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a) \(\left(x+8\right)\left(x+6\right)=104+x^2\Leftrightarrow x^2+6x+8x+48=104+x^2\)
\(\Leftrightarrow x^2+6x+8x-x^2=104-48\Leftrightarrow14x=56\Leftrightarrow x=\dfrac{56}{14}=4\)
vậy \(x=4\)
b) \(\left(x+1\right)\left(x+2\right)-\left(x-3\right)\left(x+4\right)=6\)
\(\Leftrightarrow x^2+2x+x+2-\left(x^2+4x-3x-12\right)=6\)
\(\Leftrightarrow x^2+2x+x+2-x^2-4x+3x+12=6\)
\(\Leftrightarrow2x+14=6\Leftrightarrow2x=6-14=-8\Leftrightarrow x=\dfrac{-8}{2}=-4\)
vậy \(x=-4\)
c) \(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=5\)
\(\Leftrightarrow4x^2-20x-\left(4x^2-3x-4x+3\right)=5\)
\(\Leftrightarrow4x^2-20x-4x^2+3x+4x-3=5\)
\(\Leftrightarrow-13x-3=5\Leftrightarrow-13x=5+3=8\Leftrightarrow x=\dfrac{8}{-13}=\dfrac{-8}{13}\)
vậy \(x=\dfrac{-8}{13}\)
d) \(\left(3x-4\right)\left(x-2\right)=3x\left(x-9\right)-3\)
\(\Leftrightarrow3x^2-6x-4x+8=3x^2-27x-3\)
\(\Leftrightarrow3x^2-6x-4x-3x^2+27x=-3-8\)
\(\Leftrightarrow17x=-11\Leftrightarrow x=\dfrac{-11}{17}\) vậy \(x=\dfrac{-11}{17}\)
e) câu này đề bị thiếu rồi nha bn
f) \(5x\left(x-3\right)=\left(x-2\right)\left(5x-1\right)-5\)
\(\Leftrightarrow5x^2-15x=5x^2-x-10x+2-5\)
\(\Leftrightarrow5x^2-15x-5x^2+x+10x=2-5\)
\(\Leftrightarrow-4x=-3\Leftrightarrow x=\dfrac{-3}{-4}=\dfrac{3}{4}\) vậy \(x=\dfrac{3}{4}\)
a) \(\left(x+8\right)\left(x+6\right)=104+x^2\)
\(\Leftrightarrow x^2+14x+48=104+x^2\)
\(\Leftrightarrow14x=56\)
\(\Rightarrow x=4\)
b) \(\left(x+1\right)\left(x+2\right)-\left(x-3\right)\left(x+4\right)=6\)
\(\Leftrightarrow x^2+3x+2-x^2-7x+12=6\)
\(\Leftrightarrow-4x=-8\)
\(\Rightarrow x=2\)
c) \(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=5\)
\(\Leftrightarrow4x^2-20x-4x^2+3x+4x-3=5\)
\(\Leftrightarrow-13x=8\)
\(\Rightarrow x=\dfrac{-8}{13}\)
d) \(\left(3x-4\right)\left(x-2\right)=3x\left(x-9\right)-3\)
\(\Leftrightarrow3x^2-10x+8=3x^2-27x-3\)
\(\Leftrightarrow17x=-11\)
\(\Rightarrow x=\dfrac{-11}{17}\)
e) \(\left(x-5\right)\left(x-4\right)-\left(x+1\right)\left(x-2\right)=7\)
\(\Leftrightarrow x^2-9x+20-x^2+x+2=7\)
\(\Leftrightarrow-8x=-15\)
\(\Rightarrow x=\dfrac{15}{8}\)
f) \(5x\left(x-3\right)=\left(x-2\right)\left(5x-1\right)-5\)
\(\Leftrightarrow5x^2-15x=5x^2-11x+2-5\)
\(\Leftrightarrow-4x=-3\)
\(\Rightarrow x=\dfrac{3}{4}\)
Bài 4.
1) ( x + 3 )( x2 - 3x + 9 ) - x( x2 - 3 ) = 8( 5 - x )
<=> x3 + 27 - x3 + 3x = 40 - 8x
<=> 27 + 3x = 40 - 8x
<=> 3x + 8x = 40 - 27
<=> 11x = 13
<=> x = 13/11
2) ( 2x + 1 )3 + ( 2x + 3 )3 = 0
<=> [ ( 2x + 1 ) + ( 2x + 3 ) ][ ( 2x + 1 )2 - ( 2x + 1 )( 2x + 3 ) + ( 2x + 3 )2 ] = 0
<=> ( 2x + 1 + 2x + 3 )[ 4x2 + 4x + 1 - ( 4x2 + 8x + 3 ) + 4x2 + 12x + 9 ] = 0
<=> ( 4x + 4 )( 8x2 + 16x + 10 - 4x2 - 8x - 3 ) = 0
<=> ( 4x + 4 )( 4x2 + 8x + 7 ) = 0
<=> \(\orbr{\begin{cases}4x+4=0\\4x^2+8x+7=0\end{cases}}\)
+) 4x + 4 = 0
<=> 4x = -4
<=> x = -1
+) 4x2 + 8x + 7 = 0 (*)
Ta có 4x2 + 8x + 7 = ( 4x2 + 8x + 4 ) + 3 = ( 2x + 2 )2 + 3 ≥ 3 > 0 ∀ x
=> (*) không xảy ra
Vậy x = -1
Bài 5.
1) A = x2 - 2x + 2 = ( x2 - 2x + 1 ) + 1 = ( x - 1 )2 + 1 ≥ 1 ∀ x
Đẳng thức xảy ra <=> x - 1 = 0 => x = 1
=> MinA = 1 <=> x = 1
2) A = 4x2 + 4x + 5 = ( 4x2 + 4x + 1 ) + 4 = ( 2x + 1 )2 + 4 ≥ 4 ∀ x
Đẳng thức xảy ra <=> 2x + 1 = 0 => x = -1/2
=> MinA = 4 <=> x = -1/2
3) A = 2x2 + 3x + 3 = 2( x2 + 3/2x + 9/16 ) + 15/8 = 2( x + 3/4 )2 + 15/8 ≥ 15/8 ∀ x
Đẳng thức xảy ra <=> x + 3/4 = 0 => x = -3/4
=> MinA = 15/8 <=> x = -3/4
4) A = 3x2 + 5x = 3( x2 + 5/3x + 25/36 ) - 25/12 = 3( x + 5/6 )2 - 25/12 ≥ -25/12 ∀ x
Đẳng thức xảy ra <=> x + 5/6 = 0 => x = -5/6
=> MinA = -25/12 <=> x = -5/6
5) B = 2x - x2 - 4 = -( x2 - 2x + 1 ) - 3 = -( x - 1 )2 - 3 ≤ -3 ∀ x
Đẳng thức xảy ra <=> x - 1 = 0 => x = 12
=> MaxB = -3 <=> x = 1
6) -x2 - 4x = -( x2 + 4x + 4 ) + 4 = -( x + 2 )2 + 4 ≤ 4 ∀ x
Đẳng thức xảy ra <=> x + 2 = 0 => x = -2
=> MaxB = 4 <=> x = -2
7) B = 3x - 2x2 - 2 = -2( x2 - 3/2x + 9/16 ) - 7/8 = -2( x - 3/4 )2 - 7/8 ≤ -7/8 ∀ x
Đẳng thức xảy ra <=> x - 3/4 = 0 => x = 3/4
=> MaxB = -7/8 <=> x = 3/4
8) B = x( 3 - x ) = -x2 + 3x = -( x2 - 3x + 9/4 ) + 9/4 = -( x - 3/2 )2 + 9/4 ≤ 9/4 ∀ x
Đẳng thức xảy ra <=> x - 3/2 = 0 => x = 3/2
=> MaxB = 9/4 <=> x = 3/2
9) A = ( x - 1 )( x + 1 )( x + 2 )( x + 4 )
= [ ( x - 1 )( x + 4 ) ][ ( x + 1 )( x + 2 ) ]
= ( x2 + 3x - 4 )( x2 + 3x + 2 ) (*)
Đặt t = x2 + 3x - 4
(*) <=> t( t + 6 )
= t2 + 6t
= ( t2 + 6t + 9 ) - 9
= ( t + 3 )2 - 9
= ( x2 + 3x - 4 + 3 )2 - 9
= ( x2 + 3x - 1 )2 - 9 ≥ -9 ∀ x
=> MinA = -9 ( chỗ này mình không xét giá trị của x vì nghiệm nó xấu lắm '-' )
rút gọn biểu thức
a) \(4x^2-\left(x+3\right).\left(x-5\right)+x\)
\(=4x^2-\left(x^2-5x+3x-15\right)+x\)
\(=4x^2-x^2+5x-3x+15+x\)
\(=3x^2+3x+15.\)
b) \(x.\left(x-5\right)-3x.\left(x+1\right)\)
\(=x^2-5x-\left(3x^2+3x\right)\)
\(=x^2-5x-3x^2-3x\)
\(=-2x^2-8x.\)
d) \(\left(x+3\right).\left(x-1\right)-\left(x-7\right).\left(x-6\right)\)
\(=x^2-x+3x-3-\left(x^2-6x-7x+42\right)\)
\(=x^2-x+3x-3-x^2+6x+7x-42\)
\(=15x-45.\)
Chúc bạn học tốt!
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+...+\dfrac{1}{\left(x+6\right)\left(x+7\right)}+\dfrac{1}{\left(x+7\right)\left(x+8\right)}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+7}-\dfrac{1}{x+8}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+8}=\dfrac{8}{x\left(x+8\right)}\)
a) `4x(x-5)-(x-1)(x-3)=23`
`<=> 4x^2-20x-x^2+4x-3=23`
`<=>3x^2-16x-26=0`
`<=> x=(8+-\sqrt142)/3
*Nếu đề là: `4x(x-5)-(x-1)(4x-3)=23`
`<=> 4x^2-20x-4x^2+7x-3=23`
`<=>-13x=26`
`<=>x=-2`
b) `(x-5)(x-4)-(x+1)(x-2)=7`
`<=>x^2-9x+20-x^2+x+2=7`
`<=>-8x=-15`
`<=>x=15/8`
\(\text{a) 4x(x-5)-(x-1)(1x-3)=23}\)
\(\Leftrightarrow4x^2-20x-\left(x^2-3x-x+3\right)=23\)
\(\Leftrightarrow4x^2-20x-x^2+3x+x-3=23\)
\(\Leftrightarrow3x^2-16x-26=0\)
Giải phương trình bậc 2 một ẩn ta được :
\(x_1=\dfrac{8-\sqrt{142}}{3};x_2=\dfrac{8+\sqrt{142}}{3}\)