Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=x^2-2x+10\)
\(A=\left(x^2-2x+1\right)+9\)
\(A=\left(x-1\right)^2+9\)
Mà \(\left(x-1\right)^2\ge0\)
\(\Rightarrow A\ge9\)
Dấu "=" xảy ra khi :
\(x-1=0\Leftrightarrow x=1\)
Vậy Min A = 9 khi x = 1
\(B=x^2-5x-7\)
\(B=\left(x^2-5x+\frac{25}{4}\right)-\frac{53}{4}\)
\(B=\left(x-\frac{5}{2}\right)^2-\frac{53}{4}\)
Mà \(\left(x-\frac{5}{2}\right)^2\ge0\)
\(\Rightarrow B\ge-\frac{53}{4}\)
Dấu "=" xảy ra khi :
\(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)
Vậy \(B_{Min}=-\frac{53}{4}\Leftrightarrow x=\frac{5}{2}\)
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 1:
a) (x+1)^2-(x-1)^2-3(x+1)(x-1)
=(x+1+x-1)(x+1-x+1)-3x^2-3
=2x^2-3x^2-3
=-x^2-3
a) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)
\(\Leftrightarrow\left(x^2+6x+9\right)-\left(x^2+4x-32\right)-1=0\)
\(\Leftrightarrow2x=-40\)
\(\Rightarrow x=-20\)
b) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)
\(\Leftrightarrow x^3+27-x^3+4x=15\)
\(\Leftrightarrow4x=-12\)
\(\Rightarrow x=-3\)
c) \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)
\(\Leftrightarrow\left(x^2-4x+4\right)-\left(x^2+6x+9\right)-\left(4x+4\right)=5\)
\(\Leftrightarrow-14x=14\)
\(\Rightarrow x=-1\)
d) \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)
\(\Leftrightarrow4x^2-9-\left(x^2-2x+1\right)-\left(3x^2-15x\right)=-44\)
\(\Leftrightarrow17x=-34\)
\(\Rightarrow x=-2\)
e) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)
\(\Leftrightarrow24x=24\)
\(\Rightarrow x=1\)