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\(\left(x+3\right)^3-x\left(3x-1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-x\left(9x^2-6x+1\right)+8x^3-4x^2+2x+4x^2-2x+1=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3+6x^2-x+8x^3-4x^2+2x+4x^2-2x+1-28=0\)
\(\Leftrightarrow15x^2+26x=0\)
\(\Leftrightarrow15x\left(x+\frac{26}{15}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}15x=0\\x+\frac{26}{15}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{26}{15}\end{cases}}}\)
\(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow2x^3+2x+3x^2+3=0\)
\(\Leftrightarrow2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x^2+1\right)=0\) vì \(x^2+1>0\) nên \(2x+3=0\Rightarrow x=-\frac{3}{2}\)
8x^3 - 8x^2 + 20x^2 - 20x + 26x - 26 =0
<=> 8x^2 ( x-1) + 20x(x-1) + 26(x-1)=0
<=>( 8x^2 + 20x + 26)(x-1)=0
<=> ( x-1)= 0 ( Vì 8x^2 + 20x + 26 >=13,5)
<=> x=1
\(a,\\ A=25x^2-10x+11\\ =\left(5x\right)^2-2.5x.1+1^2+10\\ =\left(5x+1\right)^2+10\ge10\forall x\in R\\ Vậy:min_A=10.khi.5x+1=0\Leftrightarrow x=-\dfrac{1}{5}\\ B=\left(x-3\right)^2+\left(11-x\right)^2\\ =\left(x^2-6x+9\right)+\left(121-22x+x^2\right)\\ =x^2+x^2-6x-22x+9+121=2x^2-28x+130\\ =2\left(x^2-14x+49\right)+32\\ =2\left(x-7\right)^2+32\\ Vì:2\left(x-7\right)^2\ge0\forall x\in R\\ Nên:2\left(x-7\right)^2+32\ge32\forall x\in R\\ Vậy:min_B=32.khi.\left(x-7\right)=0\Leftrightarrow x=7\\Tương.tự.cho.biểu.thức.C\)
b:
\(D=-25x^2+10x-1-10\)
\(=-\left(25x^2-10x+1\right)-10\)
\(=-\left(5x-1\right)^2-10< =-10\)
Dấu = xảy ra khi x=1/5
\(E=-9x^2-6x-1+20\)
\(=-\left(9x^2+6x+1\right)+20\)
\(=-\left(3x+1\right)^2+20< =20\)
Dấu = xảy ra khi x=-1/3
\(F=-x^2+2x-1+1\)
\(=-\left(x^2-2x+1\right)+1=-\left(x-1\right)^2+1< =1\)
Dấu = xảy ra khi x=1
(x-3)(x2 + 3x +9)- ((x-4)((x+4)=21
x3 - 27 - x2 + 4 = 21
x2 + x - 27 -x2 + 4 =21
x=27 -4 + 21
x= 44
2)
( x _ 3 ) ( x^2 + 3x + 9 ) - ( x - 4 ) . ( x + 4 ) = 21
= x^3 - 9 - x^2 - 2^2 = 21
= x - 9 - x^2 - 4
= x^3 - x^2 - 9 - 4
x = - 9 - 4 = - 13
a) \(\left(x-3\right)^2-4=0\)
\(\left(x-7\right)\left(x+1\right)=0\)
\(\orbr{\begin{cases}x=7\\x=-1\end{cases}}\)
b) \(x^2-2x=24\)
\(x^2-2x-24=0\)
\(\left(x-6\right)\left(x+4\right)=0\)
\(\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)
c) \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(4x^2+4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
\(5x^2+10x+10-5x^2+245=0\)
\(10x+255=0\)
\(x=-25.5\)
A) \(\left(x-3\right)^2-4=0\)
\(\left(x-3\right)^2=4\Rightarrow\left(x-3\right)^2=\left(-2\right)^2;2^2\)
th1\(\left(x-3\right)^2=2^2\)
\(\Rightarrow x-3=2\)
\(\Rightarrow x=2+3\)
\(\Rightarrow x=5\)
th2: \(\left(x-3\right)^2=\left(-2\right)^2\)
\(\Rightarrow x-3=-2\)
\(\Rightarrow x=-2+3\)
\(\Rightarrow x=1\)
\(\Leftrightarrow x\in\left\{1;5\right\}\)
bài này mình chưa học nên xin lỗi bạn
(2x - 3)2 = (x - 2)3 - x(3 + x2 - 10x)
4x2 - 12x + 9 = x3 - 4x2 + 4x - 2x2 + 8x - 8 - 3x - x3 + 10x2
4x2 - 12x + 9 = 4x2 + 9x - 8
-12x + 9 = 9x - 8
9 = 9x - 8 + 12x
9 = 21x - 8
9 + 8 = 21x
17 = 21x
17/21 = x
=> x = 17/21