a)IxI+I-xI=3-x
b)x/6-1/y=1/2
tính x,y
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a) \(\left|x\right|\) + \(\left|-x\right|\) = 3 - \(x\)
\(\Rightarrow\) \(x+x\) = 3 - \(x\)
\(\Rightarrow\) \(x+x+x\) = 3
\(\Rightarrow\) 3\(x\) = 3
\(\Rightarrow\) \(x\) = 1
b) Ta có : \(\frac{x}{6}\) - \(\frac{1}{y}\) = \(\frac{1}{2}\)
\(\Rightarrow\) \(\frac{1}{y}\) = \(\frac{x}{6}\) - \(\frac{1}{2}\)
\(\Rightarrow\) \(\frac{1}{y}\) = \(\frac{x}{6}\) - \(\frac{3}{6}\)
\(\Rightarrow\) \(\frac{1}{y}\) = \(\frac{x-3}{6}\)
\(\Rightarrow\) 1 . 6 = \(y\)( \(x\) - 3)
\(\Rightarrow\) 6 = \(y\)(\(x\) - 3)
\(\Rightarrow\) \(y\)(\(x\) - 3) ϵ Ư(6)
\(y\) | 1 | 2 | 3 | 6 | -1 | -2 | -3 | -6 |
\(x-3\) | 6 | 3 | 2 | 1 | -6 | -3 | -2 | -1 |
\(x\) | 9 | 6 | 5 | 4 | -3 | 0 | 1 | 2 |
a: (x+1)^3-x(x-2)^2+x-1=0
=>x^3+3x^2+3x+1-x(x^2-4x+4)+x-1=0
=>x^3+3x^2+4x-x^3+4x^2-4x=0
=>7x^2=0
=>x=0
b: =>x^3-3x^2+3x-1-x^3-27+3x^2-12=2
=>3x=2+1+27+12=39+3=42
=>x=14
Lời giải:
$(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=2$
$\Leftrightarrow (x+\sqrt{x^2+1})(x-\sqrt{x^2+1})(y+\sqrt{y^2+1})=2(x-\sqrt{x^2+1})$
$\Leftrightarrow -(y+\sqrt{y^2+1})=2(x-\sqrt{x^2+1})$
$\Leftrightarrow 2x+\sqrt{y^2+1}=2\sqrt{x^2+1}-y$
$\Rightarrow (2x+\sqrt{y^2+1})^2=(2\sqrt{x^2+1}-y)^2$
$\Leftrightarrow 4x^2+y^2+1+4x\sqrt{y^2+1}=4(x^2+1)+y^2-4y\sqrt{x^2+1}$
$\Leftrightarrow 4(x\sqrt{y^2+1})+y\sqrt{x^2+1})=3$
$\Leftrightarrow 4Q=3$
$\Leftrightarrow Q=\frac{3}{4}$
Bài 2:
C=A-B
\(=2x^2-6xy+4y^2+5x^2-4xy-7y^2\)
\(=7x^2-10xy-3y^2\)
\(=7\cdot1^2-10\cdot1\cdot\dfrac{1}{2}-3\cdot\dfrac{1}{4}=7-5-\dfrac{3}{4}=2-\dfrac{3}{4}=\dfrac{5}{4}\)
\(x=3\ge2\Leftrightarrow y=3+1=4\\ x=-1< 2\Leftrightarrow y=\left(-1\right)^2-2=1-2=-1\\ x=2\ge2\Leftrightarrow y=2+1=3\)