thay xyz=2017, ta có:

\(D=\frac{xyzx}{xy+xyzx+xyz}+\frac{y}{yz+y+xzy}+\frac{z}{xz+z+1}\)

\(D=\frac{xz}{1+xz+z}+\frac{1}{x+1+xz}+\frac{z}{xz+x+1}=1\)

\(\text{Bài làm }\)

\(\text{ Gọi xyz = 2017}\)

\(\text{Ta có:}\) \(D=\frac{xyzx}{xy+xyzx+xyz}+\frac{y}{yz+y+xzy}+\frac{z}{xz+z+1}\)

           \(D=\frac{xz}{1+xz+z}+\frac{1}{x+1+xz}+\frac{z}{xz+x+1}=1\)

\(\text{# Chúc bạn học tốt #}\)

Không chắc nữa chắc sai -.-

a) \(A=\frac{3}{2}\left(-x^2y^3\right)^2\left(-x\right)^2\)

\(=\frac{3}{2}.[\left(-x\right)^{2^2}.\left(-x\right)^2].y^{3^2}\)

\(=\frac{3}{2}.\left[\left(-x\right)^4.\left(-x^2\right)\right].y^6\)

\(=\frac{3}{2}.x^6.y^6\)

b) A.B 

\(\)\((\frac{3}{2}.x^6.y^6).x^6y^6\)

\(=\frac{3}{2}.\left(x^6.x^6\right).\left(y^6.y^6\right)\)

\(=\frac{3}{2}.x^{12}.y^{12}\)

A + B

\((\frac{3}{2}.x^6.y^6)+\left(x^6y^6\right)\)

\(=\frac{3}{2}.\left(x^6.x^6\right)+\left(y^6.y^6\right)\)

\(=\frac{3}{2}.x^{12}+y^{12}\)

2) P-Q

\(\left(4x^2-5xy+3x-4\right)-(y^2+4+5xy-2y+x)\)

\(=4x^2-5xy-3x-4-y^2+4x^2+5xy-2y+x\)

\(=\left(4x^2-4x^2\right)+\left(5xy-5xy\right)+\left(3x-x\right)+4-y\)

\(=2x+4-y\)

b) Không biết làm :v

tu ve hinh :

a, xet tamgiac MBA va tamgiac MDC co :

goc BMA = goc DMC (doi dinh)

BM = CM do M la trung diem cua BC (GT)

MA = MD (GT)

=> tamgiac MBA = tamgiac MDC (c - g - c)

=> AB = DC (dn) 

tamgiac MBA = tamgiac MDC => goc CDM = goc MAB ma 2 goc nay slt

=> AB // CD (dh)

b, co tamgiac ABC vuong tai A => AB | AC (dn) ; AB // DC (cau a)

=> AC | DC (dl) => tamgiac ACD vuong tai C (dn) 

tamgiac MBA = tamgiac MDC => AB = CD (dn)

goc BAC = goc DCA = 90^o do tamgiac ABC vuong tai A va tamgiac DCA vuong tai C

xet tamgiac ACB va tamgiac CAD co AC chung

=> tamgiac ACB = tamgiac CAD (2cgv)

=> BC = AD (dn)

M la trung diem cua BC => M la trung diem cua AD => AM = AD/2 (tc)

=> AM = BC/2

Xét \(\Delta BDC\)có:

\(\widehat{DBC}+\widehat{DCB}+\widehat{CDB}=180^0\)

\(\Rightarrow\widehat{DCB}=\widehat{CDB}=45^0\)

\(\Rightarrow\widehat{BDE}=135^0\)(kề bù)

Xét \(\Delta BDE\)có:

\(\widehat{BDE}+\widehat{BED}+\widehat{DBE}=180^0\)

\(\Rightarrow135^0+2\cdot\widehat{DBE}=180^0\)(vì \(\Delta BDE\)cân tại B)

\(\Rightarrow\widehat{DBE}=22,5^0\)

Trong tam giác BDC có: \(\widehat{BDC}+\widehat{DCB}+\widehat{CBD}=180^o\)

Thay  \(\widehat{BDC}+\widehat{DCB}+90^o=180^o\)

\(\Rightarrow\widehat{BDC}+\widehat{DCB}=90^o\)

Mà \(\widehat{BDC}=\widehat{DCB}\)(Tam giác BDC vuông cân tại B)

Nên  \(\widehat{BDC}=\widehat{DCB}=\frac{90^o}{2}=45^o\)

Ta có:  \(\widehat{BDC}+\widehat{BDE}=180^o\)(2 góc kề bù)

\(\Rightarrow\widehat{BDE}=180^o-\widehat{BDC}=180^o-45^o=135^o\)

Vì DE = DB (gt) nên tam giác BDE cân tại D

\(\Rightarrow\widehat{BED}=\widehat{EBD}\)

Trong tam giác BDE có: \(\widehat{BDE}+\widehat{BED}+\widehat{EBD}=180^o\)

Thay \(135^o+\widehat{BED}+\widehat{EBD}=180^o\)

\(\Rightarrow\widehat{BED}+\widehat{EBD}=45^o\)

\(\Rightarrow\widehat{DBE}+\widehat{DBE}=45^o\)\(\left(\widehat{BED}=\widehat{BDE}\right)\)

\(\Rightarrow2\widehat{DBE}=45^o\)

\(\Rightarrow\widehat{DBE}=22,5^o\)

tu ve hinh :

a, tamgiac ADE can tai A (gt)

=> AD = AE va goc ADE = goc AED (dn)

xet  tamgiac ADB va tamgiac AEC co : DB = CE (gt)

=>  tamgiac ADB = tamgiac AEC (c - g - c)

=> AB = AC (dn)

=> tamgiac ABC can tai A (dn)

b, xet tamgiac DMB va tamgiac ENC co :

goc DMB = goc ENC = 90^o do MB | AD va CN | AE (gt) 

goc ADE = goc AED (cau a)

DB = CE (gt)

=>  tamgiac DMB =  tamgiac ENC (ch - gn)

=> BM = CN (dn)

khi kết quả sai

quá vô lí

Ví dụ:

Khi giữ nguyện:

\(\frac{x_1+x_2+x_3+....+x_n}{n}=Z\)

Khi tăng thêm:

\(\frac{x_1+20+x_2+20+x_3+20+.......+x_n+20}{n}\)\(=\frac{\left(x_1+x_2+x_3+.....+x_n\right)+20n}{n}=Z+20\)

Chúc bạn học tốt!

ủa sao ko thấy liên quan