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a ) x=0; x = -(căn bậc hai(7)*i-3)/8;x = (căn bậc hai(7)*i+3)/8;
b ) -(y-x-3)*(y+x+3)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left(3x+2\right)\left(9x^2-6x+4\right)-9x\left(3x^2+1\right)\)
\(=27x^3+8-27x^3-9x\)
=8-9x
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a: \(=\dfrac{3\left(x-2\right)}{\left(x-2\right)^3}=\dfrac{3}{\left(x-2\right)^2}\)
b: \(=\dfrac{x^2\left(x+2\right)}{\left(x+2\right)^3}=\dfrac{x^2}{\left(x+2\right)^2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
( 3 x – 4 ) ( 9 x 2 + 12 x + 16 ) = ( 3 x – 4 ) ( ( 3 x ) 2 + 3 x . 4 + 4 2 ) = ( 3 x ) 3 – 4 3
Đáp án cần chọn là: D
![](https://rs.olm.vn/images/avt/0.png?1311)
Với \(x\ge\dfrac{1}{6}\Leftrightarrow A=5x^2-6x+1-1=5x^2-6x\)
\(A=5\left(x^2-2\cdot\dfrac{3}{5}x+\dfrac{9}{25}\right)-\dfrac{9}{5}=5\left(x-\dfrac{3}{5}\right)^2-\dfrac{9}{5}\ge-\dfrac{9}{5}\\ A_{min}=-\dfrac{9}{5}\Leftrightarrow x=\dfrac{3}{5}\left(1\right)\)
Với \(x< \dfrac{1}{6}\Leftrightarrow A=5x^2+6x-1-1=5x^2+6x-2\)
\(A=5\left(x^2+2\cdot\dfrac{3}{5}x+\dfrac{9}{25}\right)-\dfrac{19}{5}=5\left(x+\dfrac{3}{5}\right)^2-\dfrac{19}{5}\ge-\dfrac{19}{5}\\ A_{min}=-\dfrac{19}{5}\Leftrightarrow x=-\dfrac{3}{5}\left(2\right)\\ \left(1\right)\left(2\right)\Leftrightarrow A_{min}=-\dfrac{19}{5}\Leftrightarrow x=-\dfrac{3}{5}\)
Với \(x\ge\dfrac{1}{3}\Leftrightarrow B=9x^2-6x-4\left(3x-1\right)+6=9x^2-18x+10\)
\(B=9\left(x^2-2x+1\right)+1=9\left(x-1\right)^2+1\ge1\\ B_{min}=1\Leftrightarrow x=1\left(1\right)\)
Với \(x< \dfrac{1}{3}\Leftrightarrow B=9x^2-6x+4\left(3x-1\right)+6=9x^2+6x+2\)
\(B=\left(9x^2+6x+1\right)+1=\left(3x+1\right)^2+1\ge1\\ B_{min}=1\Leftrightarrow x=-\dfrac{1}{3}\left(2\right)\)
\(\left(1\right)\left(2\right)\Leftrightarrow B_{min}=1\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, (3x)^2-12x+4
=(3x-2)^2
b,= -(-2xy-16+x^2+y^2)
=-(x^2-2xy+y^2-16)
=-((x-y)^2-4^2)
=-((x-y-4)(x-y+4))
c, = 2x^2+3x-2
=2x^2+x-4x-2
=x(2x+1)-2(2x+1)
=(x-2)(2x+1)
2,
a,= x^3+3x^2.2+3x2^2+2^3
=(x+3)^3, thay vào ta có
=(-3+3)^3=0
b, chịu ko biết làm
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 2:
a: \(\Leftrightarrow3x^2+2x-1=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)
b: \(\Leftrightarrow x^3-4x-x^3-8=4\)
hay x=-3
\(M=\left(3x-4\right)\left(9x^2-12x+16\right)+\left(6x+8\right)^2\)
\(\Rightarrow M=\left(3x-4\right)\left(3x-4\right)^2+4\left(3x-4\right)^2\)
\(\Rightarrow M=\left(3x-4\right)^2\left(3x-4+4\right)\)
\(\Rightarrow M=\left(3x-4\right)^2.3x\)
Ko ghi lại đề
\(=>M=\left(3X-4\right)\left(3x-4\right)^2+4\left(3x-4\right)^2\)
\(=>M=\left(3X-4\right)^2\left(3x-4+4\right)\)
\(=>M=\left(3X-4\right)^2\left(3x\right)\)
~Study well~ :)