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11 tháng 2 2020

I think that we have to prove \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=-2\)

We have \(a+b+c=abc\)

\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)

We have \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=0\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=0\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=0\)( Because \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\))

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=-2\)

So...

Đề thi đánh giá năng lực

11 tháng 2 2020

Ta có: 1+1x1

= 1+1

=2

=>1+1x1=2 (điều phải chứng minh)

Vậy 1+1x1=2

Chúc bạn học tốt!

11 tháng 2 2020

are you bình thường

9 tháng 2 2020

Chọn D

Phép đối xứng qua mặt phẳng (P) biến đường thẳng d thành chính nó khi và chỉ khi d nằm trên (P) hoặc d⊥(P)d⊥(P). Có đúng ko mn, mk nghĩ là thế, ai hc lớp 12 tl hộ mk vs.

9 tháng 2 2020

a) Trục Ox là đường thẳng đi qua O(0, 0, 0) và nhận i→=(1,0,0) làm vectơ chỉ phương nên có phương trình tham số là:

Giải Toán 12 nâng cao | Giải bài tập Toán lớp 12 nâng cao

* Tương tự, trục Oy có phương trình

Giải Toán 12 nâng cao | Giải bài tập Toán lớp 12 nâng cao

Trục Oz có phương trình

Giải Toán 12 nâng cao | Giải bài tập Toán lớp 12 nâng cao

b) Đường thẳng đi qua M0 (x0,y0,z0) song song với trục Ox sẽ có vectơ chỉ phương là i→(1,0,0) nên có phương trình tham số là:

Giải Toán 12 nâng cao | Giải bài tập Toán lớp 12 nâng cao

tương tự ta có Phương trình của đường thẳng đi qua M0 (x0,y0,z0) và song song với Oy là:

Giải Toán 12 nâng cao | Giải bài tập Toán lớp 12 nâng cao

phương trình đường thẳng đi qua M0 (x0,y0,z0) và song song với Oz là

Giải Toán 12 nâng cao | Giải bài tập Toán lớp 12 nâng cao

c) Đường thẳng đi qua M(2, 0, -1) và có vectơ chỉ phương u→(-1,3,5) có phương trình tham số là

Giải Toán 12 nâng cao | Giải bài tập Toán lớp 12 nâng cao

có phương trình chính tắc là

Giải Toán 12 nâng cao | Giải bài tập Toán lớp 12 nâng cao

d) Đường thẳng đi qua N(-2, 1, 2) và có vectơ chỉ phương u→(0,0,-3) có phương trình tham số là

Giải Toán 12 nâng cao | Giải bài tập Toán lớp 12 nâng cao

Đường thẳng này không có Phương trình chính tắc.

e) Đường thẳng đi qua N(3, 2, 1) và vuông góc với mặt phẳng: 2x- 5y + 4= 0 nên nó nhận vectơ pháp tuyến của mặt phẳng này làn→(2,-5,0) là vectơ chỉ phương, nên ta có phương trình tham số là

Giải Toán 12 nâng cao | Giải bài tập Toán lớp 12 nâng cao

Đường thẳng này không có Phương trình chính tắc.

f) Đường thẳng đi qau P(2, 3, -1) và Q(1, 2, 4) sẽ nhận PQ→(-1,-1,5) là vectơ chỉ phương, nên có phương trình tham số là

Giải Toán 12 nâng cao | Giải bài tập Toán lớp 12 nâng cao

và có phương tình chính tắc là

Giải Toán 12 nâng cao | Giải bài tập Toán lớp 12 nâng cao

9 tháng 2 2020

ÔI THÔI CHẾT LM SAI

8 tháng 2 2020

bùn mún gớt  nc mắt lun T~T

Sao buồn thế bạn xênh gái 

6 tháng 2 2020

Gọi O là tâm hình bình hành; MN cắt AC tại J

Kẻ PE//SO thì E là trung điểm của OC suy ra \(IO=\frac{1}{2}PE=\frac{1}{4}SO\)

Gọi thể tích khối chóp là V

Ta có : \(\frac{V_{S.B'D'P}}{V_{S.BCD}}=\frac{SB'}{SB}.\frac{SD'}{SD}.\frac{SP}{SC}=\frac{3}{4}.\frac{3}{4}.\frac{1}{2}=\frac{9}{32}\)

suy ra \(V_{S.B'D'P}=\frac{9}{32}V_{S.BCD}=\frac{9}{64}V\)

Suy ra \(V_{BDD'BPC}=\frac{1}{2}V-\frac{9}{64}V=\frac{23}{64}V\)

pcm \(V_{MNDD'B'B}=\frac{9}{64}V\)

Ta có : \(V_{MNDD'B'B}=V_{J.BB'D'D}+V_{M.BB'J}+V_{N.DD'J}=V_{J.BB'D'D}+2.V_{M.BB'J}\)

Với \(V_{J.BB'D'D}=\frac{1}{2}V_{A.BB'D'D}=\frac{1}{2}\left[1-\left(\frac{3}{4}\right)^2\right].V_{A.SBD}\)\(=\frac{1}{2}.\frac{7}{16}.\frac{1}{2}V=\frac{7}{64}V\)

\(V_{M.BB'J}=V_{B'.BMJ}=\frac{1}{4}V_{S.BMJ}=\frac{1}{4}.\frac{1}{8}V_{S.ABD}\)\(=\frac{1}{4}.\frac{1}{8}.\frac{1}{2}V=\frac{1}{64}V\)

Vậy \(V_{MNDD'B'B}=V_{J.BB'D'D}+2.V_{M.BB'J}=\frac{7}{64}V+2\frac{1}{64}V=\frac{9}{64}V\left(đpcm\right)\)

20 tháng 3 2020

Gọi H là khối đa diện nằm bên dưới mp(MNP)
Gọi h,S,V lần lượt là chiều cao, diện tích đáy, thể tích của khối chóp S.ABCD
Dễ thấy:

\(\hept{\begin{cases}S_{DNU}=S_{BMT}=S_{AMN}=\frac{1}{4}S_{ABD}=\frac{1}{8}S\\d\left(p;\left(ABCD\right)\right)=\frac{1}{2}h;d\left(q;\left(ABCD\right)\right)=d\left(r;\left(ABCD\right)\right)=\frac{1}{4}h\end{cases}}\)

Ta có: \(S_{CTU}=S+\frac{1}{8}S=\frac{9}{8}S\)

\(V_{P\cdot CTU}=\frac{1}{3}\cdot\frac{1}{2}h\cdot\frac{9}{8}S=\frac{9}{16}V\)

\(V_{Q\cdot UDN}=V_{R\cdot BMT}=\frac{1}{3}\cdot\frac{1}{4}\cdot\frac{1}{8}S=\frac{1}{32}V\)

\(V_H=V_{P\cdot CTU}-V_{Q\cdot UDN}-V_{R\cdot BMT}=\frac{1}{2}V\)

=> đpcm
Nguồn:  Chú lùn thứ 8 

4 tháng 2 2020

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4 tháng 2 2020

Thui khỏi bt lm òi * nhắn lun ko bt cho nhanh lại cn lm màu *