(3x-2)^2-2(x-1)^2
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\(\left(3x-2\right)^2-2\left(x-1\right)^2\)
\(=9x^2-12x+4-2\left(x^2-2x+1\right)\)
\(=9x^2-12x+4-2x^2+4x-2\)
\(=7x^2-16x+2\)
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:
$(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=2$
$\Leftrightarrow (x+\sqrt{x^2+1})(x-\sqrt{x^2+1})(y+\sqrt{y^2+1})=2(x-\sqrt{x^2+1})$
$\Leftrightarrow -(y+\sqrt{y^2+1})=2(x-\sqrt{x^2+1})$
$\Leftrightarrow 2x+\sqrt{y^2+1}=2\sqrt{x^2+1}-y$
$\Rightarrow (2x+\sqrt{y^2+1})^2=(2\sqrt{x^2+1}-y)^2$
$\Leftrightarrow 4x^2+y^2+1+4x\sqrt{y^2+1}=4(x^2+1)+y^2-4y\sqrt{x^2+1}$
$\Leftrightarrow 4(x\sqrt{y^2+1})+y\sqrt{x^2+1})=3$
$\Leftrightarrow 4Q=3$
$\Leftrightarrow Q=\frac{3}{4}$
\(x=3\ge2\Leftrightarrow y=3+1=4\\ x=-1< 2\Leftrightarrow y=\left(-1\right)^2-2=1-2=-1\\ x=2\ge2\Leftrightarrow y=2+1=3\)
P(x) = 9x4 + 5x3 + 8x2 - 15x3 - 4x2 - x4 + 15 - 7x2
= (9x4 - x4) + (5x3 - 15x3) + (8x2 - 4x2 - 7x2) + 15
= 8x4 - 10x3 - 3x2 + 15
Ta có: P(1) = 8. 14 - 10. 13 - 3. 12 + 15 = 8 - 10 - 3 + 15 = 10
P(0) = 8. 04 - 10. 03 - 3. 02 + 15 = 0 - 0 - 0 + 15 = 15
P(-1) = 8.(-1)4 - 10(-1)3 - 3(-1)2 + 15 = -8 - (-10) - (-3) + 15 = 20
\(A+B=-x^2-5yz+z^2+7yz-z^2+5x^2=4x^2+2yz\)
\(A-B=-x^2-5yz+z^2+z^2-7yz-5x^2=-6x^2-12yz+2z^2\)
\(\left(3x-2\right)^2-2\left(x-1\right)^2\)
\(=9x^2-18x+4-2\left(x^2-2x+1\right)\)
\(=9x^2-18x+4-2x^2+4x-2\)
\(=7x^2-14x+2\)
\(\left(3x-2\right)^2-2\left(x-1\right)^2=9x^2-12x+4-2x^2+4x-2=7x^2-8x+2\)