nhận thấy: \(\left(a^2+1^2\right)\left(1^2+b^2\right)\ge\left(a\cdot1+b\cdot1\right)^2=\left(a+b\right)^2\)

CMTT: \(\Rightarrow\left(b^2+1\right)\left(1+c^2\right)\ge\left(b\cdot1+c\cdot1\right)^2=\left(b+c\right)^2\)

\(\left(c^2+1\right)\left(1^2+a^2\right)\ge\left(c\cdot1+a\cdot1\right)^2=\left(c+a\right)^2\)

=> \(\left\lbrack\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\right\rbrack^2\ge\left\lbrack\left(a+b\right)\left(b+c\right)\left(c+a\right)\right\rbrack^2\)

vì hai vế đều dương nên căn bậc cả hai vế ta dc

\(\Rightarrow\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

dấu "=" xảy ra khi: \(\frac{a}{1}=\frac{1}{b}\Rightarrow ab=1\)

\(\frac{b}{1}=\frac{1}{c}\Rightarrow bc=1\)

\(\frac{c}{1}=\frac{1}{a}\Rightarrow ac=1\)

=> a=b=c=1

a) \(\Rightarrow\left(\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\right)^2\le\left(1^2+1^2+1^2\right)\left\lbrack\left(\sqrt{a+1}\right)^2+\left(\sqrt{b+1}\right)^2+\left(\sqrt{c+1}\right)^2\right\rbrack\)

\(VT^2\le3\left(a+b+c+3\right)=12\)

=> \(VT\le2\sqrt3<3,5\)

b) => \(VT^2\le3\left(2a+2b+2c\right)=6\)

=> \(VT\le\sqrt6\)

dấu "=" xảy ra khi và chỉ khi:

\(\frac{\sqrt{a+b}}{1}=\frac{\sqrt{b+c}}{1}=\frac{\sqrt{c+a}}{1}\)

=> a=b=c=\(\frac13\)

cậu ơi:), có bài gì cậu tung hết một thể cho tớ làm chứ giờ ăn cơm cứ đi hành tớ v

nhận thấy: \(\frac{a+2b+c}{1+a^2}=\frac{\left(a+2b+c\right)\left(1+a^2\right)-a^2\left(a+2b+c\right)}{1+a^2}=a+2b+c-\frac{a^2\left(a+2b+c\right)}{1+a^2}\)

mà \(\frac{a^2\left(a+2b+c\right)}{1+a^2}\le\frac{a^2\left(a+2b+c\right)}{2a}=\frac{a\left(a+2b+c\right)}{2}=\frac{a^2+2ab+ac}{2}\)

=> \(\frac{a+2b+c}{1+a^2}\ge a+2b+c-\frac{a^2+2ab+ac}{2}\)

CMTT:=> \(\frac{b+2c+d}{1+b^2}\ge b+2c+d-\frac{b^2+2bc+bd}{2}\)

\(\frac{c+2d+a}{1+c^2}\ge c+2d+a-\frac{c^2+2dc+ac}{2}\)

\(\frac{d+2a+b}{1+d^2}\ge d+2a+b-\frac{d^2+2ad+bd}{2}\)

=> \(VT\ge4\left(a+b+c+d\right)-\frac12\left(a^2+b^2+c^2+d^2+2ab+2bc+2dc+2ad+ac+bd+ac+bd\right)\) \(VT\ge4\cdot4-\frac12\left(a+b+c+d\right)^2=16-\frac12\cdot16=8\left(đpcm\right)\)

cảm giác bạn Nguyễn Tuấn Dũng spam mỗi bài này hay thôi sao ấy, hôm trước vừa giải cho xong:v

a) AEMF là hình chữ nhật, ABMF là hình thang

b) dễ CM: △ADE=△DCF

=> DE=CF và góc ADE= góc DCF

mà góc DCF+ góc DFC= 90 độ

=> góc ADE+ góc DFC= 90 độ

=> DE⊥FC

c) CMTT:=> △ABF=△BCE

=> BF=CE và BF⊥CE

d) kẻ CM cắt EF tại I, cắt AB tại G

CM dc: △BAM=△BCM

=> góc BCG= góc BAM

mà trong hình hcn AEMF=> góc BAM=góc AEF

xét tam giác BCG vuông tại B có:

góc BCG+ góc BGC= 90 độ

=> góc AEF+góc BGC= 90 độ hay góc IEG+ góc IGE= 90 độ

=> CG⊥EF

xét tam giác EFC có:

ED⊥FC

CG⊥EF

BF⊥EC

=> DE,BF,CM đồng quy

d) ta có: \(S_{AEMF}=AE\cdot AF\)

mà À=EB

=> \(S_{AEMF}=AE\cdot EB=AE\left(AB-AE\right)=AE\cdot AB-AE^2\)

\(=\frac{AB^2}{4}-\left(\frac{AB^2}{4}-AE\cdot AB+AE^2\right)\)

\(=\frac{AB^2}{4}-\left(\frac{AB}{2}-AE\right)^2\le\frac{AB^2}{4}\)

dấu "=" xảy ra khi \(\frac{AB}{2}-AE=0\Rightarrow\frac{AB}{2}=AE\)

=> E là trung điểm AB

xét tam giác BAD có:

E là trung điểm

EM//AD

=> M là trung điểm BD thì \(S_{AEMF}\) lớn nhất

ta có: \(\frac{a^2}{a+b^2}=a-\frac{ab^2}{a+b^2}\ge a-\frac{b\sqrt{a}}{2}\)

=> \(VT\ge9a+b+c)-\frac12\left(b\sqrt{a}+c\sqrt{b}+a\sqrt{c}\right)\)

ta có: \(b\sqrt{a}=\sqrt{b\cdot ba}\le\frac{b+ab}{2}\)

=> \(b\sqrt{a}+c\sqrt{b}+a\sqrt{c}\le\frac{a+b+c+ab+bc+ca}{2}\)

cần CM: \(\frac{a+b+c+ab+bc+ca}{2}\le a+b+c\Rightarrow ab+bc+ca\le a+b+c\)

ta có: \(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)=9\Rightarrow a+b+c\le3\)

=> \(\frac{1}{a+b+c}\ge\frac13\)

=> \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\le\frac{\left(a+b+c\right)^2}{a+b+c}=a+b+c\)

vậy bt dc CM

dấu "=" xảy ra khi a=b=c=1

ta có: \(\frac{a}{1+b^2c}=a-\frac{ab^2c}{1+b^2c}\)

mà \(1+b^2c\ge2\sqrt{1\cdot b^2\cdot c}=2b\sqrt{c}\)

\(\Rightarrow\frac{ab^2c}{1+b^2c}\le\frac{ab^2c}{2b\sqrt{c}}=\frac{ab\sqrt{c}}{2}\)

=> \(\frac{a}{1+b^2c}\ge a-\frac{ab\sqrt{c}}{2}\)

=> \(VT\ge\left(a+b+c+d\right)-\frac12\left(ab\sqrt{c}+bc\sqrt{d}+cd\sqrt{a}+da\sqrt{b}\right)\)

\(VT\ge4-\frac12\left(ab\sqrt{c}+bc\sqrt{d}+cd\sqrt{a}+da\sqrt{b}\right)\)

Để VT\(\ge2\) => \(ab\sqrt{c}+bc\sqrt{d}+cd\sqrt{a}+da\sqrt{b}\le4\)

mà \(\sqrt{c}\le\frac{c+1}{2}\)

=> \(ab\sqrt{c}+bc\sqrt{d}+cd\sqrt{a}+da\sqrt{b}\le\frac12\left(abc+bcd+cda+dab+ab+bc+cd+da\right)\)

ta có: \(ab+bc+cd+da=b\left(a+c\right)+d\left(c+a\right)=\left(a+c\right)\left(b+d\right)\)

ta có bđt: \(\left(a+c\right)\left(b+d\right)\le\left(\frac{a+c+b+d}{2}\right)^2=4\)

ta có: \(abc+bcd+cda+dab=ac\left(b+d\right)+bd\left(c+a\right)\)

mà \(ac\le\frac{\left(a+c\right)^2}{4},bd\le\frac{\left(b+d\right)^2}{4}\)

=> \(abc+bcd+cda+dab\le\frac{\left(a+c\right)^2}{4}\left(b+d\right)+\frac{\left(b+d\right)^2}{4}\left(c+a\right)\)

\(\Rightarrow abc+bcd+cda=dab\le\frac{\left(a+c\right)\left(b+d\right)}{4}\left(a+b+c+d\right)=\left(a+c\right)\left(b+d\right)\le4\)

=> \(ab\sqrt{c}+bc\sqrt{d}+cd\sqrt{a}+da\sqrt{b}\le\frac12\left(4+4\right)=4\)

=> \(VT\ge4-\frac12\cdot4=2\)

ta có: \(\frac{a^3}{a^2+b}=a-\frac{ab}{a^2+b}\ge a-\frac{\sqrt{b}}{2}\)

=> \(VT\ge\left(a+b+c\right)-\frac12\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)

ta có \(\sqrt{a}=\sqrt{a\cdot1}\le\frac{a+1}{2}\)

=> \(VT\ge\left(a+b+c\right)-\frac{a+b+c+3}{4}=\frac{3\left(a+b+c\right)-3}{4}\)

\(VT\ge\frac{3\cdot3-3}{4}=\frac64=\frac32\)

dấu "=" xảy ra khi a=b=c=1

ta có: \(\frac{a^4}{a^3+2b^3}=a-\frac{2ab^3}{a^3+2b^3}\)

mà \(a^3+2b^3=a^3+b^3+b^3\ge3^3\sqrt{a^3\cdot b^3\cdot b^3}=3ab^2\)

=> \(\frac{2ab^3}{a^3+2b^3}\le\frac{2ab^3}{3ab^2}=\frac{2b}{3}\)

\(\Rightarrow\frac{a^4}{a^3+2b^3}\ge a-\frac{2b}{3}\)

=> \(VT\ge\left(a+b+c+d\right)-\frac23\left(a+b+c+d\right)\)

\(VT\ge\frac{a+b+c+d}{3}\)

dấu "=" xảy ra khi a=b=c=d

A=99...9 x \(10^{n+2}\) +8 x \(10^{n+1}+1\)

ta có \(99\ldots9=10^{n}-1\)

=> \(A=\left(10^{n}-1\right)\cdot100\cdot10^{n}+80\cdot10^{n}+1\)

\(A=100\left(10^{n}\right)^2-100\cdot10^{n}+80\cdot10^{n}+1\)

\(A=100\left(10^{n}\right)^2-20\cdot10^{n}+1=\left(10\cdot10^{n}-1\right)^2\)

\(A=\left(10^{n+1}-1\right)^2=\left(99\ldots9\right)^2\)

KL:...

B= \(99\ldots9\cdot10^{n+2}+25=\left(10^{n}-1\right)\cdot100\cdot10^{n}+25\)

\(B=100\left(10^{n}\right)^2-100\cdot10^{n}+25\)

\(B=\left(10\cdot10^{n}-5\right)^2=\left(10^{n+1}-5\right)=\left(99\ldots95\right)^2\)

C= \(11\ldots1\cdot10^{n+2}+22\ldots2\cdot10+5\)

đặt a=11...1=> \(a=\frac{10^{n}-1}{9}\Rightarrow10^{n}=9a+1\)

=> 22...2( với n+1 số)=\(2\cdot\frac{10^{n+1}-1}{9}=\frac{20\cdot10^{n}-2}{9}=\frac{20\left(9a+1\right)-2}{9}=20a+2\)

=> \(C=a\cdot100\cdot\left(9a+1\right)+\left(20a+2\right)\cdot10+5\)

\(C=900a^2+300a+25=\left(30a+5\right)^2\)

D= \(a\cdot10^{n}+a-2a=a\left(10^{n}-1\right)=a\cdot9a=\left(3a\right)^2\)

câu 3: dễ vẽ hình

gọi O là tâm hcn AFHE và K là giao của đường cao từ A cắt EF

xét tam giác KAE vuông tại K

=> góc MAB+ góc AEO=90 độ

dễ CM: góc OAE= góc OEA

xét tam giác ABH vuông tại H

=> góc EAO+ góc B=90 độ

=>góc MAB= góc B

=> △MAB cân tại M

=> MB=MA

CMTT: ta có góc KAF+ góc AFE = 90 độ

mà góc AFE= góc HAC

ta có góc HAC+ góc ACH= 90 độ

=> góc KAF= góc HCA hay góc MAC= góc MCA

=> △MAC cân tại M

=> MA=MC=MB

=> M là trung điểm BC

bài 4:

a) \(\Rightarrow A+B+C=\left(4x^2-5xy+3y^2\right)+\left(3x^2+2xy+y^2\right)+\left(-x^2+3xy+2y\right)\)

\(A+B+C=6x^2+4y^2+2y\)

b) \(\Rightarrow B-C-A=\left(3x^2+2xy+y^2\right)-\left(-x^2+3xy+2y\right)-\left(4x^2-5xy+3y^2\right)\)

\(B-C-A=4xy-2y^2-2y\)

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

\(2A-3B-C=\left(8x^2-10xy+6y^2\right)-\left(9x^2+6xy+3y^2\right)+\left(x^2-3xy-2y\right)\)

\(2A-3B-C=3y^2-19xy-2y\)

bài 5:

a)gọi 4 số lẻ liên tiếp là: 2n+1,2n+3,2n+5,2n+7(n ∈ Z)

=> \(\left(2n+5\right)\left(2n+7\right)-\left(2n+1\right)\left(2n+3\right)\)

\(=16\left(n+2\right)\) ⋮16(đpcm)

b) ta có: \(3\left(5x+y\right)+\left(4x-3y\right)=19x\) ⋮19

=> \(4x-3y\) ⋮19(đpcm)

c) a chia 3 dư 1=> a=3k+1

b chia 3 dư 2=> b=3m+2

=> \(ab=\left(3k+1\right)\left(3m+2\right)=9\operatorname{km}+6k+3m+2=3\left(3\operatorname{km}+2k+m\right)+2\)

vì \(3\left(3\operatorname{km}+2k+m\right)\) ⋮3

=> ab chia 3 dư 2(đpcm)