Đề là chứng minh hay tìm GTNN zậy -_-

Ta có : 

\(\frac{a^2+b^2}{2}\ge ab\)

\(\Leftrightarrow\)\(a^2+b^2\ge2ab\)

\(\Leftrightarrow\)\(a^2-2ab+b^2\ge0\)

\(\Leftrightarrow\)\(\left(a-b\right)^2\ge0\) ( luôn đúng với mọi a, b ) 

Vậy ...

Chúc bạn học tốt ~ 

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ABCHKIEF

a) 

Xét \(\Delta\)ABC và \(\Delta\)HBA có: 

^BAC = ^BHA ( = 90 độ ) 

^ABC = ^HBA ( ^B chung ) 

=> \(\Delta\)ABC ~ \(\Delta\)HBA 

b) AB = 3cm ; AC = 4cm 

Theo định lí pitago ta tính được BC = 5 cm 

Từ (a) => \(\frac{AB}{BH}=\frac{BC}{AB}\Rightarrow BH=\frac{AB^2}{BC}=1,8\)m 

c) Xét \(\Delta\)AHC và \(\Delta\)AKH có: ^AKH = ^AHC = 90 độ 

và ^HAC = ^HAK ( ^A chung ) 

=> \(\Delta\)AHC ~ \(\Delta\)AKH 

=> \(\frac{AH}{AK}=\frac{AC}{AH}\Rightarrow AH^2=AC.AK\)

d) Bạn kiểm tra lại đề nhé!

3 cm 4cm 5 cm 1cm 1cm O H I K O'

XÉT 2 TAM GIÁC HIM VÀ HOM CÓ

+,HI=HO(=3CM)

+,∠MHI=∠MHO

+,CHUNG CẠNH HM

SUY RA:▲HIM=▲HOM(C.G.C)

⇒MI=MO(2 CẠNH TƯƠNG ỨNG)

⇒∠HIM=∠HOM(2 GÓC TƯƠNG ỨNG)

⇒∠MOK=∠MIO'(VÌ CÙNG BÙ VỚI ∠HIM VÀ ∠HOM)

XÉT 2 TAM GIÁC KOM VÀ O'IM CÓ

OI=OM

IO'=OK=1CM

∠MOK=∠MIO'

⇒▲KOM=▲O'MI(C.G.C)

⇒IM=IK(2 CẠNH TƯƠNG ỨNG)

TA CÓ:IM+IK=5CM

⇒IM✖2=IK✖2=5CM

⇒IM=IK=2,5

VẬY IM=IK=2.5 CM

4x²-12x-y²+2y+8

= (4x²-12x+9)-(y²-2y+1)

= (2x-3)²-(y-1)²

sai roi cho 4x phai la 2x ma

Bài này cần có công thức:

Ta có:\(x+\frac{1}{x}=3=>x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2=9-2=7\)

Lại có: \(x^5+\frac{1}{x^5}=\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)-\left(x+\frac{1}{x}\right)\)

=\(7\left(x+\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}-1\right)-3=7.3.6-3=123\)

Vậy \(x^5+\frac{1}{x^5}=123\)

con này không nhầm có lời giửi rồi!

\(\left(x+\frac{1}{x}\right)=3\Rightarrow x^2+\frac{1}{x^2}=7\Rightarrow x^4+\frac{1}{x^4}=47\)

\(3.7=\left(x+\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}\right)=\left(x^3+\frac{1}{x^3}\right)+\left(x+\frac{1}{x}\right)\)

\(\Rightarrow\left(x^3+\frac{1}{x^3}\right)=3.7-3=3.6\)

\(3.47=\left(x+\frac{1}{x}\right)\left(x^4+\frac{1}{x^4}\right)=\left(x^5+\frac{1}{x^5}\right)+x^3+\frac{1}{x^3}\\ \)

\(x^5+\frac{1}{x^5}=3.47-3.6=3\left(47-6\right)=3.41=123\)

\(\left(\dfrac{2x-1}{2x+1}-\dfrac{2x+1}{2x-1}\right):\dfrac{8x}{8x-4}\)

\(=(\dfrac{\left(2x-1\right)^2}{4x^2-1}-\dfrac{\left(2x+1\right)^2}{4x^2-1}):\dfrac{8x}{4\left(2x-1\right)}\)

\(=\left(\dfrac{4x^2-4x+1}{4x^2-1}-\dfrac{4x^2+4x+1}{4x^2-1}\right):\dfrac{8x}{4\left(2x-1\right)}\)

\(=\dfrac{4x^2-4x+1-4x^2-4x-1}{4x^2-1}:\dfrac{8x}{4\left(2x-1\right)}\)

\(=\dfrac{-8x}{4x^2-1}:\dfrac{8x}{4\left(2x-1\right)}\)

\(=\dfrac{-8x.4\left(2x-1\right)}{8x\left(2x-1\right)\left(2x+1\right)}\)

\(=\dfrac{-8x.4}{8x\left(2x+1\right)}\)

\(=\dfrac{-4}{2x+1}\)

a)(2x-3)(x+2)

=2x²+4x-3x-6

=2x²+x-6

b)(x-3)(2x+1)-(x-1)²

=(2x²+x-6x-3) - (x²-2x+1)

=(2x²-5x-3) - (x²-2x+1)

=2x²-5x-3-x²+2x-1

=x²-3x-4

Chúc bạn học tốt!

\(a)(2x-3)(x+2)\\=2x^2+4x-3x-6\\=2x^2+x-6\\ b)(x-3)(2x+1)-(x-1)^2\\=2x^2+x-6x-3-x^2+2x-1\\=x^2-3x-4\)

Theo câu a) ta có: \(AH^2=AI.AB\left(1\right)\)

Xét tam giác AHK và tam giác ACH có:

góc A chung; góc AKH = góc AHC = 90⁰

=> tam giác AHK đồng dạng với tam giác ACH (g-g)

=>\(\dfrac{AK}{AH}=\dfrac{AH}{AC}\Rightarrow AK.AC=AH^2\left(2\right)\)

Từ (1)(2) => \(AI.AB=AK.AC\Rightarrow\dfrac{AI}{AC}=\dfrac{AK}{AB}\)

Xét tam giác AIK và tam giác ABC có:

góc A chung; \(\dfrac{AI}{AC}=\dfrac{AK}{AB}\)

=> Tam giác AIK đồng dạng với tam giác ACB (c-g-c)

a) Xét tam giác AIH và tam giác AHB có:

góc BAH chung; góc AIH = góc AHB (= 90⁰)

=> tam giác AIH = tam giác AHB (g-g)

\(\Rightarrow\dfrac{AH}{AI}=\dfrac{AB}{AH}\Rightarrow AH^2=AI.AB\)

a: Xét ΔAHE có

AC là đường cao

AC là đường trung tuyến

Do đó: ΔAHE cân tại A

=>AC là phân giác của góc HAE

Xét ΔAHC và ΔAEC có

AH=AE
góc HAC=góc EAC
AC chung

Do đó: ΔAHC=ΔAEC

Suy ra: góc AEC=90 độ

=>AE vuông góc với CE

b:Ta có; AB<AC

nên góc C<góc B

=>90 độ-góc C>90 độ-góc B

=>góc CAH>góc BAH