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Bài làm:
a) \(x^2-7=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)
b) \(4x^2-5=\left(2x-\sqrt{5}\right)\left(2x+\sqrt{5}\right)\)
c) \(3x^2-1=\left(x\sqrt{3}-1\right)\left(x\sqrt{3}+1\right)\)
d) \(x-1=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
e) \(x-4=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)
f) \(9x-4=\left(3\sqrt{x}-2\right)\left(3\sqrt{x}+2\right)\)
a. \(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)\left(x+1\right)\left(2x-9\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\2x+5=0\\x+1=0\\2x-9=0\end{matrix}\right.\) \(\Rightarrow x=\)
b. \(\Leftrightarrow x^3+x+3x^2+3=0\)
\(\Leftrightarrow x\left(x^2+1\right)+3\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+1=0\left(vn\right)\end{matrix}\right.\)
c. \(\Leftrightarrow2x\left(3x-1\right)^2-\left(9x^2-1\right)=0\)
\(\Leftrightarrow\left(6x^2-2x\right)\left(3x-1\right)-\left(3x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(6x^2-5x-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-1\right)\left(6x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-1=0\\6x+1=0\end{matrix}\right.\)
d.
\(\Leftrightarrow x^3-3x^2+2x-3x^2+9x-6=0\)
\(\Leftrightarrow x\left(x^2-3x+2\right)-3\left(x^2-3x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2-3x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-1=0\\x-2=0\end{matrix}\right.\)
e.
\(\Leftrightarrow x^3+2x^2+x+3x^2+6x+3=0\)
\(\Leftrightarrow x\left(x^2+2x+1\right)+3\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+1=0\end{matrix}\right.\)
sử dụng dấu căn trong thanh công cụ này để soạn thảo câu hỏi rõ ràng nha
a, \(x^2-49x-50=0\Leftrightarrow\left(x-1\right)\left(x+50\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-50\end{cases}}\)
b, \(3x^2-7x-10=0\Leftrightarrow3x\left(x+1\right)-10\left(x+1\right)=0\Leftrightarrow\left(3x-10\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-10=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=10\\x=-1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{10}{3}\\x=-1\end{cases}}}\)
c, \(x^2-4x-5=0\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)
d, \(x^2+2x-3=0\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)
e, \(x^2+2020x-2021=0\)
=> vô nghiệm
f, \(x^2+9x-10=0\Leftrightarrow\left(x-1\right)\left(x+10\right)\Leftrightarrow\orbr{\begin{cases}x=1\\x=-10\end{cases}}\)
g, \(-5x^2+4x+1=0\Leftrightarrow5x^2+x-5x-1=0\Leftrightarrow x\left(5x+1\right)-1\left(5x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{5}\end{cases}}\)
h, \(4x^2+3x-7=0\Leftrightarrow x\left(4x+7\right)-1\left(4x+7\right)=0\Leftrightarrow\left(x-1\right)\left(4x+7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{7}{4}\end{cases}}\)
a) (x-50)(x+1)=0
<=>x=50 hoặc x=1
b) (x+1)(x-10/3)=0
<=>x=-1 hoặc x=10/3
c) (x-5)(x+1)=0
<=>x=5 hoặc x=-1
d) (x+3)(x-1)=0
<=>x=-3 hoặc x=1
e) (x-1)(x+2021)=0
<=>x=1 hoặc x=-2021
f) (x-1)(x+10)=0
<=> x=1 hoặc x=-10
g) (x+1/5)(x-1)=0
<=>x=1 hoặc x=-1/5
h) (x-1)(x+7/4)=0
<=> x=1 hoặc x=-7/4
Học tốt. tk vs ạ
a) \(2\sqrt{x^2}=2.\left|x\right|=-2x\)(vì x<0)
b) \(\frac{1}{2}\sqrt{x^{10}}=\frac{1}{2}\sqrt{\left(x^5\right)^2}\frac{1}{2}\left|x^5\right|=-\frac{1}{2}x^5\)(vì x>0)
c) \(x-4+\sqrt{x^2-8x+16}=x-4+\sqrt{\left(x-4\right)^2}=x-4+\left|x-4\right|=x-4+4-x=0\)(vì x<4 nên x-4<0)
d) \(\frac{3-\sqrt{x}}{x-9}=\frac{-\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{-1}{\sqrt{x}+3}\)
a, \(x^2=\frac{1}{9}\)
=> \(x=\pm\frac{1}{3}\)
b, \(x^2=\frac{1}{3}\)
=> \(x=\pm\frac{1}{\sqrt{3}}\)
c, \(x^2=\frac{2}{7}\)
=> \(x=\pm\sqrt{\frac{2}{7}}\)
d, Vô nghiệm vì \(x^2+2019\ge2019>0\forall x\)
e, \(x=\pm\sqrt{3}\)
g, Vô nghiệm vì -2 < 0
h, \(x=0\)
\(a\text{)}\:36x^2-5=\left(6x\right)^2-\left(\sqrt{5}\right)^2\\ =\left(6x-\sqrt{5}\right)\left(6x+\sqrt{5}\right)\)
\(b\text{)}\:25-3x^2=5^2-\left(\sqrt{3}x\right)^2\\ =\left(5-\sqrt{3}x\right)\left(5+\sqrt{3}\right)\)
\(c\text{)}\:x-4=\left(\sqrt{x}\right)^2-2^2\\ =\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)
\(d\text{)}\:11+9x=9.\dfrac{11}{9}+9x\\ =9\left(\dfrac{11}{9}+x\right)\)
\(e\text{)}\:31+7x=7.\dfrac{31}{7}+7x\\ =7\left(\dfrac{31}{7}+x\right)\)