8x(3x-5)-12x+20=0
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a) \(\sqrt{8x^3}\cdot2x\)
\(=\sqrt{8x^3\cdot2x}\)
\(=\sqrt{16x^4}\)
\(=\sqrt{\left(4x^2\right)^2}\)
\(=4x^2\)
b) \(\sqrt{12x^5}\cdot\sqrt{3x}\)
\(=\sqrt{12x^5\cdot3x}\)
\(=\sqrt{36x^6}\)
\(=\sqrt{\left(6x^3\right)^2}\)
\(=\left|6x^3\right|\)
\(=6x^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2: ĐKXĐ: x>=0
\(\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\cdot\sqrt{27x}=-4\)
=>\(\sqrt{3x}-2\cdot2\sqrt{3x}+\dfrac{1}{3}\cdot3\sqrt{3x}=-4\)
=>\(\sqrt{3x}-4\sqrt{3x}+\sqrt{3x}=-4\)
=>\(-2\sqrt{3x}=-4\)
=>\(\sqrt{3x}=2\)
=>3x=4
=>\(x=\dfrac{4}{3}\left(nhận\right)\)
3:
ĐKXĐ: x>=0
\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)
=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-20-3\sqrt{2}=0\)
=>\(13\sqrt{2x}=20+3\sqrt{2}\)
=>\(\sqrt{2x}=\dfrac{20+3\sqrt{2}}{13}\)
=>\(2x=\dfrac{418+120\sqrt{2}}{169}\)
=>\(x=\dfrac{209+60\sqrt{2}}{169}\left(nhận\right)\)
4: ĐKXĐ: x>=-1
\(\sqrt{16x+16}-\sqrt{9x+9}=1\)
=>\(4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>\(\sqrt{x+1}=1\)
=>x+1=1
=>x=0(nhận)
5: ĐKXĐ: x<=1/3
\(\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)
=>\(2\sqrt{1-3x}+3\sqrt{1-3x}=10\)
=>\(5\sqrt{1-3x}=10\)
=>\(\sqrt{1-3x}=2\)
=>1-3x=4
=>3x=1-4=-3
=>x=-3/3=-1(nhận)
6: ĐKXĐ: x>=3
\(\dfrac{2}{3}\sqrt{x-3}+\dfrac{1}{6}\sqrt{x-3}-\sqrt{x-3}=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}\cdot\left(\dfrac{2}{3}+\dfrac{1}{6}-1\right)=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}\cdot\dfrac{-1}{6}=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}=\dfrac{2}{3}:\dfrac{1}{6}=\dfrac{2}{3}\cdot6=\dfrac{12}{3}=4\)
=>x-3=16
=>x=19(nhận)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ 3x(12x-5)-6x(6x-5)=0
<=>36x2-15x-36x2+30x=0
<=>15x=0
<=>x=0
b/ x2-8x+6=0
Nghiệm lẻ xem lại câu b
a) \(3x\left(12x-5\right)-6x\left(6x-5\right)=0\)
\(\Rightarrow36x^2-15x-36x^2+30x=0\)
\(\Rightarrow15x=0\)
\(\Rightarrow x=0\)
b) \(x^2-8x+6=0\)
\(\Rightarrow x^2-8x+16-10=0\)
\(\Rightarrow x^2-8x+4^2=10\)
\(\Rightarrow\left(x-4\right)^2=10\)
\(\Rightarrow x-4=\sqrt{10}\)
\(\Rightarrow x=4+\sqrt{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 4 :
\(\left(5x-20\right)+\left(3x^2-12x\right)=0\)
\(\Leftrightarrow5\left(x-4\right)+3x\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(5+3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5+3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{5}{3}\end{matrix}\right.\)
Vậy \(x=4\) hoặc \(x=-\dfrac{5}{3}\)
Bài 5 :
\(\left(1-x\right)-3x^2+3x=0\)
\(\Leftrightarrow\left(1-x\right)-\left(3x^2-3x\right)=0\)
\(\Leftrightarrow\left(1-x\right)-3x\left(x-1\right)=0\)
\(\Leftrightarrow\left(1-x\right)+3x\left(1-x\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(1+3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}1-x=0\\1+3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x=1\) hoặc \(x=-\dfrac{1}{3}\)
Bài 6 :
\(\left(4x+20\right)-\left(x+5\right)^2=0\)
\(\Leftrightarrow4\left(x+5\right)-\left(x+5\right)^2=0\)
\(\Leftrightarrow\left(x+5\right)\left(4-x-5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(-x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\-x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=-1\end{matrix}\right.\)
Vậy \(x=-5\) hoặc \(x=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(1.6x\left(x-10\right)-2x+20=0\)
⇔\(6x\left(x-10\right)-2\left(x-10\right)=0\)
⇔ \(2\left(x-10\right)\left(3x-1\right)=0\)
⇔ x = 10 hoặc x = \(\dfrac{1}{3}\)
KL....
\(2.3x^2\left(x-3\right)+3\left(3-x\right)=0\)
⇔ \(3\left(x-3\right)\left(x^2-1\right)=0\)
⇔ \(x=+-1\) hoặc \(x=3\)
KL....
\(3.x^2-8x+16=2\left(x-4\right)\)
⇔ \(\left(x-4\right)^2-2\left(x-4\right)=0\)
⇔ \(\left(x-4\right)\left(x-6\right)=0\)
⇔ \(x=4\) hoặc \(x=6\)
KL.....
\(4.x^2-16+7x\left(x+4\right)=0\)
\(\text{⇔}4\left(x+4\right)\left(2x-1\right)=0\)
⇔ \(x=-4hoacx=\dfrac{1}{2}\)
KL.....
\(5.x^2-13x-14=0\)
⇔ \(x^2+x-14x-14=0\)
\(\text{⇔}\left(x+1\right)\left(x-14\right)=0\)
\(\text{⇔}x=14hoacx=-1\)
KL......
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