a) \(x^4-4x^3+x^2-4x=0\)

\(\Leftrightarrow\left(x^4-4x^3\right)+\left(x^2-4x\right)=0\)

\(\Leftrightarrow x^3\left(x-4\right)+x\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x^3+x\right)=0\)

\(\Leftrightarrow x\left(x-4\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x^2+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x^2=-1\left(loai\right)\end{matrix}\right.\)

Vậy x=0; x=4

b) \(x^3-5x^2+4x-20=0\)

\(\Leftrightarrow\left(x^3-5x^2\right)+\left(4x-20\right)=0\)

\(\Leftrightarrow x^2\left(x-5\right)+4\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x^2+4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x^2=-4\left(loai\right)\end{matrix}\right.\)

Vậy x=5

\(\frac{x_1-1}{3}=\frac{x_2-2}{2}=\frac{x_3-3}{1}vàx_1+x_2+x_3=30\)

-Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{x_1-1}{3}=\frac{x_2-2}{2}=\frac{x_3-3}{1}=\frac{x_1+x_2+x_3-6}{6}=\frac{30-6}{6}=\frac{24}{6}=4\)

-Suy ra: \(\Rightarrow\frac{x_1-1}{3}=4\Rightarrow x_1=13\)

\(\Rightarrow\frac{x_2-2}{2}=4\Rightarrow x_2=10\)

\(\Rightarrow\frac{x_3-3}{1}=4\Rightarrow x_3=7\)

Vậy \(x_1=13;x_2=10;x_3=7\)

1) \(x^2+x-6\)

\(=x^2-2x+3x-6\)

\(=\left(x^2-2x\right)+\left(3x-6\right)\)

\(=x\left(x-2\right)+3\left(x-2\right)\)

\(=\left(x-2\right)\left(x+3\right)\)

2) \(x^2-x-6\)

\(=x^2+2x-3x-6\)

\(=\left(x^2+2x\right)-\left(3x+6\right)\)

\(=x\left(x+2\right)-3\left(x+2\right)\)

\(=\left(x+2\right)\left(x-3\right)\)

3) \(x^2+2x-48\)

\(=x^2+8x-6x-48\)

\(=\left(x^2+8x\right)-\left(6x+48\right)\)

\(=x\left(x+8\right)-6\left(x+8\right)\)

\(=\left(x+8\right)\left(x-6\right)\)

4) \(x^2-2x-48\)

\(=x^2-8x+6x-48\)

\(=\left(x^2-8x\right)+\left(6x-48\right)\)

​\(=x\left(x-8\right)+6\left(x-8\right)\)​

\(=\left(x-8\right)\left(x+6\right)\)

5) \(x^2+x-42\)

\(=x^2+7x-6x-42\)

\(=\left(x^2+7x\right)-\left(6x+42\right)\)

\(=x\left(x+7\right)-6\left(x+7\right)\)

\(=\left(x+7\right)\left(x-6\right)\)

6) \(x^2-x-42\)

\(=x^2-7x+6x-42\)

\(=\left(x^2-7x\right)+\left(6x-42\right)\)

\(=x\left(x-7\right)+6\left(x-7\right)\)

\(=\left(x-7\right)\left(x+6\right)\)

1) \(x^4+64\)

\(=x^4+64+16x^2-16x^2\)

\(=\left(x^2\right)^2+2.x^2.8+8^2-16x^2\)

\(=\left(x^2+8\right)^2-\left(4x\right)^2\)

\(=\left(x^2+8-4x\right)\left(x^2+8+4x\right)\)

1)

a) \(0,7\left(6\right)=\dfrac{76-7}{90}=\dfrac{69}{90}=\dfrac{23}{30}\)

b) ta có: \(0,2\left(148\right)=\dfrac{2148-2}{9990}=\dfrac{2146}{9990}=\dfrac{29}{135}\)

do đó: \(1,2\left(148\right)=1+0,2\left(148\right)=1+\dfrac{29}{135}=\dfrac{164}{135}\)

\(4x^4+81\)

\(=4x^4+81+36x^2-36x^2\)

\(=\left(2x^2\right)^2+2.2x^2.9+9^2-36x^2\)

\(=\left(2x^2+9\right)^2-\left(6x\right)^2\)

\(=\left(2x^2+9-6x\right)\left(2x^2+9+6x\right)\)

1)

a) \(E=x^2-2x+y^2+4y+8\)

\(\Leftrightarrow E=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+3\)

\(\Leftrightarrow E=\left(x-1\right)^2+\left(y+2\right)^2+3\)

Vậy GTNN của E=3 khi \(\left\{{}\begin{matrix}x-1=0\Leftrightarrow x=1\\y+2=0\Leftrightarrow y=-2\end{matrix}\right.\)

b) \(F=x^2-4x+y^2-8y+6\)

\(\Leftrightarrow F=\left(x^2-4x+4\right)+\left(y^2-8y+16\right)-14\)

\(\Leftrightarrow F=\left(x-2\right)^2+\left(y-4\right)^2-14\)

Vậy GTNN của \(F=-14\) khi \(\left\{{}\begin{matrix}x-2=0\Leftrightarrow x=2\\y-4=0\Leftrightarrow y=4\end{matrix}\right.\)

2)

a)\(3x^2-3y^2-2\left(x-y\right)^2\)

\(=3\left(x^2-y^2\right)-2\left(x-y\right)^2\)

\(=3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\)

\(=\left(x-y\right)\left[3\left(x+y\right)-2\left(x-y\right)\right]\)

\(=\left(x-y\right)\left[3x+3y-2x+2y\right]\)

\(=\left(x-y\right)\left(x+5y\right)\)

b) \(x^3-4x^2-9x+36\)

\(=\left(x^3-4x^2\right)-\left(9x-36\right)\)

\(=x^2\left(x-4\right)-9\left(x-4\right)\)

\(=\left(x-4\right)\left(x^2-9\right)\)

\(=\left(x-4\right)\left(x-3\right)\left(x+3\right)\)

c) \(3x^2-6xy+3y^2-12z^2\)

\(=3\left(x^2-2xy+y^2-4z^2\right)\)

\(=3\left[\left(x^2-2xy+y^2\right)-4z^2\right]\)

\(=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)

\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)

d) \(5x^2-10xy+5y^2-20x^2\)

\(=5\left(x^2-2xy+y^2-4x^2\right)\)

\(=5\left[\left(x^2-2xy+y^2\right)-4x^2\right]\)

\(=5\left[\left(x-y\right)^2-\left(2x\right)^2\right]\)

\(=5\left(x-y-2x\right)\left(x-y+2x\right)\)

\(=5\left(-x-y\right)\left(3x-y\right)\)

đề là gì

\(5^x.125^x=625\)

\(\Leftrightarrow5^x.\left(5^3\right)^x=5^4\)

\(\Leftrightarrow5^x.5^{3.x}=5^4\)

\(\Leftrightarrow5^{x+3x}=5^4\)

\(\Leftrightarrow x+3x=4\)

\(\Leftrightarrow4x=4\)

\(\Leftrightarrow x=1\)

Vậy x=1