2014mm=?m
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Giá tiền hai món mà Hà mua là:
`30000 - 2000 = 28000 ` (đồng)
Giá bành mì là:
`(28000 + 12000) : 2 = 20 000 ` (đồng)
Giá hộp sữa là:
`20000 - 12000 = 8000` (đồng)
Đáp số: ...
![](https://rs.olm.vn/images/avt/0.png?1311)
a.
\(\sqrt{x^2-4x+1}=x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2-4x+1=x^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-4x+1=0\end{matrix}\right.\)
\(\Rightarrow x=\dfrac{1}{4}\)
b.
\(\sqrt{5x^2-2x+2}=x+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\5x^2-2x+2=\left(x+1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\4x^2-4x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow x=\dfrac{1}{2}\)
c.
\(\sqrt{x^2-8x+16}=4-x\)
\(\Leftrightarrow\sqrt{\left(4-x\right)^2}=4-x\)
\(\Leftrightarrow\left|4-x\right|=4-x\)
\(\Leftrightarrow4-x\ge0\)
\(\Rightarrow x\le4\)
d.
\(\sqrt{3x+1}=\sqrt{4x-3}\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x-3\ge0\\3x+1=4x-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\end{matrix}\right.\)
\(\Rightarrow x=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1)A=x^2-7x+2\\ =\left(x^2-2\cdot x\cdot\dfrac{7}{2}+\dfrac{49}{4}\right)-\dfrac{41}{4}\\ =\left(x-\dfrac{7}{2}\right)^2-\dfrac{41}{4}\)
Ta có: `(x-7/2)^2>=0` với mọi x
`=>A=(x-7/2)^2-41/4>=-41/4` với mọi x
Dấu "=" xảy ra: `x-7/2=0<=>x=7/2`
\(2)B=9x^2-12x+5\\ =\left(9x^2-12x+4\right)+1\\ =\left[\left(3x\right)^2-2\cdot3x\cdot2+2^2\right]+1\\ =\left(3x-2\right)^2+1\)
Ta có: `(3x-2)^2>=0` với mọi x
`=>B=(3x-2)^2+1>=1` với mọi x
Dấu "=" xảy ra: `3x-2=0<=>x=2/3`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3\left(x^2+2x-1\right)-2\left(x^2+3x-1\right)+5x^2=0\)
=>\(3x^2+6x-3-2x^2-6x+2+5x^2=0\)
=>\(6x^2-1=0\)
=>\(6x^2=1\)
=>\(x^2=\dfrac{1}{6}\)
=>\(x=\pm\dfrac{\sqrt{6}}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1: \(\left(x-y\right)^2-\left(x+y\right)^2\)
\(=x^2-2xy+y^2-x^2-2xy-y^2\)
=-4xy
2: \(\left(7n-2\right)^2-\left(2n-7\right)^2\)
\(=\left(7n-2+2n-7\right)\left(7n-2-2n+7\right)\)
\(=\left(9n-9\right)\left(5n+5\right)\)
\(=9\left(n-1\right)\left(5n+5\right)⋮9\)
3: \(P=-x^2+6x+1\)
\(=-\left(x^2-6x-1\right)\)
\(=-\left(x^2-6x+9-10\right)\)
\(=-\left(x-3\right)^2+10< =10\forall x\)
Dấu '=' xảy ra khi x-3=0
=>x=3
4: \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)
=>\(a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+b^2y^2+2abxy\)
=>\(a^2y^2-2abxy+b^2x^2=0\)
=>\(\left(ay-bx\right)^2=0\)
=>ay-bx=0
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x-1\right)\left(3x+1\right)+\left(3x+4\right)\left(3x-2\right)=5\)
=>\(6x^2+2x-3x-1+9x^2-6x+12x-8=5\)
=>\(15x^2+5x-9-5=0\)
=>\(15x^2+5x-14=0\)
\(\Delta=5^2-4\cdot15\cdot\left(-14\right)=25+60\cdot14=25+840=865>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left[{}\begin{matrix}x=\dfrac{-5-\sqrt{865}}{2\cdot15}=\dfrac{-5-\sqrt{865}}{30}\\x=\dfrac{-5+\sqrt{865}}{30}\end{matrix}\right.\)
`2014 mm = 2014/1000 m = 2,014m`