a) (x - 1)⁵ = -243

<=> (x - 1)⁵ = (-3)⁵

=> x - 1 = -3

=> x = -2

b) \(x-2\sqrt{x}=0\)

\(\sqrt{x^2}-2\sqrt{x}=0\)

\(\sqrt{x}.\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{x}=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

a, \(\left(x-1\right)^5=-243\)

\(\Leftrightarrow x-1=-3\Leftrightarrow x=-2\)

Vậy x = -2

b, \(x-2\sqrt{x}=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{2}\end{matrix}\right.\)

Vậy...

a, \(\left(x-1\right)^5=-243\)

\(\Leftrightarrow\left(x-1\right)^5=-3^5\)

\(\Leftrightarrow x-1=-3\Leftrightarrow x=-2\)

b,\(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)

\(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}-\dfrac{x+2}{14}-\dfrac{x+2}{15}=0\)

\(\Leftrightarrow\left(x+2\right).\left(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\right)=0\)

\(do\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\ne0\)

\(\Rightarrow x+2=0\Leftrightarrow x=-2\)

c, \(x-2\sqrt{x}=0\Leftrightarrow\sqrt{x^2}-2\sqrt{x}=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\\sqrt{x}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\sqrt{2}\end{matrix}\right.\)

\(x-2\sqrt{x}=0\)

\(\Leftrightarrow\left(\sqrt{x}\right)^2-2\sqrt{x}=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)

\(\Rightarrow\sqrt{x}=0\) và \(\sqrt{x}-2=0\)

\(\Rightarrow x=0\) và \(\sqrt{x}=2\)

\(\Rightarrow x=0\) và \(x=4\)

\(\sqrt{x}\left(\sqrt{x}-2\right)=0\)

\(x-2\sqrt{x}=0\)

\(\Leftrightarrow\left(\sqrt{x}\right)^2-2\sqrt{x}=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)

\(\Rightarrow\sqrt{x}=0\)và \(\sqrt{x}-2=0\)

\(\Rightarrow x=0\)và \(\sqrt{x}=2\)

\(\Rightarrow x=0\)và \(x=4\)

\(x=\sqrt{31-8\sqrt{15}}=\sqrt{\left(4-\sqrt{15}\right)^2}=4-\sqrt{15}\)

Biểu thức nghịch đảo của x là \(\dfrac{1}{4-\sqrt{15}}=4+\sqrt{15}\)

\(\Rightarrow x=4\pm\sqrt{15}\) là nghiệm PT \(x^2+bx+c\left(1\right)\)

Đặt \(\left\{{}\begin{matrix}S=x_1+x_2\\P=x_1x_2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}S=8\\P=1\end{matrix}\right.\) (x₁ và x₂ là nghiệm của (1))

Áp dụng Viet đảo thì x là nghiệm của PT \(x^2-8x+1\)

Vậy \(b=-8;c=1\)

Bạn có thể làm thế này:

\(x=4-\sqrt{15}\)

Giả sử \(x=4-\sqrt{15}\) là nghiệm của PT \(x^2+bx+c=0\)

\(\Leftrightarrow\left(4-\sqrt{15}\right)^2+b\left(4-\sqrt{15}\right)+c=0\\ \Leftrightarrow31-8\sqrt{15}+4b-b\sqrt{15}+c=0\\ \Leftrightarrow\sqrt{15}\left(b+8\right)=-\left(4b+c+31\right)\)

Vì b,c hữu tỉ nên \(\sqrt{15}\left(b+8\right)\) hữu tỉ

\(\Leftrightarrow\left\{{}\begin{matrix}b+8=0\\4b+c+31=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-8\\c=1\end{matrix}\right.\)

ĐK : \(x\ge0\)

\(x-2\sqrt{x}=0\Rightarrow x=2\sqrt{x}\)

Bình phương hai vế ta có :

\(x^2=4x\Leftrightarrow x^2-4x=0\)

\(\Rightarrow x(x-4)=0\Rightarrow\hept{\begin{cases}x=0\\x=4\end{cases}}\)

\(x-2\sqrt{x}=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)

Vậy x = 0 hoặc x = 4

\(x-2\sqrt{x}=0\)

<=> \(\sqrt{x}.\sqrt{x}-2\sqrt{x}=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)

Đk:\(x\ge0\)

\(x-2\sqrt{x}=0\Leftrightarrow x=2\sqrt{x}\)

Bình phương 2 vế ta có:

\(x^2=4x\Leftrightarrow x^2-4x=0\)

\(\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)

x - 2\(\sqrt{x}\) = 0

<=> \(\sqrt{x}\)(\(\sqrt{x}\)- 2) = 0

<=> x = 0 hoặc x = 4

\(x-2\sqrt{x}=0\)

\(\Leftrightarrow\sqrt{x}^2-2\sqrt{x}=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\\sqrt{x}=2\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=4\end{cases}}}\)

\(x-2\sqrt{x}=0\)\(\Leftrightarrow x=2\sqrt{x}\)

\(\Leftrightarrow x^2=\left(2\sqrt{x}\right)^2\)\(\Leftrightarrow x^2=4x\)

\(\Leftrightarrow x^2-4x=0\)\(\Leftrightarrow x\left(x-4\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)( thoả mãn điều kiện )

Vậy \(x=0\)hoặc \(x=4\)