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

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

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

\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=1\)

3, \(x^4-9x^2=0\Leftrightarrow x^2\left(x^2-9\right)=0\)

\(\Leftrightarrow x^2\left(x-3\right)\left(x+3\right)=0\Leftrightarrow x=0;\pm3\)

4, \(x^2-6x+9=81\Leftrightarrow\left(x-3\right)^2=9^2\)

\(\Leftrightarrow\left(x-3-9\right)\left(x-3+9\right)=0\Leftrightarrow\left(x-12\right)\left(x+6\right)=0\Leftrightarrow x=-6;x=12\)

5, em xem lại đề nhé

à lag tý @@

5, \(x^3+6x^2+9x-4x=0\Leftrightarrow x^3+6x^2+5x=0\)

\(\Leftrightarrow x\left(x^2+6x+5\right)=0\Leftrightarrow x\left(x^2+x+5x+5\right)=0\)

\(\Leftrightarrow x\left(x+1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=-1;x=0\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x+2\right)^2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}}\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+3-2=0\\x+3+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-5\end{cases}}}\)

c)\(x^4-9x^2=0\)

\(\Leftrightarrow x^2\left(x^2-9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}}\)

d)\(x^2-6x+9=81\)

\(\Leftrightarrow\left(x-3\right)^2=81\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=9\\x-3=-9\end{cases}\Leftrightarrow\orbr{\begin{cases}x=12\\x=-6\end{cases}}}\)

e)\(x^3+6x^2+9x-4x=0\)

\(\Leftrightarrow x^3+6x^2+5x=0\)

\(\Leftrightarrow\left(x^2+5x\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+5x=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0;x=-5\\x=-1\end{cases}}}\)

#H

1. \(x^6-2x^3+1=0\Leftrightarrow\left(x^3-1\right)^2=0\Leftrightarrow x=1\)

2. \(x^6+\dfrac{1}{4}x^3+\dfrac{1}{64}=0\Leftrightarrow\left(x^3\right)^2+2.x^3.\dfrac{1}{8}+\left(\dfrac{1}{8}\right)^2=0\Leftrightarrow\left(x+\dfrac{1}{8}\right)^2=0\Leftrightarrow x=-\dfrac{1}{2}\)4. \(x^3-10x^2+25x=0\Leftrightarrow x^3-5x^2-5x^2+25x=0\)

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

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

5. \(\dfrac{1}{4}x^3-3x^2+9x=0\)

\(\Leftrightarrow x\left(\dfrac{1}{4}x^2-3x+9\right)=0\)

\(\Leftrightarrow x\left[\left(\dfrac{1}{2}x\right)^2-2.\dfrac{1}{2}x.3+3^2\right]=0\)

\(\Leftrightarrow x\left(\dfrac{1}{2}x-3\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

6. \(x^5-16x=0\Leftrightarrow x\left(x^4-16\right)=0\Leftrightarrow x\left(x^2-4\right)\left(x^2+4\right)=0\)

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

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

7. \(4x^2+4x-3=0\Leftrightarrow4x^2-2x^2-6x-3=0\)

\(\Leftrightarrow2x\left(2x-1\right)-3\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

8. \(4x^2+28x+48=0\Leftrightarrow4x^2+12x+14x+48=0\)

\(\Leftrightarrow4x\left(x+3\right)+12\left(x+4\right)=0\)

\(\Leftrightarrow\left(4x+12\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-4\end{matrix}\right.\)

9. \(9x^2-12x+3=0\Leftrightarrow9x^2-9x-3x+3=0\Leftrightarrow9x\left(x-1\right)-3\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(9x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

|2 - x|² + 6x - 3 = 0

<=> (x - 2)² + 6x - 3 = 0

<=> x² - 4x + 4 + 6x - 3 = 0

<=> x² + 2x + 1 = 0

<=> (x + 1)² = 0

<=> x + 1 = 0

<=> x = -1

Bắt phải thể hiện -_-

f) = x²( x - 4 ) - 9( x - 4 ) = ( x - 4 )( x - 3 )( x + 3 )

g) = 4( x - y ) + ( x - y )² = ( x - y )( x - y + 4 )

h) = x³( x + 1 ) + ( x - 1 )( x + 1 ) = ( x + 1 )( x³ + x - 1 )

i) = ( x - y )( x + y ) - 4( x + y ) = ( x + y )( x - y - 4 )

j) = ( x - y )( x² + xy + y² ) - 3( x - y ) = ( x - y )( x² + xy + y² - 3 )

Trả lời:

f, x³ - 4x² - 9x + 36 = ( x³ - 4x² ) - ( 9x - 36 ) = x² ( x - 4 ) - 9 ( x - 4 ) = ( x - 4 )( x² - 9 ) = ( x - 4 )( x - 3 )( x + 3 )

g, 4x - 4y + x² - 2xy + y² = ( 4x - 4y ) + ( x² - 2xy + y² ) = 4 ( x - y ) + ( x - y )² = ( x - y ) ( 4 + x - y )

h, x⁴ + x³ + x² - 1 = ( x⁴ + x³ ) + ( x² - 1 ) =  x³ ( x + 1 ) + ( x - 1 )( x + 1 ) = ( x + 1 )( x³ + x - 1 ) 

i, x² - y² - 4x - 4y = ( x² - y² ) - ( 4x + 4y ) = ( x - y )( x + y ) - 4 ( x + y ) = ( x + y )( x - y - 4 )

j, x³ - y³ - 3x + 3y = ( x³ - y³ ) - ( 3x - 3y ) = ( x - y )( x² + xy + y² ) - 3 ( x - y ) = ( x - y )( x² + xy + y² - 3 ) 

f) x³-4x²-9x+36

=x²(x-4)-9(x-4)

=(x-4)(x²-9)

=(x-4)(x-3)(x+3)

g) 4x-4y+x²-2xy+y²

=4(x-y)+(x-y)²

=(x-y)(4+x-y)

h) x⁴+x³+x²-1

=x³(x+1)+(x-1)(x+1)

=(x+1)(x³+x-1)

i) x²-y²-4x-4y

=(x-y)(x+y)-4(x+y)

=(x+y)(x-y-4)

j) x³-y³-3x+3y

=(x-y)(x²+xy+y²)-3(x-y)

=(x-y)(x²+xy+y²-3)

#H

\(a,\left(-5\right).\left|x\right|=-75\)

\(\left|x\right|=\frac{-75}{-5}=15\)

\(\Rightarrow\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)

Vậy....

\(b,\left(-6\right)^3.x^2=-1944\)

\(-216.x^2=-1944\)

\(x^2=9\)

\(\Rightarrow x=\pm3\)

Vậy....

\(d,\left|9-x\right|=-7+64\)

\(\left|9-x\right|=57\)

\(\Rightarrow\orbr{\begin{cases}9-x=57\\9-x=-57\end{cases}\Rightarrow\orbr{\begin{cases}x=-48\\x=66\end{cases}}}\)

Vậy...

\(e,\left|x+101\right|-\left(-16\right)=\left(-43\right).\left(-5\right)\)

\(\left|x+101\right|+16=215\)

\(\left|x+101\right|=199\)

\(\Rightarrow\orbr{\begin{cases}x+101=199\\x+101=-199\end{cases}\Rightarrow\orbr{\begin{cases}x=98\\x=-300\end{cases}}}\)

Vậy..

hok tốt!!

a,\(\left(-5\right).\left|x\right|=-75\)

\(=>\left|x\right|=-75:\left(-5\right)=15\)

\(=>\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)

b,\(\left(-6\right)^3.x^2=-1944\)

\(=>\frac{1944}{216}=x^2\)

\(=>x=\sqrt{\frac{1944}{216}}=3\)

CẢM ƠN MỌI NGƯỜI NHIỀU Ạ 

a, \(5x\left(x-1\right)+\left(x+17\right)=0\)

\(\Leftrightarrow5x^2-5x+x+17=0\Leftrightarrow5x^2-4x+17=0\)

\(\Leftrightarrow5\left(x^2-\frac{4}{5}x\right)+17=0\Leftrightarrow5\left(x^2-2.\frac{2}{5}x+\frac{4}{25}-\frac{4}{25}\right)+17=0\)

\(\Leftrightarrow5\left(x-\frac{2}{5}\right)^2-\frac{4}{5}+17=0\Leftrightarrow5\left(x-\frac{2}{5}\right)^2+81\ge81>0\)

Vậy pt vô nghiệm 

b, \(3x\left(x-3\right)^2-3x\left(x+3\right)^2=0\)

\(\Leftrightarrow3x\left[\left(x-3\right)^2-\left(x+3\right)^2\right]=0\)

\(\Leftrightarrow3x\left(x-3-x-3\right)\left(x-3+x+3\right)=0\Leftrightarrow x.2x=0\Leftrightarrow x=0\)

c, \(2x^2-9x+7=0\Leftrightarrow2x^2-7x-2x+7=0\)

\(\Leftrightarrow x\left(2x-7\right)-\left(2x-7\right)=0\Leftrightarrow\left(x-1\right)\left(2x-7\right)=0\Leftrightarrow x=1;x=\frac{7}{2}\)

Trả lời:

a, \(5x\left(x-1\right)+\left(x+17\right)=0\)

\(\Leftrightarrow5x^2-5x+x+17=0\)

\(\Leftrightarrow5x^2-4x+17=0\)

\(\Leftrightarrow5\left(x^2-\frac{4}{5}x+\frac{17}{5}\right)=0\)

\(\Leftrightarrow x^2-\frac{4}{5}x+\frac{17}{5}=0\)

\(\Leftrightarrow x^2-2.x.\frac{2}{5}+\frac{4}{25}+\frac{81}{25}=0\)

\(\Leftrightarrow\left(x-\frac{2}{5}\right)^2+\frac{81}{25}=0\)

Vì \(\left(x-\frac{2}{5}\right)^2+\frac{81}{25}\ge\frac{81}{25}>0\forall x\)

nên pt vô nghiệm 

b, \(3x\left(x-3\right)^2-3x\left(x+3\right)^2=0\)

\(\Leftrightarrow3x\left[\left(x-3\right)^2-\left(x+3\right)^2\right]=0\)

\(\Leftrightarrow3x\left(x-3-x-3\right)\left(x-3+x+3\right)=0\)

\(\Leftrightarrow3x.\left(-9\right).2x=0\)

\(\Leftrightarrow-54x^2=0\)

\(\Leftrightarrow x^2=0\)

\(\Leftrightarrow x=0\)

Vậy x = 0 là nghiệm của pt.

c, \(7-9x+2x^2=0\)

\(\Leftrightarrow2x^2-7x-2x+7=0\)

\(\Leftrightarrow x\left(2x-7\right)-\left(2x-7\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=1\end{cases}}}\)

Vậy x = 7/2; x = 1 là nghiệm của pt.

d, trùng ý c