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x2+4x-5=0
<=> x2-5x+x-5=0
<=> x(x-5)+(x-5)=0
<=> (x-5)(x+1)=0
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}}\)
\(12x^3=3x\)
\(\Leftrightarrow12x^3-3x=0\)
\(\Leftrightarrow3x\left(4x^2-1\right)=0\)
\(\Leftrightarrow3x\left(2x+1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow x\in\left\{0;\pm\frac{1}{2}\right\}\)
<=> \(3x\left(4x^2-1\right)=0\)
<=> \(\orbr{\begin{cases}x=0\\4x^2=1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=\frac{1}{4}\end{cases}}}\)
<=> \(\orbr{\begin{cases}x=0\\x=\pm\frac{1}{2}\end{cases}}\)
Vay \(x\in\left\{-\frac{1}{2};0;\frac{1}{2}\right\}\)
Hoc tot
\(x^3-2x=-x^2+2\)
\(\Leftrightarrow x^3+x^2-2x-2=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-2\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Ta có: \(x^3-2x=-x^2+2\)
\(\Leftrightarrow\left(x^3+x^2\right)-\left(2x+2\right)=0\)
\(\Leftrightarrow x^2.\left(x+1\right)-2.\left(x+1\right)=0\)
5\(\Leftrightarrow\left(x+1\right).\left(x^2-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x^2=2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\pm\sqrt{2}\end{cases}}\)
Vậy \(S=\left\{-\sqrt{2};-1;\sqrt{2}\right\}\)
<=> \(x^2-25=10x+35-2x^2-7x\)
<=> \(3x^2-3x-60=0\)
<=> \(x^2-x-20=0\)
<=> \(\left(x-5\right)\left(x+4\right)=0\)
<=> \(\orbr{\begin{cases}x=5\\x=-4\end{cases}}\)
Vay \(x\in\left\{-4;5\right\}\)
Chuc ban hoc tot
Cách 1: \(5x^2-60x-5600=0\)\(\Leftrightarrow x^2-12x-1120=0\)\(\Leftrightarrow x^2-40x+28x-1120=0\)
\(\Leftrightarrow x\left(x-40\right)+28\left(x-40\right)=0\)\(\Leftrightarrow\left(x-40\right)\left(x+28\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x-40=0\\x+28=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=40\\x=-28\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=40\\x=-28\end{cases}}\)
Cách 2: \(5x^2-60x-5600=0\)\(\Leftrightarrow x^2-12x-1120=0\)\(\Leftrightarrow x^2-2x.6+6^2-1156=0\)
\(\Leftrightarrow\left(x-6\right)^2-34^2=0\)\(\Leftrightarrow\left(x-6-34\right)\left(x-6+34\right)=0\)
\(\Leftrightarrow\left(x-40\right)\left(x+28\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x-40=0\\x+28=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=40\\x=-28\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=40\\x=-28\end{cases}}\)
tính ra ik Pé Linh Miu Ly Ly
667;666;665