Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
c \(\frac{\left(2x-4\right)\left(x-3\right)}{\left(x-2\right)\left(3x^2-27\right)}=\frac{2\left(x-2\right)\left(x-3\right)}{3\left(x-2\right)\left(x^2-9\right)}\)
\(=\frac{2\left(x-2\right)\left(x-3\right)}{3\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{2}{3\left(x+3\right)}\)
d, \(\frac{x^2+5x+6}{x^2+4x+4}=\frac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\frac{x+3}{x+2}\)
Tương tự với a ; b
a)Ta có : 9(a + b)2 - 4(a - 2b)2
= [3(a + b) - 2(a - 2b)].[3(a + b) + 2(a - 2b)]
= (3a + 3b - 2a + 4b)(3a + 3b + 2a - 4b)
= (a + 7b)(5a - b)
Bài 1.
\(a\Big) 9(4x+3)^2=16(3x-5)^2\\\Leftrightarrow 9[(4x)^2+2\cdot 4x\cdot3+3^2]=16[(3x)^2-2\cdot3x\cdot5+5^2]\\\Leftrightarrow9(16x^2+24x+9)=16(9x^2-30x+25)\\\Leftrightarrow 144x^2+216x+81=144x^2-480x+400\\\Leftrightarrow (144x^2-144x^2)+(216x+480x)=400-81\\\Leftrightarrow 696x=319\\\Leftrightarrow x=\dfrac{11}{24}\\Vậy:x=\dfrac{11}{24}\\---\)
\(b\Big)(x-3)^2=4x^2-20x+25\\\Leftrightarrow(x-3)^2=(2x)^2-2\cdot2x\cdot5+5^2\\\Leftrightarrow(x-3)^2=(2x-5)^2\\\Leftrightarrow (x-3)^2-(2x-5)^2=0\\\Leftrightarrow (x-3-2x+5)(x-3+2x-5)=0\\\Leftrightarrow (-x+2)(3x-8)=0\\\Leftrightarrow \left[\begin{array}{} -x+2=0\\ 3x-8=0 \end{array} \right.\\\Leftrightarrow \left[\begin{array}{} -x=-2\\ 3x=8 \end{array} \right.\\\Leftrightarrow \left[\begin{array}{} x=2\\ x=\dfrac{8}{3} \end{array} \right.\\Vậy:...\)
a. \(x^2+4x+4=x^2+2\cdot x\cdot2+2^2=\left(x+2\right)^2\)
b. \(4x^2-4x+1=\left(2x\right)^2-2\cdot2x\cdot1+1^2=\left(2x-1\right)^2\)
c. \(4x^2+12x+9=\left(2x\right)^2+2\cdot2x\cdot3+3^2=\left(2x+3\right)^2\)
d. \(9x^2+30x+25=\left(3x\right)^2+2\cdot3x\cdot5+5^2=\left(3x+5\right)^2\)
e. \(4x^2-20x+25=\left(2x\right)^2-2\cdot2x\cdot5+5^2=\left(2x+5\right)^2\)
Ta có : 9x2 + 12x + 15
= (3x)2 + 2.3x.2 + 4 + 11
= (3x + 2)2 + 11
Mà (3x + 2)2 \(\ge0\forall x\)
Nên (3x + 2)2 + 11 \(\ge11\forall x\)
Vậy Bmin = 11 dấu "=" sảy ra khi và chỉ khi x = \(-\frac{2}{3}\)
Ta có : A = x2 - 4x - 6
= x2 - 4x + 4 - 10
= (x - 2)2 - 10
Mà (x - 2)2 \(\ge0\forall x\)
=> (x - 2)2 - 10 \(\ge-10\forall x\)
Vậy Amin = -10 dấu "=" sảy ra khi và chỉ khi x = 2
\(a,\Rightarrow4x^2+20x+25=0\Rightarrow\left(2x+5\right)^2=0\Rightarrow2x+5=0\Rightarrow x=-\dfrac{5}{2}\\ b,\Rightarrow x^3-6x^2+12x-8=0\Rightarrow\left(x-2\right)^3=0\Rightarrow x-2=0\Rightarrow x=2\)
a) \(\Rightarrow4x^2+20x+25=0\)
\(\Rightarrow\left(2x+5\right)^2=0\Rightarrow2x+5=0\)
\(\Rightarrow x=-\dfrac{5}{2}\)
b) \(\Rightarrow x^3-6x^2+12x-8=0\)
\(\Rightarrow\left(x-2\right)^3=0\Rightarrow x=2\)