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A = -x2 + 2x + 7 = -( x2 - 2x + 1 ) + 8 = -( x - 1 )2 + 8 ≤ 8 ∀ x
Dấu "=" xảy ra <=> x = 1 => MinA = 8
B = 5x - 3x2 + 6 = -3( x2 - 5/3x + 25/36 ) + 97/12 = -3( x - 5/6 )2 + 97/12 ≤ 97/12 ∀ x
Dấu "=" xảy ra <=> x = 5/6 => MinB = 97/12
a) 4x(x-1) = x-1
4x(x-1) - ( x-1) = 0
(4x - 1) (x-1 ) = 0
=> 4x -1=0 x-1=0
x=1/4 x=1
b) 2x3 -50x =0
2x( x2 - 25 ) = 0
=> 2x = 0 x2 -25 = 0
x=0 x2 =25 => x =-5
x=5
c) 5x(2x-7)-14x = -49
5x(2x-7) -(14x -49 ) =0
5x(2x-7) - 7(2x-7) = 0
(5x-7)(2x-7)=0
=> 5x-7 =0 2x-7=0
x=7/5 x=7/2
Vậy .....
hok tốt
Lời giải:
a) \(4x(x-1)=x-1\)
\(\Leftrightarrow 4x(x-1)-(x-1)=0\Leftrightarrow (x-1)(4x-1)=0\)
\(\Rightarrow \left[\begin{matrix} x-1=0\rightarrow x=1\\ 4x-1=0\rightarrow x=\frac{1}{4}\end{matrix}\right.\)
b) \(2x^3-50x=0\Leftrightarrow 2x(x^-25)=0\)
\(\Leftrightarrow 2x(x-5)(x+5)=0\)
\(\Rightarrow \left[\begin{matrix} x=0\\ x-5=0\rightarrow x=5\\ x+5=0\rightarrow x=-5\end{matrix}\right.\)
c) \(5x(2x-7)-14x=-49\)
\(\Leftrightarrow 5x(2x-7)-7(2x-7)=0\)
\(\Leftrightarrow (5x-7)(2x-7)=0\Rightarrow \left[\begin{matrix} 5x-7=0\rightarrow x=\frac{7}{5}\\ 2x-7=0\rightarrow x=\frac{7}{2}\end{matrix}\right.\)
a, \(12-2\left(1-x\right)^2=\left(3x-2\right)\left(2x-3\right)\)
\(< =>12-2\left(1-2x+x^2\right)=6x^2-9x-4x+6\)
\(< =>12-2+4x-2x^2=6x^2-13x+6\)
\(< =>10+4x-2x^2-6x^2+13x-6=0\)
\(< =>-8x^2+17x+4=0< =>\orbr{\begin{cases}x=\frac{17-\sqrt{417}}{16}\\x=\frac{17+\sqrt{417}}{16}\end{cases}}\)
b, \(10x+3-5x=4x+12< =>5x+3-4x-12=0\)
\(< =>x-9=0< =>x=9\)
c, \(11x+42-2x=100-9x-22< =>9x+42-100+9x+22=0\)
\(< =>18x+64-100=0< =>18x-36=0< =>x=\frac{36}{18}=2\)
d, \(2x-\left(3-5x\right)=4\left(x+3\right)< =>2x-3+5x=4x+12\)
\(< =>7x-3-4x-12=0< =>3x-15=0< =>x=\frac{15}{3}=5\)
e, \(2\left(x-3\right)+5x\left(x-1\right)=5x^2< =>2x-6+5x^2-5=5x^2\)
\(< =>2x-11+5x^2-5x^2=0< =>2x-11=0< =>x=\frac{11}{2}\)
f, \(-6\left(1,5-2x\right)=3\left(-15+2x\right)< =>-6\left(\frac{3}{2}-2x\right)=3\left(2x-15\right)\)
\(< =>-9+12x-6x+45=0< =>6x+36=0< =>x=-6\)
g, \(14x-\left(2x+7\right)=3x+12x-13< =>14x-2x-7=15x-13\)
\(< =>12x-7-15x+13=0< =>-3x+6=0< =>x=-2\)
h, \(\left(x-4\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\)
\(< =>x^2-16-6x+4=x^2-8x+16\)
\(< =>x^2-6x-12-x^2+8x-16=0\)
\(< =>2x-28=0< =>x=\frac{28}{2}=14\)
q, \(4\left(x-2\right)-\left(x-3\right)\left(2x-5\right)=?\)thiếu đề
a) -4x2 + 8x - 4
= - (4x2 - 8x + 4)
= - (2x - 2)2
b) -x52 + 10 x - 5
= - 5(x2 - 2x + 1)
= - 5(x - 1)2