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1,
a) 3,2x +(-12)x +2,7=-4,9 b)-5,6x+2,9x-3,86=-9,8
=>3,2x+(-12)x=-4,9-2,7 =>-5,6x+2,9x=-9,8+3,86
=>3,2x+(-12)x=-7,6 =>-5,6x+2,9x=-5,94
=>x[3,2+(-12)]=-7,6 =>x[(-5,6)+2,9]=-5,94
=>2x=-7,6=>x=-7,6:2=-3,8 =>-2,7x=-5,94=>x=(-5,94):(-2,7)=2,2
Gọi \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{x}=\dfrac{1023}{1024}\)
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{x}=\dfrac{1023}{1024}\)
VẬy x là một lũy thừa của 2. Đặt x = 2y , ta có:
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^y}\)
\(\Rightarrow2A=1+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{y-1}}\)
\(\Rightarrow2A-A=1+\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{y-1}}-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^8}\right)\)
\(=A-\dfrac{1}{2^y}\)
Vậy \(1-\dfrac{1}{2^y}=\dfrac{1023}{1024}\Leftrightarrow\dfrac{1}{2^y}=\dfrac{1}{1024}\Leftrightarrow2^y=1024\Rightarrow x=1024\)
Vậy x = 1024
ta có :
\(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}=\dfrac{bxz-cyz}{ax}=\dfrac{cxy-azy}{by}=\dfrac{ayz-bxz}{cz}\)áp dụng tính chất dãy tỉ số bằng nhau, ta có:\(\dfrac{bzx-cyx}{ax}=\dfrac{cxy-azy}{by}=\dfrac{ayz-bxz}{cz}=\dfrac{bzx-cyx+cxy-azy+ayz-bxz}{ax+by+cz}=0\)
suy ra : bz-cy=0 \(\Rightarrow bz=cy\Rightarrow\dfrac{y}{b}=\dfrac{z}{c}\left(1\right)\)
cx-az=0\(\Rightarrow cx=az\Rightarrow\dfrac{x}{a}=\dfrac{z}{c}\left(2\right)\)
ay-bx=0\(\Rightarrow ay=bx\Rightarrow\dfrac{x}{a}=\dfrac{y}{b}\left(3\right)\)
từ (1), (2) và (3) suy ra: \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\)
Ta có: \(\dfrac{bz-cy}{a}\)= \(\dfrac{cx-az}{b}\)=\(\dfrac{ay-bx}{c}\)
\(\Rightarrow\dfrac{abz-acy}{a^2}=\dfrac{bcx-baz}{b^2}=\dfrac{cay-cbx}{c^2}\)(nhân cả tử và mẫu với mẫu của phân số)
\(=\dfrac{abz-acy+bcx-baz+cay-cbx}{a^2+b^2+c^2}\) (t/c của dãy tỉ số bằng nhau)
(đến đây ta thấy tử = 0 vì chúng là các số đối nhau, abz-baz; acy-cay; bcx-cbx)
\(=\dfrac{0}{a^2+b^2+c^2}\)
\(\Rightarrow\)\(\dfrac{bz-cy}{a}\)= 0 (mà để là một phân số thì mẫu phải khác 0) suy ra a khác 0 vậy bz-cy=0 \(\Rightarrow bz=cy\Rightarrow\dfrac{z}{c}=\dfrac{y}{b}\)(1)
tương tự \(\dfrac{cx-az}{b}\)=0 suy ra \(\dfrac{x}{a}=\dfrac{z}{c}\) từ 1 và 2 suy ra điều phải c/m\(\)
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3.Ao Dai waswas frequentlyfrequently worn by both manman=>men and womenwomen.
4.Wearing uniformsWearing uniforms helphelp=>helps students feelfeel equal in many waysways.
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8.A newnew style of jeans havehave=>has just been introducedintroduced in in the USA.
9.InIn the 18th18thcentury, jean cloth waswas made completelycompletely byby=>from cotton.
10.Mr.Brown is interestedinterested in to go shopping to go shopping=>going shopping everyevery Sunday.
11.I wishwish I havehave=>had a new watch. MineMine is too too old.
12.Emma used toused to lie onon the beach when she liveslives=>lived by the seaby the sea.
13.WearingWearing uniforms encourageencourage=>encourages the students to beto be proud of theirtheir schools.
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15.Ao dai waswas frequentfrequent=>frequently worn worn by both men and womenboth men and women.
Error Identification.
1.Jean cloth was made complete=>completely from cotton.
2.Would you like come=>to come and visit me next summer?
3.Ao Dai was frequently worn by both man=>men and women.
4.Wearing uniforms help=>helps students feel equal in many ways.
5.Wearing uniforms encourages students to be proud about =>of their school.
6.She wishes she can =>could speak English as well as her teacher.
7.I seldom go to a football match.I'm not very interesting=>interested in football.
8.A new style of jeans have=>has just been introduced in the USA.
9.In the 18th century, jean cloth was made completely by=>from cotton.
10.Mr.Brown is interested in to go shopping=>going shopping every Sunday.
11.I wish I have=>had a new watch. Mine is too old.
12.Emma used to lie on the beach when she lives=>lived by the sea.
13.Wearing uniforms encourage=>encourages the students to be proud of their schools.
14.How long have you learning=>learnt English?
15.Ao dai was frequent=>frequently worn by both men and women.