Đề bài thiếu hở :??

cần gấp

\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}\)

\(=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n-1}-\frac{1}{2n+1}\right)\)

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

\(=\frac{1}{2}\cdot\frac{2n+1-1}{2n+1}\)

\(=\frac{1}{2}\cdot\frac{2n}{2n+1}=\frac{n}{2n+1}\)

Ta có : 2x² + 2x + 3

= 2( x² + x + 1/4 ) + 5/2

= 2( x + 1/2 )² + 5/2 ≥ 5/2 ∀ x

hay \(2x^2+2x+3\ge\frac{5}{2}\)

=> \(\frac{1}{2x^2+2x+3}\le\frac{2}{5}\)

=> \(\frac{3}{2x^2+2x+3}\le\frac{6}{5}\)

Đẳng thức xảy ra khi x = -1/2

Vậy MaxA = 6/5, đạt được khi x = -1/2

\(A=\frac{3}{2x^2+2x+3}=\frac{3}{2\left(x^2+2x.\frac{1}{2}+\frac{1}{4}\right)+\frac{5}{2}}\)

\(=\frac{3}{2\left(x+\frac{1}{2}\right)^2+\frac{5}{2}}\le\frac{3}{2.0+\frac{5}{2}}=\frac{6}{5}\)

Dấu '' = '' xảy ra khi \(x=\frac{-1}{2}\)

a) a³ - a²c + a²b - abc

= a( a² - ac + ab - bc )

= a[ a( a - c ) + b( a - c ) ]

= a( a - c )( a + b )

b) ( x² + 1 )² - 4x²

= ( x² + 1 )² - ( 2x )²

= ( x² - 2x + 1 )( x² + 2x + 1 )

= ( x - 1 )²( x + 1 )²

c) x² - 10x - 9y² + 25

= ( x² - 10x + 25 ) - 9y²

= ( x - 5 )² - ( 3y )²

= ( x - 3y - 5 )( x + 3y - 5 )

d) 4x² - 36x + 56

= 4( x² - 9x + 14 )

= 4( x² - 7x - 2x + 14 )

= 4[ x( x - 7 ) - 2( x - 7 ) ]

= 4( x - 7 )( x - 2 )

a,\(a^3-a^2c+a^2b-abc\)

\(=a\left(a^2-ac+ab-bc\right)\)

\(=a\left[a\left(a-c\right)+b\left(a-c\right)\right]\)

\(=a\left(a-b\right)\left(a-c\right)\)

b,\(\left(x^2+1\right)^2-4x^2\)

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

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

c,\(x^2-10x-9y^2+25\)

\(=\left(x^2-10x+25\right)-9y^2\)

\(=\left(x-5\right)^2-\left(3y\right)^2\)

\(=\left(x-5-3y\right)\left(x-5+3y\right)\)

d,\(4x^2-36x+56\)

\(=4\left(x^2-9x+14\right)\)

\(=4\left(x^2-7x-2x+14\right)\)

\(=4\left(x-7\right)\left(x-2\right)\)

a) Chứng minh $\frac{A{H}^{\prime }}{AH}$ = $\frac{B^{\prime }{C}^{\prime }}{BC}$ 

Vì B’C’ // với BC => $\frac{B^{\prime }{C}^{\prime }}{BC}$ = $\frac{A{B}^{\prime }}{AB}$            (1)

Trong ∆ABH có BH’ // BH => $\frac{A{H}^{\prime }}{AH}$ = $\frac{A{B}^{\prime }}{BC}$  (2)

Từ 1 và 2 => $\frac{B^{\prime }{C}^{\prime }}{BC}$ = $\frac{A{H}^{\prime }}{AH}$

b) B’C’ // BC mà AH ⊥ BC nên AH’ ⊥ B’C’ hay AH’ là đường cao của tam giác AB’C’.

Áp dụng kết quả câu a) ta có: AH’ = $\frac{1}{3}$ AH

$\frac{B^{\prime }{C}^{\prime }}{BC}$ = $\frac{A{H}^{\prime }}{AH}$ = $\frac{1}{3}$ => B’C’ = $\frac{1}{3}$ BC

=> SAB’C’= $\frac{1}{2}$ AH’.B’C’ = $\frac{1}{2}$.$\frac{1}{3}$AH.$\frac{1}{3}$BC

=>SAB’C’= ($\frac{1}{2}$AH.BC)$\frac{1}{9}$

mà SABC= $\frac{1}{2}$AH.BC = 67,5 cm²

Vậy SAB’C’= $\frac{1}{9}$.67,5= 7,5 cm²

\(\frac{2x-5}{6}-x+2=\frac{5x-3}{3}-\frac{6x-7}{4}+x\)

\(\Leftrightarrow\frac{1}{3}x-\frac{5}{6}-x+2=\frac{5}{3}x-1-\frac{3}{2}x+\frac{7}{4}+x\)

\(\Leftrightarrow\left(\frac{1}{3}-1-\frac{5}{3}+\frac{3}{2}-1\right)x=-1+\frac{7}{4}+\frac{5}{6}-2\)

\(\Leftrightarrow\frac{-11}{6}x=\frac{-5}{12}\)

\(\Leftrightarrow x=\frac{5}{22}\)

\(\frac{2x-5}{6}-x+2=\frac{5x-3}{3}-\frac{6x-7}{4}+x\)

\(\Leftrightarrow\frac{2\left(2x-5\right)}{12}-\frac{12x}{12}+\frac{24}{12}=\frac{4\left(5x-3\right)}{12}-\frac{3\left(6x-7\right)}{12}+\frac{12x}{12}\)

\(\Leftrightarrow4x-10-12x+24=20x-12-18x+21+12x\)

\(\Leftrightarrow4x-10-12x+24-20x+12+18x-21-12x=0\)

\(\Leftrightarrow-22x+5=0\)

\(\Leftrightarrow-22=-5\)

\(\Leftrightarrow x=\frac{5}{22}\)

1.\(\left(x+1\right)\left(x+4\right)=\left(2-x\right)\left(2+x\right)\)

\(\Leftrightarrow x^2+4x+x+4=4-x^2\)

\(\Leftrightarrow x^2+5x+4=4-x^2\)

\(\Leftrightarrow x^2+5x+4-4+x^2=0\)

\(\Leftrightarrow2x^2+6x=0\)

\(\Leftrightarrow2x\left(x+3\right)=0\)

\(\Rightarrow2x=0\)hoặc \(x+3=0\)

Giải 2 pt:

\(2x=0\Leftrightarrow x=0\)

\(x+3=0\Leftrightarrow x=-3\)

Vậy \(S=\left\{0;-3\right\}\)

1)\(\left(x+1\right)\left(x+4\right)=\left(2-x\right)\left(2+x\right)\)

\(\Leftrightarrow x^2+5x+4=4-x^2\)

\(\Leftrightarrow x^2+5x+4-4+x^2=0\)

\(\Leftrightarrow2x^2+5x=0\)

\(\Leftrightarrow x\left(2x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{5}{2}\end{cases}}}\)

b,\(x^3-x^2=1-x\)

\(\Leftrightarrow x^3-x^2+x-1=0\)

\(\Leftrightarrow x^2\left(x-1\right)+\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-1\\x=1\end{cases}\Leftrightarrow}x=1}\)

3)\(2x\left(x+1\right)=x^2-1\)

\(\Leftrightarrow2x\left(x+1\right)-\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x-x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2=0\)

\(\Leftrightarrow x=-1\)

4)\(\left(x-2\right)\left(2x+5\right)=\left(2x-4\right)\left(x+1\right)\)

\(\Leftrightarrow\left(x-2\right)\left(3x+5\right)-2\left(x-2\right)\left(x+1\right)\)

\(\Leftrightarrow\left(x-2\right)\left(3x+5-2x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)