Cho a,b,c,d là 4 số khác 0 thoả mãn\(b^2=ac,c^2=bd\) và\(b^3+c^3+d^3\)khác 0. Chứng minh rằng:\(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{\left(a+b-c\right)^3}{\left(b+c-d\right)^3}=\frac{a}{d}\)
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1. In all the world, there (be) ____are______ only 14 mountains that (reach) ____reach______above 8,000 meters.
2. He sometimes (come) ______comes____ to see his parents.
3. When I (come) _____came_____, she (leave) ____had left______for Dalat ten minutes ago.
4. My grandfather never (fly) ____has never flown______ in an airplane, and he has no intention of ever doing so.
5. We have just (decide) ____decided_____ that we (undertake) _____would undertake_______ the job.
6. He told me that he (take) _____would take_____ a trip to California the following week.
7. I knew that this road (be) ____was______ too narrow.
8. Right now I (attend) ______am attending____ class. Yesterday at this time I (attend) __was attending________class.
9. Tomorrow I'm going to leave for home. When I (arrive) __arrive________at the airport, Mary (wait) ____will be waiting______ for me.
10. Margaret was born in 1950. By last year, she (live) __had lived________on this earth for 55 years .
11. The traffic was very heavy. By the time I (get) ______got____to Mary's party, everyone had already (arrive) __arrived________
12. I will graduate in June. I (see) ___will see_______ you in July. By the time I (see) ___see_______ you , I (graduate) ___will have graduate_______.
13. I (visit) _____visited_____ my uncle's home regularly when I (be) ____was______ a child.
14. That book (be) ____has been______ on the table for weeks. Haven't You (not read) ____read______ it yet ?
15. David (wash) ___is washing_______ his hands. He has just (repair) ____repaired______ the TV set.
16. Have You (be) ______been____here before? Yes, I (spend) ____spent______ my holidays here last year.
17. We have never (meet) _____met_____ him. We don't know what he (look) _____looks_____ .
18. The car (be) ____will be______ ready for him by the time he (come) ___comes_______tomorrow.
19. On arriving at home I (find) ____found______that she had just (leave) ____left______a few minutes before.
20. When we (arrive) ____arirve______ in London tonight, it will be probably (rain) __raining________.
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Câu 1:
a) \(\left(x-\frac{1}{2}\right)^3=-27\) \(\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\left(-3\right)^3\)
\(\Leftrightarrow x-\frac{1}{2}=-3\)\(\Leftrightarrow x=-\frac{5}{2}\)
Vậy \(x=-\frac{5}{2}\)
b( \(2x^2+x=0\)\(\Leftrightarrow x\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\2x=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{2}\end{cases}}\)
Vậy \(x=0\)hoặc \(x=-\frac{1}{2}\)
c) \(5^{x+2}=625\)\(\Leftrightarrow5^{x+2}=5^4\)
\(\Leftrightarrow x+2=4\)\(\Leftrightarrow x=2\)
Vậy \(x=2\)
d) \(\frac{7^x+7^{x+1}+7^{x+2}+7^{x+3}}{2^2.5^2.7^2}=2^2\)
\(\Leftrightarrow7^x+7^{x+1}+7^{x+2}+7^{x+3}=2^2.2^2.5^2.7^2\)
\(\Leftrightarrow7^x+7^x.7+7^x.7^2+7^x.7^3=\left(2.2.5.7\right)^2\)
\(\Leftrightarrow7^x+7^x.7+7^x.49+7^x.343=140^2\)
\(\Leftrightarrow7^x.\left(1+7+49+343\right)=19600\)
\(\Leftrightarrow7^x.400=19600\)
\(\Leftrightarrow7^x=49=7^2\)
\(\Leftrightarrow x=2\)
Vậy \(x=2\)
Câu 2:
a) \(C=1+4+4^2+4^3+.......+4^{48}\)
\(\Rightarrow4C=4+4^2+4^3+4^4+........+4^{49}\)
\(\Rightarrow4C-C=4^{49}-1\)
\(\Rightarrow3C=4^{49}-1\)
\(\Rightarrow C=\frac{4^{49}-1}{3}\)
b) Ta có: \(3C+1=4^{49}-1+1=4^{49}=4^{7.7}=\left(4^7\right)^7⋮4^7\)( đpcm )
c) \(C=1+4+4^2+4^3+........+4^{48}\)
\(=\left(1+4+4^2\right)+\left(4^3+4^4+4^5\right)+........+\left(4^{46}+4^{47}+4^{48}\right)\)
\(=\left(1+4+4^2\right)+4^3.\left(1+4+4^2\right)+........+4^{46}.\left(1+4+4^2\right)\)
\(=\left(1+4+4^2\right).\left(1+4^3+....+4^{46}\right)\)
\(=\left(1+4+16\right).\left(1+4^3+........+4^{46}\right)\)
\(=21.\left(1+4^3+.....+4^{46}\right)⋮21\)
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Ta có
\(\frac{x}{3}=\frac{y}{4};\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{28}=\frac{3y}{60}=\frac{2x}{30}\)
Ap dụng tính chất DTSBN ta có
\(\frac{2x}{30}=\frac{3y}{60}=\frac{z}{28}=\frac{2x+3y+z}{30+60+28}=\frac{186}{118}=\frac{93}{59}\)
\(\hept{\begin{cases}\frac{x}{15}=\frac{93}{59}\\\frac{y}{20}=\frac{93}{59}\\\frac{z}{28}=\frac{93}{59}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1395}{59}\\y=\frac{1860}{59}\\z=\frac{2604}{59}\end{cases}}\)
Ta có : \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{3}.\frac{1}{5}=\frac{y}{4}.\frac{1}{5}\Rightarrow\frac{x}{15}=\frac{y}{20}\)
\(\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{y}{5}.\frac{1}{4}=\frac{z}{7}.\frac{1}{4}\Rightarrow\frac{y}{20}=\frac{z}{28}\)
\(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\)
Theo tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}=\frac{2x+3y+z}{2.15+3.20+28}=\frac{186}{118}=\frac{93}{59}\)
\(\Rightarrow\frac{x}{15}=\frac{93}{59}\Rightarrow x=\frac{93}{59}.15=\frac{1395}{59}\)
\(\frac{y}{20}=\frac{93}{59}\Rightarrow y=\frac{93}{59}.20=\frac{1860}{59}\)
\(\frac{z}{28}=\frac{93}{59}\Rightarrow z=\frac{93}{59}.28=\frac{2604}{59}\)
Vậy : \(\left(x;y;z\right)=\left(\frac{1395}{59};\frac{1860}{59};\frac{2604}{59}\right)\)
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Tìm x,y
1+3y/12 =1+5y/5x =1+7y/4x
Giải:Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{1+3y}{12}=\frac{1+5y}{5x}=\frac{1+7y}{4x}=\frac{1+5y-1-7y}{x}=\frac{-2y}{x}\)
\(\Rightarrow x+3xy=-24y\Rightarrow x+3xy+24y=0\Rightarrow x\left(3y+1\right)+8\left(3y+1\right)=8\)
\(\Rightarrow\left(x+8\right)\left(3y+1\right)=8\)
Đến đây đơn giản rồi.Bạn tự làm nha.....................................
Ta có:1+3y/12=1+5y/5x=1+7y/4z=1+3y+1+7y/12+4x=2+10y
=> 1+5y/5x=2+10y/12+4x=>2+10y/10x=2+10y/12+4x
=>10x=12+4x
6x=12
=>x=12
bạn thấy x để tìm ý nhé
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