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bai nay dung chia cho 13 va 7 moi ca dg hang dang thuc a^n-b^n

1. x^3-x^2-x=1/3tuong duong voi 3x^3-3x^2-3x=1  tuong duong voi 4x^3=(x+1)^3ta tim ra dc x ok nhe ban
2.                                

ko nham thi bai nay giai vay

to chua quen go

$\Leftrightarrow \overline{ab}^{3}=\overline{abcde}$$\Leftrightarrow \overline{ab}^{3}=\overline{abcde}$$\Leftrightarrow $Đặt\overline{ab}=x,\overline{cde}=y (10\leq x\leq 99),100\leq y\leq 999$$ta co x^{3}=1000x+y,y\geq 0\rightarrow x^{3}\geq 1000x\rightarrow x(x^{2})-1000)>0\rightarrow x^{2}> 1000\rightarrow x> 31$$.Tip theo ta co y< 1000\rightarrow x^{3}< 1000x+1000\rightarrow x^{3}-1000x< 1000\rightarrow x(x^{2}-1000)< 1000.Neu x\geq 33\rightarrow x(x^{2}-1000)\geq 1000\rightarrow x< 33\rightarrow x=32$$vay ta dc so can tim la 32768 va ab=32$

voi moi n chinh phuong ma ban

bai hinh cau b cm a p q thang hang la xong

$\bigtriangleup ABC nt\left ( 0 \right ).P nằm trong \bigtriangleup ABC.Lấy Q sao cho BQ song song voi AC,PQ song song voi AB.Lấy R thuộc BC sao cho R,Q khac phia A,P và \angle PRQ=\angle ACB.CMR(PQR)tx (O)$

ban thu noi xem nao

$\sqrt{\frac{b+c}{a}} +\sqrt{\frac{c+a}{b}}+\sqrt{\frac{a+b}{c}}\geq \sqrt{\frac{16(a+b+c)^{3}}{3(a+b)(b+c)(a+c)}}$

$x^{2}+y^{3}

=7$

$x^{2}+y^{3}=7$

$ta co :4^{a}-2008=4^a-1-2007.Ta co4^a-1luon chia het cho 3,2007 chia het cho 3\rightarrow 4^a-2008 luon chia het cho 3.Mat khac 4^a-2008 chia het cho 2\rightarrow 4^a-2008 chia het cho 6.ta co 4^a-2008=4^a+a+b-(a+1+b+2007).Tu day suy ra4^a+a+b chia het cho 6(dpcm)$

$\bigtriangleup ABC$ nt(O,r).các đường cao $AD,BE,CF$.$H$ la trực tâm.CM

a)$(AH+BH+CH)^2\leq a^2+b^2+c^2$ ($a,b,c$ la 3 canh tam giac)

b) $a^2+b^2+c^2\geq 8\sqrt{3}p{R}'$ ($p$ là nửa chu vi tam giac $DEF$,${R}')$ la bán kính $(DEF)$

$ap dụng bdt cauchy-schward ta coVT\geq \frac{a^2+b^2+c^2}^2{\sum a^3+\sum 2a^2b^2}.Cần Cm Bdt\geq 1\rightarrow phai cm \sum a^4\geq \sum a^3.vi a+b+c=3\rightarrow dpcm$

$cho a,b,c la 3 số thực ko am thoa man a^2+4b^2+9c^2=14.CMR:3b+8c+abc\leq 12$$cho a,b,c la 3 số thực ko am thoa man a^2+4b^2+9c^2=14.CMR:3b+8c+abc\leq 12$

$Vơi x1,x2,....,xn;a1,a2,.....an\geq 0.Ta co:x1a1+x2a2+....xnan\geq (x1+x2+....+xn)(a1^x1.a2^x2.....an^xn)^{\frac{1}{a1+a2+....an}}$

$Đặt P la biểu thức ở vế trái cua BDT cần cm Áp dụng BDT \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z},với mọi x,y,z>0 ta có: P=\frac{1}{x(1-x)}+\frac{1}{y(1-y)}+\frac{1}{z(1-z)}=(\frac{1}{x}+\frac{1}{1-x})+(\frac{1}{y}+\frac{1}{1-y})+(\frac{1}{z}+\frac{1}{1-z})=(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})+(\frac{1}{1-x}+\frac{1}{1-y}+\frac{1}{1-z})\geq \frac{9}{x+y+z}+\frac{9}{3-(x+y+z)} Chú y rang $0\leq x,y,z\leq \frac{1}{2}.Đăt S=x+y+z ta co P\geq \frac{9}{S}+\frac{9}{3-S}=\frac{27}{S(3-S)}= \frac{27}{\frac{9}{4}-(\frac{3}{2}-S)^2}.

roi danh gia trong doan 0 ,1/2 la ra

bai toan nay con len dc bao ba co

$Ta có: x^2+y^2=1\rightarrow 0\leq x,y\leq 1\rightarrow x\geq x^2,y\geq y^2\rightarrow x+y\geq x^2+y^2=1.Từ đó suy ra P^2=8+5(x+y)+2\sqrt{(4+5x)(4+5y)}\geq 13.Dau bang xay ra \Leftrightarrow x=1,y=o hoac x=0 y=1$

uk minh nham

sao a,b,c lai tim min x,y,z the kia

$Coi phương trình 1 la phuong trinh bac hai an x , ta co: \Delta (1)=4-4y^4\geq 0\rightarrow -1\leq y\leq 1 Phương trinh (2)\Leftrightarrow 2(x-1)^2+1+y^3=0 Do y\geq 1\rightarrow y^3+1\geq 0,lai co 2(x-1)^2\geq \rightarrow 2(x-1)^2=0 va y^3+1=0\rightarrow x=1,y=-1$