6x+2^3=2x-12
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d: \(\dfrac{x^4-2x^3+2x-1}{x^2-1}\)
\(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}\)
\(=x^2-2x+1\)
\(=\left(x-1\right)^2\)
\(=\left(6x^3+9x^2-16x^2-24x+8x+12\right):\left(2x+3\right)\\ =\left(2x+3\right)\left(3x^2-8x+4\right):\left(2x+3\right)\\ =3x^2-8x+4\)
\(\left(6x^3-7x^2-16x+12\right):\left(2x+3\right)\\ =\left[\left(6x^3+9x^2\right)-\left(16x^2+24x\right)+\left(8x+12\right)\right]:\left(2x+3\right)\\ =\left[3x^2\left(2x+3\right)-8x\left(2x+3\right)+4\left(2x+3\right)\right]:\left(2x+3\right)\\ =\left[\left(2x+3\right)\left(3x^2-8x+4\right)\right]:\left(2x+3\right)\\ =3x^2-8x+4\)
b) \(x^3-6x^2+11x-12=0\)
\(\Leftrightarrow\)\(x^3-4x^2-2x^2+8x+3x-12=0\)
\(\Leftrightarrow\)\(x^2\left(x-4\right)-2x\left(x-4\right)+3\left(x-4\right)\)
\(\Leftrightarrow\)\(\left(x-4\right)\left(x^2-2x+3\right)=0\)
Ta có: \(x^2-2x+3\)
\(=\left(x-1\right)^2+2>0\)
\(\Rightarrow\)\(x-4=0\)
\(\Leftrightarrow\)\(x=4\)
Vậy...
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\(6x+2^3=2x-12\)
\(6x+8=2x-12\)
\(6x-2x=-12-8\)
\(4x=-20\)
\(\Rightarrow x=-5\)
học tốt
6x+2^3=2x-12
=> 6x + 8 = 2x - 12
=> 4x = -20
=> x = -5
vậy x = -5
\(\frac{6x^3-7x^2-16x+12}{2x+3}\)
\(=\frac{6x^3-12x^2+5x^2-10x-6x+12}{2x+3}\)
\(=\frac{6x^2\left(x-2\right)+5x\left(x-2\right)-6\left(x-2\right)}{2x+3}\)
\(=\frac{\left(x-2\right)\left(6x^2+5x-6\right)}{2x+3}\)
\(=\frac{\left(x-2\right)\left(6x^2+9x-4x-6\right)}{2x+3}\)
\(=\frac{\left(x-2\right)\left[3x\left(2x+3\right)-2\left(2x+3\right)\right]}{2x+3}\)
\(=\frac{\left(x-2\right)\left(2x+3\right)\left(3x-2\right)}{2x+3}\)
\(=\left(x-2\right)\left(3x-2\right)\)
a: ĐKXĐ: x>=-3/2
\(\sqrt{x^2+4}=\sqrt{2x+3}\)
=>\(x^2+4=2x+3\)
=>\(x^2-2x+1=0\)
=>\(\left(x-1\right)^2=0\)
=>x-1=0
=>x=1(nhận)
b: \(\sqrt{x^2-6x+9}=2x-1\)(ĐKXĐ: \(x\in R\))
=>\(\sqrt{\left(x-3\right)^2}=2x-1\)
=>\(\left\{{}\begin{matrix}\left(2x-1\right)^2=\left(x-3\right)^2\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(2x-1-x+3\right)\left(2x-1+x-3\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x+2\right)\left(3x-4\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>x=4/3(nhận) hoặc x=-2(loại)
c:
Sửa đề: \(\sqrt{4x+12}=\sqrt{9x+27}-5\)
ĐKXĐ: \(x>=-3\)
\(\sqrt{4x+12}=\sqrt{9x+27}-5\)
=>\(2\sqrt{x+3}=3\sqrt{x+3}-5\)
=>\(-\sqrt{x+3}=-5\)
=>x+3=25
=>x=22(nhận)
d: ĐKXĐ: \(\left[{}\begin{matrix}x< =\dfrac{3-\sqrt{5}}{4}\\x>=\dfrac{3+\sqrt{5}}{4}\end{matrix}\right.\)
\(\sqrt{4x^2-6x+1}=\left|2x-5\right|\)
=>\(\sqrt{\left(4x^2-6x+1\right)}=\sqrt{4x^2-20x+25}\)
=>\(4x^2-6x+1=4x^2-20x+25\)
=>\(-6x+20x=25-1\)
=>\(14x=24\)
=>x=12/7(nhận)
Mk gặp bài này khá nhiều rồi!
\(6x+2^3=2x-12\)
\(\Leftrightarrow6x+8=2x-12\)
\(\Leftrightarrow6x+8-2x=-12\)
\(\Leftrightarrow4x+8=-12\)
\(\Leftrightarrow4x=-20\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
Trả lời:
6x +2^3 = 2x -12
<=> 6x +8 =2x -12
<=> 6x -2x = -12 -8
<=> 4x = -20
<=> x =-5
#Học tốt:))