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\(2.\left(x-4\right)-x+3=0\)
\(2x-8-x+3=0\)
\(x-5=0\)
\(x=5\)
\(x^2-25-x-5=0\)
\(\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\)
\(\left(x+5\right)\left(x-5-1\right)=0\)
\(\left(x+5\right)\left(x-6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=6\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=-5\\x=6\end{cases}}\)
a) x = 1; x = - 1 3 b) x = 2.
c) x = 3; x = -2. d) x = -3; x = 0; x = 2.
a: \(x\in\left\{0;25\right\}\)
c: \(x\in\left\{0;5\right\}\)
a. x mũ 2 - 2x + 1 = 25
= x^2 + 2.x.1 + 1^2
= ( x + 1 ) ^2
ko bt có đúng ko nữa, mấy câu kia tui ko bt lm
b: \(\Leftrightarrow\left(x-5\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-6\end{matrix}\right.\)
\(9,\left(2x-5\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(2x-5-x-1\right)\left(2x-5+x+1\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\3x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{4}{3}\end{matrix}\right.\)
Vậy \(S=\left\{6;\dfrac{4}{3}\right\}\)
\(10,\left(x+3\right)^2-x^2=45\)
\(\Leftrightarrow x^2+6x+9-x^2-45=0\\ \Leftrightarrow6x=36\\ \Leftrightarrow x=6\)
Vậy \(S=\left\{6\right\}\)
\(11,\left(5x-4\right)^2-49x^2=0\\ \Leftrightarrow\left(5x-4\right)^2-\left(7x\right)^2=0\\ \Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\\ \Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x-4=0\\12x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(S=\left\{-2;\dfrac{1}{3}\right\}\)
\(12,16\left(x-1\right)^2-25=0\\ \Leftrightarrow4^2\left(x-1\right)^2-5^2=0\\ \Leftrightarrow\left[4\left(x-1\right)\right]^2-5^2=0\\ \Leftrightarrow\left(4x-4\right)^2-5^2=0\\ \Leftrightarrow\left(4x-4-5\right)\left(4x-4+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-9=0\\4x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{1}{4};\dfrac{9}{4}\right\}\)
\(\left(x^2-25\right)^2-\left(x+5\right)^2=0\)
\(\Leftrightarrow\left[x^2-5^2\right]^2-\left(x+5\right)^2=0\)
\(\Leftrightarrow\left[\left(x+5\right)\left(x-5\right)\right]^2-\left(x+5\right)^2=0\)
\(\Leftrightarrow\left(x+5\right)^2\left(x-5\right)^2-\left(x+5\right)^2=0\)
\(\Leftrightarrow\left(x+5\right)^2\left[\left(x-5\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x+5\right)^2\left[\left(x-5\right)+1\right]\left[\left(x-5\right)-1\right]=0\)
\(\Leftrightarrow\left(x+5\right)^2\left(x-4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+5\right)^2=0\\x-4=0\\x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x=4\\x=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=4\\x=6\end{matrix}\right.\)
Vậy: \(S=\left\{-5;6;4\right\}\)
Ta có ( x2 - 25 )2 - ( x + 5 )2 = 0
Vì ( x2 - 25 )2 ≥ 0 ; ( x + 5 )2 ≥ 0
⇒ ( x2 - 25 )2 - ( x + 5 )2 ≥ 0
Dấu " = " xảy ra khi
\(\left[{}\begin{matrix}\left(x^2-25\right)^2=0\\\left(x+5\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\pm5\\x=-5\end{matrix}\right.\Rightarrow x=-5\)
Vậy x = 5