cho tam giác ABC, có AC= 10cm. Lấy D trên cạnh BC sao cho BD=3cm. Lấy các điểm G,H trên cạnh AC sao cho AG=CH=4cm. Gọi E là giao điểm của BG và AD. Tính tỉ số của AE và AD
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\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n-1}-\frac{1}{2n+1}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{2n+1}\right)\)
\(=\frac{1}{2}\cdot\frac{2n+1-1}{2n+1}\)
\(=\frac{1}{2}\cdot\frac{2n}{2n+1}=\frac{n}{2n+1}\)
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) 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 =
Vì B’C’ // với BC => = (1)
Trong ∆ABH có BH’ // BH => = (2)
Từ 1 và 2 => =
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’ = AH
= = => B’C’ = BC
=> SAB’C’= AH’.B’C’ = .AH.BC
=>SAB’C’= (AH.BC)
mà SABC= AH.BC = 67,5 cm2
Vậy SAB’C’= .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)\(\left(x-2\right)\left(2x+5\right)=\left(2x-4\right)\left(x+1\right)\)
\(\Leftrightarrow\left(x-2\right)\left(3x+5\right)-2\left(x-2\right)\left(x+1\right)\)
\(\Leftrightarrow\left(x-2\right)\left(3x+5-2x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)
\(\frac{\left(x-2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)
\(\Leftrightarrow\frac{4\left(x-2\right)\left(x+10\right)}{12}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{3\left(x-2\right)\left(x+4\right)}{12}\)
\(\Leftrightarrow4\left(x-2\right)\left(x+10\right)-\left(x+4\right)\left(x+10\right)=3\left(x-2\right)\left(x+4\right)\)
\(\Leftrightarrow4\left(x^2-2x+10x-20\right)-\left(x^2+4x+10x+40\right)=3\left(x^2-2x+4x-8\right)\)
\(\Leftrightarrow4\left(x^2+8x-20\right)-\left(x^2+14x+40\right)=3\left(x^2+2x-8\right)\)
\(\Leftrightarrow4x^2+32x-80-x^2-14x-40=3x^2+6x-24\)
\(\Leftrightarrow3x^2+18x-120=3x^2+6x-24\)
\(\Leftrightarrow12x=96\)\(\Leftrightarrow x=8\)
Vậy tập nghiệm của phương trình là \(S=\left\{8\right\}\)
\(\frac{\left(x-2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)
\(\Leftrightarrow\frac{4\left(x-2\right)\left(x+10\right)}{12}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{3\left(x-2\right)\left(x+4\right)}{12}\)
\(\Leftrightarrow4\left(x^2+8x-20\right)-\left(x^2+14x+40\right)=3\left(x^2+2x-8\right)\)
\(\Leftrightarrow4x^2+32x-80-x^2-14x-40-3x^2-6x+24=0\)
\(\Leftrightarrow12x-96=0\)
\(\Leftrightarrow x=8\)