(x-2)^2-(x+3)^2 = 2(3x-1)
giúp mik với :>
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f(x)=9x3-1/3x+3x2-3x+1/3x2-1/9x3-3x2-9x+27+3x
= 9x3-1/9x3+3x2+1/3x2-3x2-1/3-3x-9x+3x+27
= 80/9x3+1/3x2-28/3x+27
\(\dfrac{x^3+8}{x^2+2x+1}.\dfrac{x^2+3x+2}{1-x^2}\left(x\ne\pm1\right)\\ =\dfrac{x^3+2^3}{\left(x+1\right)^2}.\dfrac{\left(x^2+x\right)+\left(2x+2\right)}{1^2-x^2}\\ =\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x+1\right)^2}.\dfrac{x\left(x+1\right)+2\left(x+1\right)}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x+1\right)^2}.\dfrac{\left(x+2\right)\left(x+1\right)}{\left(1-x\right)\left(x+1\right)}\\ =\dfrac{\left(x+2\right)^2\left(x^2-2x+4\right)}{\left(1-x\right)\left(x+1\right)^2}\)
a: \(\left(x-1\right)^3+27\)
\(=\left(x-1+3\right)\left(x^2-2x+1+3x-3+3\right)\)
\(=\left(x+2\right)\left(x^2+x+1\right)\)
b: \(\left(x-2\right)^3-8\)
\(=\left(x-2-2\right)\left(x^2-4x+4+2x-4+4\right)\)
\(=\left(x-4\right)\left(x^2-2x+4\right)\)
\(\frac{3}{x-1}+\frac{4}{x+1}=3x+\frac{2}{1-x^2}\)
<=>\(\frac{3\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{4\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{3x\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{-2}{\left(x-1\right)\left(x+1\right)}\)
=> \(3x+3+4x-4=3x\left(x^2-1\right)-2\)
<=> \(7x-1=3x^3-3x-2\)
<=> \(7x+3x-3x^3-1+2=0\)
<=> \(-3x^3+10x+1=0\)
<=> \(x=\frac{\sqrt{3}}{3};-\frac{\sqrt{3}}{3}\)
Ta có: \(\left(3x+2\right)^3-3x\left(3x+4\right)^2-17x\left(x-3\right)=-54\)
\(\Leftrightarrow27x^3+54x^2+36x+8-3x\left(9x^2+24x+16\right)-17x^2+51x=-54\)
\(\Leftrightarrow27x^3+37x^2+87x+8+54-27x^3-72x^2-48x=0\)
\(\Leftrightarrow-35x^2+39x+62=0\)
\(\Delta=39^2-4\cdot\left(-35\right)\cdot62=10201\)
Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-39-101}{-70}=\dfrac{-140}{-70}=2\\x_2=\dfrac{-39+101}{-70}=\dfrac{-62}{70}=\dfrac{-31}{35}\end{matrix}\right.\)
Giải phương trình chứa ẩn ở mẫu:a) 4x 2/3x-6-x/2-x=1 3x/2x-4b) x-3/x 3-x 3/x-3=3/x2-9Các bạn hãy giúp mik với:))
\(\left(x-2\right)^2-\left(x-3\right)^2=2\left(3x-1\right)\)
\(\Leftrightarrow\left(x-2-x+3\right)\left(x-2+x-3\right)=2\left(3x-1\right)\)
\(\Leftrightarrow\left(2x-5\right)=2\left(3x-1\right)\)
\(\Leftrightarrow2x-5=6x-2\)
\(\Leftrightarrow2x-6x=5-2\)
\(\Leftrightarrow-4x=3\)
\(\Leftrightarrow x=-\frac{3}{4}\)
Vậy phương trình trên có nghiệm là: \(S=\left\{-\frac{3}{4}\right\}\)
#hoktot<3#