Tình x biết a) x2(x-2)-x-2=0
b) 5(x-11)-x+11=0
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Khách
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5 tháng 8 2021
a) \(x^2-5=\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)
b) \(x^2-11=\left(x-\sqrt{11}\right)\left(x+\sqrt{11}\right)\)
c: \(x-2=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)
d: \(x^2-2\sqrt{5x}+5=\left(x-\sqrt{5}\right)^2\)
CM
15 tháng 11 2018
a ) x 2 – 5 = 0 ⇔ x 2 = 5 ⇔ x 1 = √ 5 ; x 2 = - √ 5
Vậy phương trình có hai nghiệm x 1 = √ 5 ; x 2 = - √ 5
Cách khác:
x 2 – 5 = 0 ⇔ x 2 – ( √ 5 ) 2 = 0
⇔ (x - √5)(x + √5) = 0
hoặc x - √5 = 0 ⇔ x = √5
hoặc x + √5 = 0 ⇔ x = -√5
b)
x 2 – 2 √ 11 x + 11 = 0 ⇔ x 2 – 2 √ 11 x + ( √ 11 ) 2 = 0 ⇔ ( x - √ 11 ) 2 = 0
⇔ x - √11 = 0 ⇔ x = √11
Vậy phương trình có một nghiệm là x = √11
16 tháng 12 2017
1 a ) \(\left|x-11\right|+11-x=0\)
\(\Leftrightarrow\left|x-11\right|=x-11\)
\(\Leftrightarrow\orbr{\begin{cases}x-11=x-11\\x-11=11-x\end{cases}\Leftrightarrow\orbr{\begin{cases}\forall x\\x=11\end{cases}}}\)
p./s tham khảo nha
6 tháng 3 2022
\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)
\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)
29 tháng 11 2015
a, (x+1) + (x+3) + (x+5) +.......+(x+99) =0
\(\Rightarrow\)x+x+x+...+x(50 số hạng) + 1+3+5+...+99=0
\(\Rightarrow\)50.x+4950=0
50.x=0-4950
50.x=(-4950)
x=(-4950):50
x=(-99)
\(a,x^2\left(x-2\right)-\left(x-2\right)=0\)
\(\Rightarrow x^2\left(x-2\right)-\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x^2-1\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\\x=1\end{matrix}\right.\)
\(b,5\left(x-11\right)-x+11=0\)
\(\Rightarrow5\left(x-11\right)-\left(x-11\right)=0\)
\(\Rightarrow\left(x-11\right)\left(5-1\right)=0\)
\(\Rightarrow4\left(x-11\right)=0\)
\(\Rightarrow x-11=0\Rightarrow x=11\)
a)fix đề : \(x^2\left(x-2\right)+x-2=0\)
\(x^2\left(x-2\right)+x-2=0\\ \Leftrightarrow\left(x-2\right)\left(x^2-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\end{matrix}\right.\)
b)
\(5\left(x-11\right)-x+11=0\\ \Leftrightarrow\left(x-11\right)\left(5-1\right)=0\\ \Leftrightarrow x-11=0\Leftrightarrow x=11\)