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B1:
a) A = \(\dfrac{1}{x+2}+\dfrac{x^2-x-2}{x^2-7x+10}-\dfrac{2x-4}{x-5}\)
= \(\dfrac{1}{x+2}+\dfrac{\left(x^2-2x\right)+\left(x-2\right)}{\left(x^2-2x\right)-\left(5x-10\right)}-\dfrac{2\left(x-2\right)}{x-5}\)
= \(\dfrac{1}{x+2}+\dfrac{\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{2\left(x-2\right)}{x-5}\) [ĐKXĐ: x ≠ -2; x ≠ 5]
= \(\dfrac{x-5}{\left(x+2\right)\left(x-5\right)}+\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-5\right)\left(x+2\right)}-\dfrac{2\left(x-2\right)\left(x+2\right)}{\left(x-5\right)\left(x+2\right)}\)
= \(\dfrac{-x^2+4x+5}{\left(x+2\right)\left(x-5\right)}\)
= \(\dfrac{-x\left(x-5\right)-\left(x-5\right)}{\left(x+2\right)\left(x-5\right)}\)
= \(\dfrac{\left(x-5\right)\left(-x-1\right)}{\left(x-5\right)\left(x+2\right)}\)
= \(-\dfrac{x+1}{x+2}\)
b) Thay x = 3 vào A, ta có:
A = \(-\dfrac{3+1}{3+2}=-\dfrac{4}{5}\)
c) A = 1
<=> \(-\dfrac{x+1}{x+2}\)= 1 <=> -(x + 1) = x + 2 <=> -x - 1 = x + 2
<=> -2x = 3 <=> x = \(\dfrac{-3}{2}\)
d) A = \(\dfrac{-\left(x+1\right)}{x+2}\)= \(\dfrac{-\left(x+2\right)+1}{x+2}\)= -1 + \(\dfrac{1}{x+2}\)
A đạt giá trị nguyên khi 1 chia hết cho x + 2 hay x + 2 ∈ Ư(1) = {1;-1}
* x + 2 = 1 <=> x = -1
* x + 2 = -1 <=> x = -3
B2: M = \(\dfrac{x^2+2x}{2x+10}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
= \(\dfrac{x\left(x+2\right)}{2\left(x+5\right)}+\dfrac{x-5}{x}+\dfrac{5\left(10-x\right)}{2x\left(x+5\right)}\)[ĐKXĐ: x ≠ 0; x ≠ -5
= \(\dfrac{x^2\left(x+2\right)+2\left(x+5\right)\left(x-5\right)+5\left(10-x\right)}{2x\left(x+5\right)}\)
= \(\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
= \(\dfrac{x^2+4x-5}{2\left(x+5\right)}\)
= \(\dfrac{\left(x^2+5x\right)-\left(x+5\right)}{2\left(x+5\right)}\)
\(\dfrac{\left(x+5\right)\left(x-1\right)}{2\left(x+5\right)}=\dfrac{x-1}{2}\)
b) Thay x = 3 vào M, ta có:
M = \(\dfrac{3-1}{2}=1\)
Thay x = 5 vào M, ta có:
M = \(\dfrac{5-1}{2}=2\)
a/ \(\left(x-4\right)^2-36=0\)
<=> \(\left(x-4-6\right)\left(x-4+6\right)=0\)
<=> \(\left(x-10\right)\left(x+2\right)=0\)
<=> \(\orbr{\begin{cases}x-10=0\\x+2=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=10\\x=-2\end{cases}}\)
b/ \(\left(x+8\right)^2=121\)
<=> \(\left(x+8\right)^2-121=0\)
<=> \(\left(x+8-11\right)\left(x+8+11\right)=0\)
<=> \(\left(x-3\right)\left(x+19\right)=0\)
<=> \(\orbr{\begin{cases}x-3=0\\x+19=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=3\\x=-19\end{cases}}\)
d/ \(4x^2-12x+9=0\)
<=> \(\left(2x\right)^2-2.2x.3+3^2=0\)
<=> \(\left(2x-3\right)^2=0\)
<=> \(2x-3=0\)
<=> \(x=\frac{3}{2}\)
a) Ta có : \(x^2-6x+10\)
\(=\left(x^2-6x+9\right)+1\)
\(=\left(x-3\right)^2+1\ge1>0\forall x\)
b) Ta có : \(4x-x^2-5\)
\(=-\left(x^2-4x+4\right)-1\)
\(=-\left(x-2\right)^2-1\le-1< 0\forall x\)
Vậy ...
1. -4x( x + 3 )( x - 4 ) - 3x( x2 - x + 1 )
= -4x( x2 - x - 12 ) - 3x( x2 - x + 1 )
= -4x3 + 4x2 + 48x - 3x3 + 3x2 - 3x
= -7x3 + 7x2 + 45x
2. a) 4x( x - 5 ) - ( x - 1 )( 4x - 3 ) = 5
<=> 4x2 - 20x - ( 4x2 - 7x + 3 ) = 5
<=> 4x2 - 20x - 4x2 + 7x - 3 = 5
<=> -13x - 3 = 5
<=> -13x = 8
<=> x = -8/13
b) 6( x - 3 )( x - 4 ) - 6x( x - 2 ) = 4
<=> 6( x2 - 7x + 12 ) - 6x2 + 12x = 4
<=> 6x2 - 42x + 72 - 6x2 + 12x = 4
<=> -30x + 72 = 4
<=> -30x = -68
<=> x = 34/15
Bài 1 :
\(-4x\left(x+3\right)\left(x-4\right)-3x\left(x^2-x+1\right)\)
\(=-7x^3+7x^2+45x\)
Bài 2 :
a, \(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=5\)
\(\Leftrightarrow4x^2-20x-\left[4x^2-7x+3\right]=5\)
\(\Leftrightarrow4x^2-20x-4x^2+7x-3=5\)
\(\Leftrightarrow-13x-8=0\Leftrightarrow x=-\frac{8}{13}\)
b, \(6\left(x-3\right)\left(x-4\right)-6x\left(x-2\right)=4\)
\(\Leftrightarrow6x^2-42x+72-6x^2+12x=4\)
\(\Leftrightarrow-30x+68=0\Leftrightarrow x=\frac{34}{15}\)
\( a)\dfrac{{3{x^4} - 2{x^3} - 2{x^2} + 4x - 8}}{{{x^2} - 2}}\\ = \dfrac{{3{x^4} - 2{x^3} - 6{x^2} + 4{x^2} + 4x - 8}}{{{x^2} - 2}}\\ = \dfrac{{3{x^2}\left( {{x^2} - 2} \right) - 2x\left( {{x^2} - 2} \right) + 4\left( {{x^2} - 2} \right)}}{{{x^2} - 2}}\\ = \dfrac{{\left( {{x^2} - 2} \right)\left( {3{x^2} - 2x + 4} \right)}}{{{x^2} - 2}}\\ = 3{x^2} - 2x + 4 \)
\( b)\dfrac{{2{x^3} - 26x - 24}}{{{x^2} + 4x + 3}}\\ = \dfrac{{2\left( {{x^3} - 13x - 12} \right)}}{{x + 3x + x + 3}}\\ = \dfrac{{2\left( {{x^3} + {x^2} - {x^2} - x - 12x - 12} \right)}}{{x\left( {x + 3} \right) + x + 3}}\\ = \dfrac{{2\left[ {{x^2}\left( {x + 1} \right) - x\left( {x + 1} \right) - 12\left( {x + 1} \right)} \right]}}{{\left( {x + 3} \right)\left( {x + 1} \right)}}\\ = \dfrac{{2\left( {x + 1} \right)\left( {{x^2} - x - 12} \right)}}{{\left( {x + 3} \right)\left( {x + 1} \right)}}\\ = \dfrac{{2\left( {{x^2} + 3x - 4x - 12} \right)}}{{x + 3}}\\ = \dfrac{{2\left[ {x\left( {x + 3} \right) - 4\left( {x + 3} \right)} \right]}}{{x + 3}}\\ = \dfrac{{2\left( {x + 3} \right)\left( {x - 4} \right)}}{{x + 3}}\\ = 2\left( {x - 4} \right)\\ = 2x - 8\)
\(4x^2-28x+49=\left(2x\right)^2-2\cdot2x\cdot7+7^2=\left(2x-7\right)^2\)
Khi x=4 thì \(4x^2-28x+49=\left(2x-7\right)^2=\left(2\cdot4-7\right)^2=1\)
Bài 1:
\(A=4x^2-28x+49\)
\(=4x^2-16x-12x+48+1\)
\(=4x\left(x-4\right)-12\left(x-4\right)+1\)
\(=\left(4x-12\right)\left(x-4\right)+1\)
Thay x = 4
\(\Leftrightarrow A=1\)
Vậy A= 1 tại x = 4
câu b
\(x^2-x=24\Leftrightarrow x^2-2\dfrac{1}{2}x+\dfrac{1}{4}=24+\dfrac{1}{4}=\dfrac{97}{4}\)
\(\left(x-\dfrac{1}{2}\right)^2=\left(\dfrac{\sqrt{97}}{2}\right)^2\)
\(\left[{}\begin{matrix}x_1=\dfrac{1+\sqrt{97}}{2}\\x_2=\dfrac{1-\sqrt{97}}{2}\end{matrix}\right.\)