Ta có: 

\(2x+y=11z\) và \(3x-y=4z\)

Chia theo vế ta có:

\(\dfrac{2x+y}{3x-y}=\dfrac{11z}{4z}=\dfrac{11}{4}\)

\(\Leftrightarrow4\left(2x+y\right)=11\left(3x-y\right)\)

\(\Leftrightarrow8x+4y=33x-11y\)

\(\Leftrightarrow15y=25x\)

\(\Leftrightarrow3y=5x\)

\(\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{5}=k\)

\(\Rightarrow x=3k,y=5k\)

 Thay vào Q ta có:

\(Q=\dfrac{2\cdot\left(3k\right)^2-3\cdot3k\cdot5k}{\left(3k\right)^2+3\cdot\left(5y\right)^2}\)

\(Q=\dfrac{18k^2-45k^2}{9k^2+75k^2}\)

\(Q=\dfrac{k^2\left(18-45\right)}{k^2\left(9+75\right)}\)

\(Q=\dfrac{-27}{84}=-\dfrac{9}{28}\)

\(\dfrac{2x+y}{3x-y}=\dfrac{11}{4}\)

=>33x-11y=8x+4y

=>25x=15y

=>5x=3y

=>x/3=y/5=k

=>x=3k; y=5k

\(Q=\dfrac{2\cdot9k^2-3\cdot3k\cdot5k}{9k^2+3\cdot25k^2}=\dfrac{18-9\cdot5}{9+3\cdot25}=\dfrac{-9}{28}\)

\(\dfrac{1}{3+0,5}+\dfrac{1}{3-0,5}\)

\(=\dfrac{3-0,5}{\left(3+0,5\right)\left(3-0,5\right)}+\dfrac{3+0,5}{\left(3+0,5\right)\left(3-0,5\right)}\)

\(=\dfrac{3-0,5+3+0,5}{3^2-\left(0,5\right)^2}\)

\(=\dfrac{6}{9-0,25}\)

\(=\dfrac{24}{35}\)

Cảm ơn!!!

Ta có: 

\(P=\dfrac{5x-4y}{5x+4y}\)

\(\Leftrightarrow P^2=\left(\dfrac{5x-4y}{5x+4y}\right)^2\)

\(\Leftrightarrow P^2=\dfrac{\left(5x-4y\right)^2}{\left(5x+4y\right)^2}\)

\(\Leftrightarrow P^2=\dfrac{\left(5x\right)^2-2\cdot5x\cdot4y+\left(4y\right)^2}{\left(5x\right)^2+2\cdot5x\cdot4y+\left(4y\right)^2}\)

\(\Leftrightarrow P^2=\dfrac{\left(25x^2+16y^2\right)-40xy}{\left(25x^2+16y^2\right)+40xy}\)

Thay \(25x^2+16y^2=50xy\) vào ta có:

\(P^2=\dfrac{50xy-40xy}{50xy+40xy}=\dfrac{10xy}{90xy}=\dfrac{1}{9}=\left(\dfrac{1}{3}\right)^2\)

Mà: \(4y< 5x< 0\)

Nên: \(P=\dfrac{5x-4y}{5x+4y}< 0\)

Vậy: \(P=-\dfrac{1}{3}\)

25x^2+16y^2=50xy

=>25x^2-50xy+16y^2=0

=>25x^2-10xy-40xy+16y^2=0

=>5x(5x-2y)-8y(5x-2y)=0

=>(5x-2y)(5x-8y)=0

=>5x=2y hoặc 5x=8y

5x>4y

=>5x=8y

=>x/8=y/5=k

=>x=8k; y=5k

\(P=\dfrac{5\cdot8k-4\cdot5k}{5\cdot8k+4\cdot5k}=\dfrac{40-20}{40+20}=\dfrac{1}{3}\)

a:

Sửa đề: A=x^4-9x^3+21x^2+x+a

A chia hết cho B

=>x^4-2x^3-7x^3+14x^2+7x^2-14x+15x-30+a+30 chia hết cho x-2

=>a+30=0

=>a=-30

b: A chia hết cho B

=>x^4+2x^3-12x^3-24x^2+45x^2+90x-82x-164+a+164 chia hết cho x+2

=>a+164=0

=>a=-164

chi tiết đc kh ạ?

a)Ta có: BE, CF là pgiac(gt)

=> ∠CBE=∠FEB\(=\dfrac{1}{2}\widehat{ABC}\)

     \(\widehat{BCF}=\widehat{ECF}=\dfrac{1}{2}\widehat{ABC}\)

Mà ∠ABC=∠ACB(tam giác ABC cân tại A); ∠BCF=∠CBE(cmt)

Ta có: xét tam giác BFC và tam giác CEB có:

+∠FBC=∠ECB (tam cân)

+BC chung

+∠BCF=∠CBE(cmt)

=> tam giác BFC=tam giác CEB (g.c.g)

=>BF=CE(2 cạnh tương ứng)

Mà AB=AC(gt)

=>AB-BC=AC-CE

=>AF=AE

=>tam giác AFE cân tại A

=> \(\widehat{AFE}=\dfrac{1}{2}\left(180^o-\widehat{A}\right)\)

Mà ∠ABC=1/2(180-A)

=>∠AFE=∠ABC

Mà 2 góc ở vị trí đồng vị

=>EF//BC

=>BFEC là hình thang 

Mà ∠CBF=BCE(tam giác cân)

=>BFEC là hình thang cân)

b) Do BFEC là hình thang cân

=>FE//BC; BF=CE(1)

=>góc FEB= góc EBC

Mà BE là pgiac góc B

=>góc FBE=FEB

=> tam giác FBE cân

=>BF=FE (2)

Từ(1);(2)=>BF=FE=EC

Xét ΔAQP có MN//PQ

nên AM/MQ=AN/NP

mà MQ=NP

nên AM=AN

Xét ΔMNQ và ΔNMP có

MN chung

NQ=MP

MQ=NP

=>ΔMNQ=ΔNMP

=>góc BMN=góc BNM

=>BM=BN

AM=AN

BM=BN

=>AB là trung trực của MN

\(\dfrac{1}{9}-\left(2x-y\right)^2\)

\(=\left(\dfrac{1}{3}\right)^2-\left(2x-y\right)^2\)

\(=\left[\dfrac{1}{3}-\left(2x-y\right)\right]\left[\dfrac{1}{3}+\left(2x-y\right)\right]\)

\(=\left(\dfrac{1}{3}-2x+y\right)\left(\dfrac{1}{3}+2x-y\right)\)

cảm ơn bạn nha

\(\dfrac{5}{9}\cdot\dfrac{7}{13}+\dfrac{5}{9}\cdot13-\dfrac{5}{9}\cdot\dfrac{3}{13}\)

\(=\dfrac{5}{9}\cdot\left(\dfrac{7}{13}-\dfrac{3}{13}+13\right)\)

\(=\dfrac{5}{9}\cdot\left(\dfrac{4}{13}+13\right)\)

\(=\dfrac{5}{9}\cdot\dfrac{173}{13}\)

\(=\dfrac{865}{117}\)

=5/9(7/13+13-3/13)

=5/9*(13+4/13)

=5/9*173/13=865/117

\(\dfrac{4}{19}\cdot-\dfrac{3}{7}+-\dfrac{3}{7}\cdot\dfrac{15}{19}+\dfrac{5}{7}\)

\(=-\dfrac{3}{7}\cdot\left(\dfrac{4}{19}+\dfrac{15}{19}\right)+\dfrac{5}{7}\)

\(=-\dfrac{3}{7}\cdot\dfrac{19}{19}+\dfrac{5}{7}\)

\(=-\dfrac{3}{7}+\dfrac{5}{7}\)

\(=\dfrac{-3+5}{7}\)

\(=\dfrac{2}{7}\)

=-3/7(4/19+5/19)+5/7

=5/7-3/7=2/7