CM : 12-4x2-11<0 với mọi x
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![](https://rs.olm.vn/images/avt/0.png?1311)
+ Ta có: ( - 4 ).5 = 4.( - 5 ) → Khẳng định ( 1 ) sai.
+ Ta có: 12 > 11 ⇒ 12.( - 7 ) < 11.( - 7 ) → Khẳng định ( 2 ) sai.
+ Ta có: x 2 ≥ 0 ⇒ - 4 x 2 ≤ 0 → Khẳng định ( 3 ) sai
Chọn đáp án A.
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: ( − 4 ) .5 = 4. ( − 5 ) nên khẳng định (1) đúng.
Vì 12 > 11 ⇒ 12. − 7 < 11. − 7 nên khẳng định (2) sai.
Vì x 2 ≥ 0 ⇒ − 4 x 2 ≤ 0 nên khẳng định (3) sai.
Chọn đáp án D.
![](https://rs.olm.vn/images/avt/0.png?1311)
c: Ta có: \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)
\(\Leftrightarrow3x^2+26x=0\)
\(\Leftrightarrow x\left(3x+26\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\)
\(a,\Leftrightarrow x^2+8x+16-x^3-12x^2=16\\ \Leftrightarrow x^3+11x^2-8x=0\\ \Leftrightarrow x\left(x^2+11x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+11x-8=0\left(1\right)\end{matrix}\right.\\ \Delta\left(1\right)=121+32=153\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11-3\sqrt{17}}{2}\\x=\dfrac{-11+3\sqrt{17}}{2}\end{matrix}\right.\\ S=\left\{0;\dfrac{-11-3\sqrt{17}}{2};\dfrac{-11+3\sqrt{17}}{2}\right\}\)
\(c,\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\\ \Leftrightarrow3x^2+26x=0\\ \Leftrightarrow x\left(3x+26\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\\ d,\Leftrightarrow x^3-6x^2+12x-8-x^3-125-6x^2=11\\ \Leftrightarrow-12x^2+12x-144=0\\ \Leftrightarrow x^2-x+12=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(8x+12-4x^2\right):\left(x-3\right)=\left[4x\left(x-3\right)+20\left(x-3\right)+72\right]:\left(x-3\right)=\left[\left(x-3\right)\left(4x+20\right)+72\right]:\left(x-3\right)=4x+20R72\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a)16x^2-24x+9=25`
`<=>(4x-3)^2=25`
`+)4x-3=5`
`<=>4x=8<=>x=2`
`+)4x-3=-5`
`<=>4x=-2`
`<=>x=-1/2`
`b)x^2+10x+9=0`
`<=>x^2+x+9x+9=0`
`<=>x(x+1)+9(x+1)=0`
`<=>(x+1)(x+9)=0`
`<=>` \(\left[ \begin{array}{l}x=-9\\x=-1\end{array} \right.\)
`c)x^2-4x-12=0`
`<=>x^2+2x-6x-12=0`
`<=>x(x+2)-6(x+2)=0`
`<=>(x+2)(x-6)=0`
`<=>` \(\left[ \begin{array}{l}x=-2\\x=6\end{array} \right.\)
`d)x^2-5x-6=0`
`<=>x^2+x-6x-6=0`
`<=>x(x+1)-6(x+1)=0`
`<=>(x+1)(x-6)=0`
`<=>` \(\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.\)
`e)4x^2-3x-1=0`
`<=>4x^2-4x+x-1=0`
`<=>4x(x-1)+(x-1)=0`
`<=>` \(\left[ \begin{array}{l}x=1\\x=-\dfrac14\end{array} \right.\)
`f)x^4+4x^2-5=0`
`<=>x^4-x^2+5x^2-5=0`
`<=>x^2(x^2-1)+5(x^2-1)=0`
`<=>(x^2-1)(x^2+5)=0`
Vì `x^2+5>=5>0`
`=>x^2-1=0<=>x^2=1`
`<=>` \(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
3/11 x 2 = 6/11
6/11 : 3/11 = 2
6/11 : 2 = 3/11
2/3 x 11 = 22/3
4/2 x 7 = 14
8/7 : 2/7 = 4
8/7 : 4 = 2/7
2/7 x 4 = 8/7
đề sai r bn ơi