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Bài này áp dụng hằng đẳng thức \(\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3\) nha bạn
Ta có :
\(x^3-6x^2+12x-8=0\)
\(\Leftrightarrow\)\(x^3-3.x^2.2+3.x.2^2-2^3=0\)
\(\Leftrightarrow\)\(\left(x-2\right)^3=0\)
\(\Leftrightarrow\)\(x-2=0\)
\(\Leftrightarrow\)\(x=2\)
Vậy \(x=2\)
Chúc bạn học tốt ~
\(x^3-6x^2+12x-8=0\) \(\Leftrightarrow x^3-3.x^2.2+3.x.2^2-2^3=0\) \(\Leftrightarrow\left(x-2\right)^3=0\Leftrightarrow x-2=0\) \(\Leftrightarrow x=2\)
Ta có \(x^3-6x^2+12-8=0\)
\(\Rightarrow x^3-6x^2+4=0\)
\(\Rightarrow x^3-6x^2=-4\)
\(\Rightarrow x^2.\left(x-6\right)=-4\)
đề bài sai sai thì phải :v
x3 - 6x2 + 12 - 8
<=> (x + 2)3 = 0
<=> x + 2 = 0
<=> x = 0 - 2
<=> x = -2
=> x = -2
a ) 2x ( x - 5 ) - x ( 3 + 2x ) = 26
2x2 - 10x - 3x - 2x2 = 26
- 13x = 26
x = 26 : ( -13 )
x = -2
b) 49x2 - 81 = 0
( 7x - 9 )( 7x + 9 ) = 0
Th1 :
7x - 9 = 0
7x = 9
x = \(\frac{9}{7}\)
Th2
7x + 9 = 0
7x = -9
x = \(-\frac{9}{7}\)
Vay x = \(\frac{9}{7}\) hoac x = \(-\frac{9}{7}\)
\(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-2\right)\left(x+1\right)+3x-2=0\)
\(x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+6x+12+3x-2=0\)
\(1+1+6x+3x+12-2=0\)
\(9x+12=0\)
\(9x=-12\)
\(x=\frac{-4}{3}\)
\(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-2\right)\left(x+1\right)+3x-2=0\)
\(\Leftrightarrow\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-2\right)\left(x+1\right)+3x=0+2\)
\(\Leftrightarrow\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-2\right)\left(x+1\right)+3x=2\)
\(\Leftrightarrow9x+14=2\)
\(\Leftrightarrow9x=2-14\)
\(\Leftrightarrow9x=-12\)
\(\Leftrightarrow x=\frac{-12}{9}=\frac{-4}{3}\)
\(\Rightarrow x=\frac{-4}{2}\)
Ta có :
\(\left(3x+2\right)\left(9x^2-6x+4\right)-\left(x-3\right)\left(x+3\right)\)
\(=\)\(\left(3x+2\right)\left[\left(3x\right)^2-3x.2+2^2\right]-\left(x^2-3^2\right)\)
\(=\)\(\left(3x\right)^3+2^3-x^2-3^2\)
\(=\)\(27x^3-x^2+8-9\)
\(=\)\(27x^3-x^2-1\)
Chúc bạn học tốt ~
a ) ( 2x + 1 )2 - 4 ( x + 2 )2 = 9
4x2 + 4x + 1 - 4 ( x2 +4x + 4 ) = 9
4x2 + 4x + 1 - 4x2 -16x -16 = 9
-12x - 15 = 9
-12x = 24
x = -2
b) 3 ( x - 1 )2 - 3x ( x - 5 ) = 1
3 ( x2 - 2x + 1 ) - 3x2 + 15x = 1
3x2 - 6x + 3 - 3x2 + 15x = 1
9x + 3 = 1
9x = -2
x = \(\frac{-2}{9}\)
\(8x^3+12x^2+6x+1=\left(2x+1\right)^3\)
\(=\left(2\cdot24.5+1\right)^3=50^3=125000\)
\(B=x^3-3x^2+3x\)
\(=x^3-3x^21+3x1^2-1^3+1\)
\(=\left(x-1\right)^3+1\)
thay x=11 vào P ta đc:
\(B=\left(11-1\right)^3+1=1001\)
Vậy B=1001
Ta có :
\(49x^2-81=0\)
\(\Leftrightarrow\)\(\left(7x\right)^2-9^2=0\)
\(\Leftrightarrow\)\(\left(7x-9\right)\left(7x+9\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}7x-9=0\\7x+9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}7x=0+9\\7x=0-9\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}7x=9\\7=-9\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{7}\\x=\frac{-9}{7}\end{cases}}}\)
Vậy \(x=\frac{9}{7}\) hoặc \(x=\frac{-9}{7}\)
Chúc bạn học tốt ~
\(49x^2-81=0\)
\(\Rightarrow\)\(\left(7x\right)^2=9^2\)
\(\Rightarrow\)\(7x=9\)
\(\Rightarrow\)\(x=\frac{9}{7}\)