x² + 1 - y² - 2x 

= x² - 2x + 1 - y²

=[x² - 2x + 1] - y²

=[x-1]²  - y²

=[x-1-y][x-1+y]

a) \(x^2+1-y^2-2x=\left(x^2-2x+1\right)-y^2=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\)

b) \(64x^4+y^4=\left(8x^2\right)^2+\left(y^2\right)^2=\left(8x^2\right)^2+16x^2y^2+\left(y^2\right)^2-16x^2y^2\)

\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2=\left(8x^2+y^2-4xy\right)\left(8x^2+y^2+4xy\right)\)

https://h.vn/hoi-dap/tim-kiem?q=cho+tam+gi%C3%A1c+ABC+vu%C3%B4ng+c%C3%A2n+t%E1%BA%A1i+A+.+g%E1%BB%8Di+M+v%C3%A0+D+l%E1%BA%A7n+l%C6%B0%E1%BB%A3t+l%C3%A0+trung+%C4%91i%E1%BB%83m+c%E1%BB%A7a+BC+v%C3%A0+AC+,+E+%C4%91%E1%BB%91i+x%E1%BB%A9ng+v%E1%BB%9Bi+M+qua+D+a)+t%E1%BB%A9+gi%C3%A1c+AEMB+v%C3%A0+AECM+l%C3%A0+h%C3%ACnh+g%C3%AC+b)+tam+gi%C3%A1c+ABC+c%C3%B3+%C4%91i%E1%BB%81u+ki%E1%BB%87n+g%C3%AC+th%C3%AC+t%E1%BB%A9+gi%C3%A1c+AECM+l%C3%A0+h%C3%ACnh+vu%C3%B4ng&id=500076

DƯƠNG PHAN KHÁNH DƯƠNG28 tháng 11 2017 lúc 18:39

ABCMDE

Từ giả thiết ta có :

MD là đường trung bình của ΔABCΔABC

⇒⎧⎩⎨MD//ABMD=12AB⇒{MD//ABMD=12AB (1)

Do E đối xứng với M qua D (2)

Từ 1 và 2

⇒{ME//ABME=AB⇒{ME//ABME=AB

Suy ra AEMB là hình bình hành

Tính GTLN , GTNN: a, A=2x²-6x. b,B=x²+y²-x+6y+10. c,C=x-x² .... 1, tìm x : a) (x+2).(x+3)-(x-2).(x+5)=0. b) (2x+3).(x-4)+(x-5).(x-2)=(x-4).(3x-5). c) (3x-5). ... Viết các biểu thức dưới dạng bình phương của một tổng hoặc hiệu:.

A = 2x² - 6x - 1

A = 2 . ( x² - 3x - 1 / 2 )

A = 2 . [ ( x² - 2 . x . 3 / 2 + ( 3 / 2 )² - ( 3 / 2 )² - 1 / 2 ) ]

A = 2 . [ ( x - 3 / 2 )² - 11 / 4 ]

A = ( x - 3 / 2 )² - 11 / 2 \(\ge\)11 / 2

Dấu " = " xảy ra \(\Leftrightarrow\)x - 3 / 2 = 0

                             \(\Rightarrow\)x             = 3 / 2

Min A = 11 / 2 \(\Leftrightarrow\)x = 3 / 2

\(\frac{1-2x}{2x}+\frac{2x}{2x-1}+\frac{1}{2x-4x^2}\)

\(=\frac{1}{2x}-1+1+\frac{1}{2x-1}+\frac{1}{2x\left(1-2x\right)}=\frac{1-2x}{2x\left(1-2x\right)}-\frac{2x}{2x\left(1-2x\right)}+\frac{1}{2x\left(1-2x\right)}\)

\(=\frac{1-2x-2x+1}{2x\left(1-2x\right)}=\frac{2}{2x\left(1-2x\right)}=\frac{1}{x\left(1-2x\right)}\)

Ta có: \(\frac{1-2x}{2x}+\frac{2x}{2x-1}+\frac{1}{2x-4x^2}\)

= \(\frac{1-2x}{2x}+\frac{2x}{2x-1}-\frac{1}{2x\left(2x-1\right)}\)

= \(\frac{\left(1-2x\right)\left(2x-1\right)}{2x\left(2x-1\right)}+\frac{2x.2x}{2x\left(2x-1\right)}-\frac{1}{2x\left(2x-1\right)}\)

= \(\frac{-\left(4x^2-4x+1\right)}{2x\left(2x-1\right)}+\frac{4x^2}{2x\left(2x-1\right)}-\frac{1}{2x\left(2x-1\right)}\)

= \(\frac{-4x^2+4x-1+4x^2-1}{2x\left(2x-1\right)}\)

= \(\frac{4x-2}{2x\left(2x-1\right)}\)

= \(\frac{2\left(2x-1\right)}{2x\left(2x-1\right)}=\frac{1}{x}\)

1+1=2

2+2=4

1+1=2

2+2=4

(3x + 1)² - 4(x - 1)² = 0

<=> 5x² + 14x - 3 = 0

<=> (5x - 1)(x + 3) = 0

<=> 5x - 1 = 0 hoặc x + 3 = 0

<=> x = 1/5     hoặc x = -3

=> x = 1/5 hoặc x = -3

\(GT\Leftrightarrow\frac{1}{1+a}-1+\frac{1}{1+b}-1+\frac{1}{1+c}-1+\frac{1}{1+d}-1\)\(\ge3-4\)

\(\Rightarrow\frac{-a}{1+a}+\frac{-b}{1+b}+\frac{-c}{1+c}+\frac{-d}{1+d}\ge-1\)

\(\Rightarrow\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}+\frac{d}{1+d}\le1\)

\(\Rightarrow\frac{a\left(1+b\right)+b\left(1+a\right)}{\left(1+a\right)\left(1+b\right)}+\frac{c\left(1+d\right)+d\left(1+c\right)}{\left(1+c\right)\left(1+d\right)}\le1\)

\(\Rightarrow\frac{a+2ab+b}{1+a+b+ab}+\frac{c+2cd+d}{1+c+d+cd}\le1\)

Áp dụng BĐT Cô - si , ta có:

\(1\ge\frac{2\sqrt{ab}+2ab}{1+2\sqrt{ab}+ab}+\frac{2\sqrt{cd}+2cd}{1+2\sqrt{cd}+cd}=\frac{2\sqrt{ab}}{1+\sqrt{ab}}+\frac{2\sqrt{cd}}{1+\sqrt{cd}}\)

\(\Rightarrow1\ge2\left[2\sqrt{\frac{\sqrt{abcd}}{1+\sqrt{ab}+\sqrt{cd}+\sqrt{abcd}}}\right]\)\(=4.\frac{\sqrt[4]{abcd}}{1+\sqrt{ab}+\sqrt{cd}+\sqrt{abcd}}\)

\(\Rightarrow1\ge\frac{4\sqrt[4]{abcd}}{1+2\sqrt[4]{abcd}+\sqrt{abcd}}=\frac{4\sqrt[4]{abcd}}{\sqrt{\left(1+\sqrt[4]{abcd}\right)^2}}\)

\(\Rightarrow4\sqrt[4]{abcd}\le\sqrt{\left(1+\sqrt[4]{abcd}\right)^2}\)

\(\Rightarrow4\sqrt[4]{abcd}\le1+\sqrt[4]{abcd}\)(vì a,b,c,d dương)

\(\Rightarrow3\sqrt[4]{abcd}\le1\)

\(\Rightarrow\sqrt[4]{abcd}\le\frac{1}{3}\)

\(\Rightarrow abcd\le\frac{1}{81}\)

(Dấu "="\(\Leftrightarrow a=b=c=d=\frac{1}{3}\))

Coll boy ! Bài này dòng 5 em áp dụng bất đẳng thức cô-si như vậy là chưa đúng nhé! Em kiểm tra lại mẫu trái dấu em nhé!