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\(\left(x-1\right)3+3x\left(x-1\right)=0\)
<=> \(3\left(x-1\right)\left(x+1\right)=0\)
<=> \(\orbr{\begin{cases}x-1=0\\x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
Vậy...
\(1,x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
\(2,\left(x+2\right)\left(x-3\right)-x-2=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+2=0\\x-4=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-2\\x=4\end{cases}}\)
\(3,36x^2-49=0\)
\(\Leftrightarrow\left(6x\right)^2-7^2=0\)
\(\Leftrightarrow\left(6x-7\right)\left(6x+7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}6x-7=0\\6x+7=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{7}{6}\\x=\frac{7}{6}\end{cases}}\)
Chúc bn học giỏi nhoa!!!
Ta có : x2 - x = 0
=> x(x - 1) = 0
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
\(x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
\(x^5-9x=0\)
\(\Leftrightarrow x\left(x^4-9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^4-9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt[4]{9}\end{cases}}\)
\(\left(9^{30}-27^{19}\right):3^{57}+\left(125^9-25^{12}\right):5^{24}\)
\(=\left(3^{60}-3^{57}\right):3^{57}+\left(5^{27}-5^{24}\right):5^{24}\)
\(=3^{57}\left(3^3-1\right):3^{57}+5^{24}\left(5^3-1\right):5^{24}\)
\(=3^3-1+5^3-1\)
\(=27-1+125-1\)
\(=150\)
2 )
\(x^2-25-\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-5\right)-\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-5-1\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)
Vậy ...
b )
\(\left(2x-1\right)^2-\left(4x^2-1\right)=0\)
\(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)
\(\Leftrightarrow2-4x=0\)
\(\Leftrightarrow4x=2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy ...
c )
\(x^2\left(x^2+4\right)-x^2-4=0\)
\(\Leftrightarrow x^2\left(x^2+4\right)-\left(4+x^2\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=0\\x^2+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=1\\x^2=-4\left(L\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy ...
\(\left(36x^2-25\right)-\left(6x+5\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(6x+5\right)\left(6x-5\right)-\left(6x+5\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(6x+5\right)\left(6x-5-x-1\right)=0\)
\(\Leftrightarrow\left(6x+5\right)\left(5x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-5}{6}\\x=\frac{6}{5}\end{cases}}\)
\(\left(36x^2-25\right)-\left(6x+5\right)\left(x+1\right)=0\Leftrightarrow\left(6x-5\right)\left(6x+5\right)-\left(6x+5\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(6x+5\right)\left(5x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{-5}{6}\\x=\frac{6}{5}\end{cases}}\)
a,(x^2+9)^2-36x^2=0
(x^2+9-6x)(x^2+9+6x=0
=>x^2+9-6x=0 hoac x^2+9+6x=0
+,x^2+9-6x=0
x^2-6x+9=0
( x-3)^2 =0
=>x-3 =0
x =3
+,x^2+9+6x=0
x^2+6x+9=0
(x+3)^2 =0
=>x+3 =0
x =-3
Ý dưới cũng tương tự..
Dấu trừ ở trước thì bạn phải đổi dấu trog ngoặc