1. a. x-1/x+2-1/x= 2/x^2 + 2x
b. 5+x/2 -3 > 1-2x/3
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a) \(\left(2x-5\right)^2-\left(2x+3\right)\left(2x-3\right)=10\Leftrightarrow\left(4x^2-20x+25\right)-\left(4x^2-9\right)-10=0\)
\(\Leftrightarrow-20x+24=0\Leftrightarrow x=\frac{6}{5}\)
b) \(\left(4x-1\right)\left(x+2\right)-\left(2x+3\right)^2-5\left(x-1\right)=9\Leftrightarrow-10x-15=0\)
\(\Leftrightarrow x=\frac{-3}{2}\)
c) \(\left(x+1\right)^3-\left(x-1\right)^3-2=6\Leftrightarrow\left(x^3+3x^2+3x+1\right)-\left(x^3-3x^2+3x-1\right)-8=0\)
\(\Leftrightarrow6x^2-6=0\Leftrightarrow x=\pm1\)
d) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x+1\right)\left(x^2-x+1\right)-3\left(-x-2\right)=5\)
\(\Leftrightarrow\left(x^3+8\right)-\left(x^3+1\right)+3x+6=5\Leftrightarrow3x+8=0\Leftrightarrow x=\frac{-8}{3}\)
a) \(\left(x+3\right)^2-\left(x-2\right)^3=\left(x+5\right)\left(x^2-5x+25\right)-108\)
\(\Leftrightarrow x^2+6x+9-x^2+4x-4=x^3-5x^2+25x+5x^2-25x+125-108\)
\(\Leftrightarrow x^3-10x+12=0\Leftrightarrow\left(x-2\right)\left(x^2+2x+6\right)=0\)
\(\Leftrightarrow x=2\)( do \(x^2+2x+6=\left(x+1\right)^2+4\ge4>0\))
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
f/ \(3xy\left(x+y\right)-\left(x+y\right)\left(x^2+y^2+2xy\right)+y^3=27\)
\(3x^2y+3xy^2-\left(x+y\right)\left(x+y\right)^2+y^3=27\)
\(3x^2y+3xy^3-\left(x+y\right)^3+y^3=27\)
\(3x^2y+3xy^3-\left(x^3+3x^2y+3xy^2+b^3\right)+y^3=27\)
\(-x^3=27\)
\(x=-3\)
a:=>x^2-1-x=2x-1
=>x^2-x-1=2x-1
=>x^2-3x=0
=>x=0(loại) hoặc x=3(nhận)
b:=>x+2=0 hoặc 5-3x=0
=>x=-2 hoặc x=5/3
c:=>20(1-2x)+6x=9(x-5)-24
=>20-40x+6x=9x-45-24
=>-34x+20=9x-69
=>-43x=-89
=>x=89/43
d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3
=>2x^2+4x-19=-2x+7
=>2x^2+6x-26=0
=>x^2+3x-13=0
=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)
e: =>(2x-3)(2x-3-x-1)=0
=>(2x-3)(x-4)=0
=>x=4 hoặc x=3/2
\(\left(x-3\right)\left(x-1\right)-3\left(x-3\right)\)
\(=\left(x-3\right)\left(x-1-3\right)\)
\(=\left(x-3\right)\left(x-4\right)\)
\(\left(x-1\right)\left(2x+1\right)+3\left(x-1\right)\left(x+2\right)\left(2x+1\right)\)
\(=\left(x-1\right)\left(2x+1\right)\left(1+3x+6\right)\)
\(=\left(x-1\right)\left(2x+1\right)\left(3x+7\right)\)
a) \(A=\left(5-x\right)\left(x+5\right)-2\left(x-1\right)\left(x-3\right)-3\left(x-2\right)^2\)
\(=\left(5-x\right)\left(5+x\right)-\left(2x-2\right)\left(x-3\right)-3\left(x^2-2.2x+2^2\right)\)
\(=\left(5^2-x^2\right)-\left[2x\left(x-3\right)-2\left(x-3\right)\right]-3\left(x^2-4x+4\right)\)
\(=25-x^2-\left[\left(2x^2-6x\right)-\left(2x-6\right)\right]-3x^2+12x-12\)
\(=25-x^2-\left(2x^2-6x-2x+6\right)-3x^2+12x-12\)
\(=25-x^2-2x^2+6x+2x-6-3x^2+12x-12\)
\(=7+20x-6x^2\)
b/ \(B=\left(3-2x\right)\left(x-2\right)+\left(2x-5\right)^2-\left(x-4\right)\)
\(=3\left(x-2\right)-2x\left(x-2\right)+\left[\left(2x\right)^2-2.2x.5+5^2\right]-x+4\)
\(=3x-6-2x^2+4x+4x^2-20x+25-x+4\)
\(=23-14x+2x^2\)
c/ \(C=\left(x-4\right)\left(x-2\right)-3\left(x-2\right)\left(3-2x\right)-\left(2x+1\right)^2\)
\(=x\left(x-2\right)-4\left(x-2\right)-\left(3x-6\right)\left(3-2x\right)-\left[\left(2x\right)^2+2.2x.1+1^2\right]\)
\(=x^2-2x-4x+8-\left[3x\left(3-2x\right)-6\left(3-2x\right)\right]-4x^2-4x-1\)
\(=x^2-2x-4x+8-\left(9x-6x^2-18+12x\right)-4x^2-4x-1\)
\(=x^2-2x-4x+8-9x+6x^2+18-12x-4x^2-4x-1\)
\(=25-31x+3x^2\)
d/ \(D=2\left(x-1\right)^2-3\left(x-1\right)\left(x+2\right)-\left(2x+1\right)^2\)
\(=2.\left(x^2-2x+1\right)-\left(3x-3\right)\left(x+2\right)-\left[\left(2x\right)^2+2.2x+1\right]\)
\(=2x^2-4x+2-\left[3x\left(x+2\right)-3\left(x+2\right)\right]-\left(4x^2+4x+1\right)\)
\(=2x^2-4x+2-\left(3x^2+6x-3x-6\right)-\left(4x^2+4x+1\right)\)
\(=2x^2-4x+2-3x^2-6x+3x+6-4x^2-4x-1\)
\(=7-11x-5x^2\)
P/s: Ko chắc ạ!
1, đk x khác -2 ; 0
\(x^2-x-x-2=2\Leftrightarrow x^2-2x-4=0\Leftrightarrow\left(x-1\right)^2-5=0\Leftrightarrow x=\pm\sqrt{5}+1\)
2, \(\dfrac{5+x-6}{2}-\dfrac{1-2x}{3}>0\Leftrightarrow\dfrac{3x-3-2+4x}{6}>0\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{5}{7}\\x\ne\dfrac{5}{7}\end{matrix}\right.\)