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a: Ta có: \(5-3x< 8\)
\(\Leftrightarrow3x>-3\)
hay x>-1
b: Ta có: \(\dfrac{2x-5}{4}\ge\dfrac{3-x}{3}\)
\(\Leftrightarrow3\left(2x-5\right)\ge4\left(3-x\right)\)
\(\Leftrightarrow6x-15\ge12-4x\)
\(\Leftrightarrow10x\ge27\)
hay \(x\ge\dfrac{27}{10}\)
c: Ta có: \(2x+5< x+7\)
\(\Leftrightarrow2x-x< 7-5\)
hay x<2
d: Ta có: \(4\left(x-3\right)\ge x+2\)
\(\Leftrightarrow4x-12-x-2\ge0\)
\(\Leftrightarrow3x\ge14\)
hay \(x\ge\dfrac{14}{3}\)
e: Ta có: \(\dfrac{2x+2}{3}< 2+\dfrac{x-2}{2}\)
\(\Leftrightarrow4x+4< 12+3x-6\)
\(\Leftrightarrow4x-3x< 6-4\)
hay x<2
f: Ta có: \(x-\dfrac{5x+2}{6}>\dfrac{7-3x}{4}\)
\(\Leftrightarrow12x-2\left(5x+2\right)>3\left(7-3x\right)\)
\(\Leftrightarrow2x-4>21-9x\)
\(\Leftrightarrow11x>25\)
hay \(x>\dfrac{25}{11}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
\(x\left(-2x+1\right)=-1\)
\(\Rightarrow x=1,-1;\left(-2x+1\right)=1;-1\\
\)
Mà \(\left(-2x+1\right)\le0\\
\left(-2x+1\right)=-1\Leftrightarrow x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=3x^2-12x+16=3\left(x^2-4x\right)+16\)
\(=3\left(x^2-4x+4-4\right)+16\)
\(=3\left(x^2-4x+4\right)-3.4+16\)
\(=3\left(x-2\right)^2+4\ge4\), với mọi x
Vì \(\left(x-2\right)^2\ge0\) với mọi x
nên \(A=3\left(x-2\right)^2+4\ge3.0+4=4\) với mọi x
dấu "=" xảy ra khi và chỉ khi: \(\left(x-2\right)^2=0\Leftrightarrow x=2\)
Vậy giá tri nhỏ nhất của A là 4 tại x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\Leftrightarrow2x^2\left(x^2-4\right)=0\\ \Leftrightarrow2x^2\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow\left(x+10\right)\left(\dfrac{2}{5}x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+10=0\\\dfrac{2}{5}x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-10\\x=\dfrac{5}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(A=x^2-20x+101\)
\(=x^2-20x+100+1\)
\(=\left(x-10\right)^2+1\ge1\forall x\)
Dấu '=' xảy ra khi x=10
\(1,\\ a,=3x\left(1-3y\right)\\ b,=9xy\left(2xy-x^2+4y\right)\\ c,=\left(x-y\right)\left(15x-5y\right)=5\left(x-y\right)\left(3x-y\right)\\ 2,\\ a,\Rightarrow2x^2\left(x^2-4\right)=0\Rightarrow2x^2\left(x-2\right)\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\\ b,\Rightarrow\dfrac{2}{5}x\left(x+10\right)-\left(x+10\right)=0\\ \Rightarrow\left(x+10\right)\left(\dfrac{2}{5}x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-10\\\dfrac{2}{5}x=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-10\\x=\dfrac{5}{2}\end{matrix}\right.\)
\(3,\)
\(a,\left\{{}\begin{matrix}AK=KD\\BI=IC\end{matrix}\right.\Rightarrow KI\) là đtb hình thang ABCD
\(b,\) Vì KI là đtb hình thang ABCD nên \(KI=\dfrac{AB+CD}{2}=\dfrac{17}{2}=8,5\left(cm\right)\)
\(c,\) \(\left\{{}\begin{matrix}AK=KD\\KE//AB\end{matrix}\right.\Rightarrow BE=ED\Rightarrow KE\) là đtb tam giác ABD
\(\Rightarrow KE=\dfrac{1}{2}AB=2,5\left(cm\right)\)
\(\left\{{}\begin{matrix}BI=IC\\IF//AB\end{matrix}\right.\Rightarrow AF=FC\Rightarrow IF\) là đtb tam giác ABC
\(\Rightarrow IF=\dfrac{1}{2}AB=2,5\left(cm\right)\)
Ta có \(EF=KI-KE-IF=8,5-2,5-2,5=3,5\left(cm\right)\)