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)

a) 0

b) 3

c) -0,5

a là 0 

b là 3

c là -0,5

a)0

b)3

c)-0,5

tick mjnh nhé

a) x=0

b) x=3

c) x= -2

tick mình đi

a) -2

b) \(\frac{-11}{3}\)

c) -0,5

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

\(=2x^2-6xy+4y^2+5x^2-4xy-7y^2\)

\(=7x^2-10xy-3y^2\)

\(=7\cdot1^2-10\cdot1\cdot\dfrac{1}{2}-3\cdot\dfrac{1}{4}=7-5-\dfrac{3}{4}=2-\dfrac{3}{4}=\dfrac{5}{4}\)

\(x=3\ge2\Leftrightarrow y=3+1=4\\ x=-1< 2\Leftrightarrow y=\left(-1\right)^2-2=1-2=-1\\ x=2\ge2\Leftrightarrow y=2+1=3\)