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a) \(7x^2-16x=2x^3-56\)
\(\Leftrightarrow\)\(2x^3-7x^2+16x-56=0\)
\(\Leftrightarrow\)\(2x\left(x^2+8\right)-7\left(x^2+8\right)=0\)
\(\Leftrightarrow\)\(\left(2x-7\right)\left(x^2+8\right)=0\)
\(\Leftrightarrow\)\(2x-7=0\)
\(\Leftrightarrow\)\(x=3,5\)
Vậy...
b) \(x^7+x^3+2x^5+2x=0\)
\(\Leftrightarrow\)\(x.\left(x^6+x^2+2x^4+2\right)=0\)
\(\Leftrightarrow\)\(x\left(x^2+2\right)\left(x^4+1\right)=0\)
\(\Leftrightarrow\)\(x=0\)
Vậy...
c) \(\left(2x+1\right)x-5\left(x+\frac{1}{2}\right)=0\)
\(\Leftrightarrow\)\(2x\left(x+\frac{1}{2}\right)-5\left(x+\frac{1}{2}\right)=0\)
\(\Leftrightarrow\)\(\left(2x-5\right)\left(x+\frac{1}{2}\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}2x-5=0\\x+\frac{1}{2}=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2,5\\x=-0,5\end{cases}}\)
Vậy...
a: \(\Leftrightarrow2x^3-56-7x^2+16x=0\)
\(\Leftrightarrow2x\left(x^2+8\right)-7\left(x^2+8\right)=0\)
=>2x-7=0
hay x=7/2
b: \(\Leftrightarrow x^5\left(x^2+2\right)+x\left(x^2+2\right)=0\)
=>x(x2+2)(x4+1)=0
=>x=0
c: \(\Leftrightarrow2x^2+x-5x-\dfrac{5}{2}=0\)
\(\Leftrightarrow2x^2-4x-\dfrac{5}{2}=0\)
hay \(x\in\left\{\dfrac{5}{2};-\dfrac{1}{2}\right\}\)
a) x2 - 5x - 6 = 0
=> x2 - 2x - 3x - 6 = 0
=> (x2 - 2x) + (-3x - 6) = 0
=> x(x - 2) - 3 (x - 2) = 0
=> (x - 2) (x - 3) = 0
=> x - 2 = 0 => x = 2
x - 3 = 0 => x = 3
còn lại tương tự nhé!! 46566578768698945635655675656788787868789789879789098089364556546
\(2x^2=x\)
\(\Rightarrow2x^2-x=0\)
\(x\left(2x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)
Vậy \(x=0\)hoặc \(x=\frac{1}{2}\)
\(x^3=x^5\)
\(\Rightarrow x^5-x^3=0\)
\(x^3.\left(x^2-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^3=0\\x^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
Vậy \(x=0\)hoặc \(x=1\)
\(x^2.\left(x+1\right)+2x\left(x+1\right)=0\)
\(\left(x+1\right)\left(x^2+2x\right)=0\)
\(x.\left(x+1\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\)hoặc \(x+2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)hoặc \(x=-2\)
Vậy \(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\) hoặc \(x=-2\)
\(x.\left(2x-3\right)-2\left(3-2x\right)=0\)
\(x.\left(2x-3\right)+2.\left(2x-3\right)=0\)
\(\left(2x-3\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=0\\x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-2\end{cases}}}\)
Vậy \(x=\frac{3}{2}\)hoặc \(x=-2\)
\(2x^2-x=0\Leftrightarrow x\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}\)
\(S\left\{0;\frac{1}{2}\right\}\)
\(d)x^3-x^5=0\Leftrightarrow x^3\left(1-x^2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^3=0\\1-x^2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{1}\end{cases}}\)
\(S=\left\{0;\pm\sqrt{1}\right\}\)
các câu sau tương tự nha bn
Chọn C