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(5x - 4)(2 + x) = 5(x - 3)2
10x + 5x2 - 8 - 4x = 5(x2 - 6x + 9)
6x + 5x2 - 8 = 5x2 - 30x + 45
36x = 53
x = 53/36
\(\left(3x+2\right)\left(x-1\right)-3\left(x+1\right)\left(x-2\right)=4\)
\(\Rightarrow3x^2-3x+2x-2-\left(3x+3\right)\left(x-2\right)=4\)
\(\Rightarrow3x^2-3x+2x-2-\left(3x^2-6x+3x-6\right)=4\)
\(\Rightarrow3x^2-3x+2x-2-3x^2+6x-3x+6=4\)
\(\Rightarrow2x+4=4\)
\(\Rightarrow x=0\)
a) x2 - 9 = 3( x - 3 )
⇔ ( x - 3 )( x + 3 ) - 3( x - 3 ) = 0
⇔ ( x - 3 )( x + 3 - 3 ) = 0
⇔ ( x - 3 ).x = 0
⇔ x - 3 = 0 hoặc x = 0
⇔ x = 3 hoặc x = 0
b) 3( 3x2 + 1 ) = 6 - 2( 3x + 2 )
⇔ 9x2 + 3 = 6 - 6x - 4
⇔ 9x2 + 6x + 3 - 6 + 4 = 0
⇔ 9x2 + 6x + 1 = 0
⇔ ( 3x + 1 )2 = 0
⇔ 3x + 1 = 0
⇔ x = -1/3
a) \(A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2x+10}{\left(x+5\right)\left(x-5\right)}\)
\(A=\dfrac{x-5+2x+10-2x-10}{\left(x+5\right)\left(x-5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}=\dfrac{1}{x+5}\)
b) \(A=-3\Rightarrow\dfrac{1}{x+5}=-3\)
\(\Leftrightarrow x+5=-\dfrac{1}{3}\Leftrightarrow x=-\dfrac{1}{3}-5=\dfrac{-16}{3}\)
\(9x^2-42x+49=\left(3x-7\right)^2=\left(3.\dfrac{-16}{3}-7\right)^2=\left(-23\right)^2=529\) \(\left(x=\dfrac{-16}{3}\right)\)
\(A=\frac{x^3-3x^2-7x-15}{x^5-x^4-10x^3-38x^2-51x-45}\)
\(=\frac{x^2\left(x-5\right)+2x\left(x-5\right)+3\left(x-5\right)}{x^4\left(x-5\right)+4x^3\left(x-5\right)+10x^2\left(x-5\right)+12x\left(x-5\right)+9\left(x-5\right)}\)
\(=\frac{\left(x-5\right)\left(x^2+2x+3\right)}{\left(x-5\right)\left(x^4+4x^3+10x^2+12x+9\right)}\)
\(=\frac{x^2+2x+3}{x^4+4x^3+10x^2+12x+9}\)
\(=\frac{x^2+2x+3}{\left(x^2\right)^2+2.x^2.2x+\left(2x\right)^2+6x^2+12x+9}\)
\(=\frac{x^2+2x+3}{\left(x^2+2x\right)^2+2.\left(x^2+2x\right).3+3^2}\)
\(=\frac{\left(x^2+2x+3\right)}{\left(x^2+2x+3\right)^2}=\frac{1}{x^2+2x+3}\)
b, \(A=\frac{1}{x^2+2x+3}=\frac{1}{\left(x+1\right)^2+2}\le\frac{1}{2}\forall x\)
Dấu "=" xảy ra khi: \(x+1=0\Rightarrow x=-1\)
Vậy GTLN của A là \(\frac{1}{2}\) khi x = -1
\(\Leftrightarrow x^2-x-2-x^2=2\)
=>-x=4
hay x=-4
Ah ơi s mà ra dược -x v ạ. Lấy gì trừ gì để ra í. E hong hỉu