c,`
`3 - 7/11`
`= 33/11 - 7/11`
`= 26/11`
`d,`
`1/4 \div 1/6 + 5`
`= 1/4 \times 6 + 5`
`= 3/2 + 5`
`= 13/2`
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15 tháng 3 2021 lúc 12:19
Rút gọn phân số:a) dfrac{2929-101}{2.1149+404}b) dfrac{6.9-2.17}{63.3-119}c) dfrac{3.13-13.18}{15.40-80}d) dfrac{-1997-1996+1}{left(-1995right).left(-1997right)+1996}e) dfrac{2.3+4.6+14.21}{3.5+6.10+21.35}g) dfrac{left(-5right)^3.40.4^3}{135.left(-3right)^{14}.left(-100right)^0} h) dfrac{18.34+left(-18right).124}{-36.17+9.left(52right)}j) dfrac{9.11+32.9}{23.15+12.23}k) dfrac{12.13+24.26+36.39}{24.26+48.52+72.78}
Đọc tiếp
Rút gọn phân số:
a) \(\dfrac{2929-101}{2.1149+404}\)
b) \(\dfrac{6.9-2.17}{63.3-119}\)
c) \(\dfrac{3.13-13.18}{15.40-80}\)
d) \(\dfrac{-1997-1996+1}{\left(-1995\right).\left(-1997\right)+1996}\)
e) \(\dfrac{2.3+4.6+14.21}{3.5+6.10+21.35}\)
g) \(\dfrac{\left(-5\right)^3.40.4^3}{135.\left(-3\right)^{14}.\left(-100\right)^0}\)
h) \(\dfrac{18.34+\left(-18\right).124}{-36.17+9.\left(52\right)}\)
j) \(\dfrac{9.11+32.9}{23.15+12.23}\)
k) \(\dfrac{12.13+24.26+36.39}{24.26+48.52+72.78}\)
b) \(\dfrac{6\cdot9-2\cdot17}{63\cdot3-119}\)
\(=\dfrac{2\left(3\cdot9-17\right)}{7\cdot\left(3\cdot9-17\right)}\)
\(=\dfrac{2}{7}\)
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giải các phương Trình sau
a) \(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)
b) \(\dfrac{x+1}{1998}+\dfrac{x+2}{1997}=\dfrac{x+3}{1996}+\dfrac{x+4}{1995}\)
c) \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)
a) \(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)
\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}=\dfrac{x+100}{96}+\dfrac{x+100}{95}\)
\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}-\dfrac{x+100}{96}-\dfrac{x+100}{95}=0\)
\(\Rightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)
Vì \(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\ne0\) nên \(x+100=0\Leftrightarrow x=-100\)
b) \(\dfrac{x+1}{1998}+\dfrac{x+2}{1997}=\dfrac{x+3}{1996}+\dfrac{x+4}{1995}\)
\(\Rightarrow\dfrac{x+1}{1998}+1+\dfrac{x+2}{1997}+1=\dfrac{x+3}{1996}+1+\dfrac{x+4}{1995}+1\)
\(\Rightarrow\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}=\dfrac{x+1999}{1996}+\dfrac{x+1999}{1995}\)
\(\Rightarrow\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}-\dfrac{x+1999}{1996}-\dfrac{x+1999}{1995}=0\)
\(\Rightarrow\left(x+1999\right)\left(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\right)=0\)
Vì \(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\ne0\) nên \(x+1999=0\Leftrightarrow x=-1999\)
c) \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)
\(\Rightarrow\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1+\dfrac{205-x}{95}+1=0\)
\(\Rightarrow\dfrac{300-x}{99}+\dfrac{300-x}{97}+\dfrac{300-x}{95}=0\)
\(\Rightarrow\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)
Vì \(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\ne0\) nên \(300-x=0\Leftrightarrow x=300\)
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\(\dfrac{x+1}{1998}\)+ \(\dfrac{x+2}{1997}\)=\(\dfrac{x+3}{1996}\)+\(\dfrac{x+4}{1995}\)
=>(x+1/1998+1)+(x+2/1997+1)=(x+3/1996+1)+(x+4/1995+1)
=>x+1999=0
=>x=-1999
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giải các phương trình sau
\(\dfrac{x+1}{1998}+\dfrac{x+2}{1997}=\dfrac{x +3}{1996}+\dfrac{x+4}{1995}\)
Vào trang cá nhân của t mà xem.T vừa làm r.Lười gõ lại lắm T^T
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\(\dfrac{x+1}{1998}+\dfrac{x+2}{1997}=\dfrac{x+3}{1996}+\dfrac{x+4}{1995}\)
\(=\dfrac{x+1}{1998}+\dfrac{x+2}{1997}-\dfrac{x+3}{1996}-\dfrac{x+4}{1995}=0\)
\(=\dfrac{x+1}{1998}+1+\dfrac{x+2}{1997}+1-\dfrac{x+3}{1996}-1-\dfrac{x+4}{1995}-1=0\)
\(=\dfrac{x+1999}{1998}+\dfrac{x+1999}{1998}-\left(\dfrac{x+3}{1996}+1\right)-\left(\dfrac{x+4}{1995}+1\right)=0\)
\(=\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}-\dfrac{x+1999}{1996}-\dfrac{x+1999}{1995}=0\)
\(=\left(x+1999\right)\left(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\right)=0\)
⇔\(x+1999=0\)
Vậy \(x=-1999\)
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17 tháng 8 2021 lúc 15:21
Cho biểu thức sau: \(A=\left(1-\dfrac{2\sqrt{a}}{a+1}\right):\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}+a+1}\right)\)
1, Rút gọn A
2, Tính giá trị của A khi \(a=1996-2\sqrt{1995}\)
1, A=\(\left(1-\dfrac{2\sqrt{a}}{a+1}\right):\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}+a+1}\right)\)
ĐKXĐ: a≥0
A=\(\left(1-\dfrac{2\sqrt{a}}{a+1}\right):\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{2\sqrt{a}}{\sqrt{a}\left(a+1\right)+1\left(a+1\right)}\right)\)
A=\(\left(\dfrac{a+1}{a+1}-\dfrac{2\sqrt{a}}{a+1}\right):\left(\dfrac{a+1}{\left(\sqrt{a}+1\right)\left(a+1\right)}-\dfrac{2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a+1\right)}\right)\)
A=\(\left(\dfrac{a+1-2\sqrt{a}}{a+1}\right):\left(\dfrac{a+1-2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a+1\right)}\right)\)
A=\(\left(\dfrac{a+1-2\sqrt{a}}{a+1}\right).\left(\dfrac{\left(a+1\right)\left(\sqrt{a}+1\right)}{a+1-2\sqrt{a}}\right)\)
A=\(\sqrt{a}+1\)
Vậy A=\(\sqrt{a}+1\)
2, a=1996-2\(\sqrt{1995}\)
a=\(1995-2\sqrt{1995}+1\)
a=\(\left(\sqrt{1995}-1\right)^2\) (TMĐKXĐ)
thay a=\(\left(\sqrt{1995}-1\right)^2\) vào A ta có:
A=\(\sqrt{\left(\sqrt{1995}-1\right)^2}+1\)
A=\(\sqrt{1995}\)
Vậy a=1996-2\(\sqrt{1995}\) thì A=\(\sqrt{1995}\)
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