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a/ \(\left(x-4\right)^2-36=0\)
<=> \(\left(x-4-6\right)\left(x-4+6\right)=0\)
<=> \(\left(x-10\right)\left(x+2\right)=0\)
<=> \(\orbr{\begin{cases}x-10=0\\x+2=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=10\\x=-2\end{cases}}\)
b/ \(\left(x+8\right)^2=121\)
<=> \(\left(x+8\right)^2-121=0\)
<=> \(\left(x+8-11\right)\left(x+8+11\right)=0\)
<=> \(\left(x-3\right)\left(x+19\right)=0\)
<=> \(\orbr{\begin{cases}x-3=0\\x+19=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=3\\x=-19\end{cases}}\)
d/ \(4x^2-12x+9=0\)
<=> \(\left(2x\right)^2-2.2x.3+3^2=0\)
<=> \(\left(2x-3\right)^2=0\)
<=> \(2x-3=0\)
<=> \(x=\frac{3}{2}\)
Ít thôi -..-
a) ( 3x + 2 )( 2x + 9 ) - ( x + 3 )( 6x + 1 ) = ( x + 1 )2 - ( x + 2 )( x - 2 )
<=> 6x2 + 31x + 18 - ( 6x2 + 19x + 3 ) = x2 + 2x + 1 - ( x2 - 4 )
<=> 6x2 + 31x + 18 - 6x2 - 19x - 3 = x2 + 2x + 1 - x2 + 4
<=> 12x + 15 = 2x + 5
<=> 12x - 2x = 5 - 15
<=> 10x = -10
<=> x = -1
b) ( 2x + 3 )( x - 4 ) + ( x - 5 )( x - 2 ) = ( 3x - 5 )( x - 4 )
<=> 2x2 - 5x - 12 + x2 - 7x + 10 = 3x2 - 17x + 20
<=> 3x2 - 12x - 2 = 3x2 - 17x + 20
<=> 3x2 - 12x - 3x2 + 17x = 20 + 2
<=> 5x = 22
<=> x = 22/5
c) ( x + 2 )3 - ( x - 2 )3 - 12x( x - 1 ) = -8
<=> x3 + 6x2 + 12x + 8 - ( x3 - 6x2 + 12x - 8 ) - 12x2 + 12x = -8
<=> x3 + 6x2 + 12x + 8 - x3 + 6x2 - 12x + 8 - 12x2 + 12x = -8
<=> 12x + 16 = -8
<=> 12x = -24
<=> x = -2
d) ( 3x - 1 )2 - 5( x + 1 ) + 6x - 3.2x + 1 - ( x - 1 )2 = 16
<=> 9x2 - 6x + 1 - 5x - 5 + 6x - 6x + 1 - ( x2 - 2x + 1 ) = 16
<=> 9x2 - 11x - 3 - x2 + 2x - 1 = 16
<=> 8x2 - 9x - 4 = 16
<=> 8x2 - 9x - 4 - 16 = 0
<=> 8x2 - 9x - 20 = 0
( Đến đây bạn có hai sự lựa chọn : 1 là vô nghiệm
2 là nghiệm vô tỉ =) )
a) (3x + 2)(2x + 9) - (x + 3)(6x + 1) = (x + 1)2 - (x + 2)(x - 2)
=> 3x(2x + 9) + 2(2x + 9) - x(6x + 1) - 3(6x + 1) = x2 + 2x + 1 - x(x - 2) - 2(x - 2)
=> 6x2 + 27x + 4x + 18 - 6x2 - x - 18x - 3 = x2 + 2x + 1 - x2 + 2x - 2x + 4
=> (6x2 - 6x2) + (27x + 4x - x - 18x) + (18 - 3) = (x2 - x2) + (2x + 2x - 2x) + (1 + 4)
=> 12x + 15 = 2x + 5
=> 12x + 15 - 2x - 5 = 0
=> 10x + 10 = 0
=> 10x = -10 => x = -1
b) (2x + 3)(x - 4) + (x - 5)(x - 2) = (3x - 5)(x - 4)
=> 2x(x - 4) + 3(x - 4) + x(x - 2) - 5(x - 2) = 3x(x - 4) - 5(x - 4)
=> 2x2 - 8x + 3x - 12 + x2 - 2x - 5x + 10 = 3x2 - 12x - 5x + 20
=> (2x2 + x2) + (-8x + 3x - 2x - 5x) + (-12 + 10) = 3x2 - 17x + 20
=> 3x2 - 12x - 2 = 3x2 - 17x + 20
=> 3x2 - 12x - 2 - 3x2 + 17x - 20 = 0
=> (3x2 - 3x2) + (-12x + 17x) + (-2 - 20) = 0
=> 5x - 22 = 0
=> 5x = 22 => x = 22/5
c) (x + 2)3 - (x - 2)3 - 12x(x - 1) = -8
=> x3 + 6x2 + 12x + 8 - (x3 - 6x2 + 12x - 8) - 12x2 + 12x = -8
=> x3 + 6x2 + 12x + 8 -x3 + 6x2 - 12x + 8 - 12x2 + 12x = -8
=> (x3 - x3) + (6x2 + 6x2 - 12x2) + (12x - 12x + 12x) + (8 + 8) = -8
=> 12x + 16 = -8
=> 12x = -24
=> x = -2
Còn bài cuối làm nốt
1) <=> x2 - 4x - x2 + 8 = 0 <=> x2 - 4x + 8 = 0
Dễ thấy phương trình vô nghiệm vì x2 - 4x + 8 = ( x - 2 )2 + 4 > 0
2) <=> ( x - 1 )3 = 0 <=> x = 1
3) <=> ( x - 2 )3 = 0 <=> x = 2
4) <=> ( 2x - 1 )3 = 0 <=> x = 1/2
\(A=x^2-12x+7=x^2-12x+36-29\)
\(=\left(x-6\right)^2-29\ge-29\)
Vậy \(A_{min}=-29\Leftrightarrow x=6\)
\(C=x-x^2-4=-\left(x^2-x+4\right)\)
\(=-\left(x^2-x+\frac{1}{4}+\frac{3}{4}\right)\)
\(=-\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right]\)
\(=-\left[\left(x-\frac{1}{2}\right)^2\right]-\frac{3}{4}\le-\frac{3}{4}\)
Vậy \(C_{min}=\frac{-3}{4}\Leftrightarrow x=\frac{1}{2}\)
a: \(=\dfrac{x+3}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x\left(x+1\right)}\)
\(=\dfrac{x^2+3x-x+1}{x\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{x\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{x\left(x-1\right)}\)
b: \(=\dfrac{24y^5}{7x^2}\cdot\dfrac{-21x}{12y^3}=2y^2\cdot\dfrac{-3}{x}=\dfrac{-6y^2}{x}\)
c: \(=\dfrac{-3\left(x-1\right)}{\left(x+1\right)^2}\cdot\dfrac{x+1}{6\left(x-1\right)\left(x+1\right)}=\dfrac{-1}{2\left(x+1\right)}\)
d: \(=\dfrac{7x+2}{3\left(2x-y\right)}\cdot\dfrac{6x\left(2x-y\right)}{2\left(7x+2\right)}=x\)
Bài 4:
a) Ta có: \(x^3+6x^2+12x+8\)
\(=x^3+2x^2+4x^2+8x+4x+8\)
\(=x^2\left(x+2\right)+4x\left(x+2\right)+4\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+4x+4\right)\)
\(=\left(x+2\right)^3\)
b) Ta có: \(x^3-3x^2+3x-1\)
\(=x^3-x^2-2x^2+2x+x-1\)
\(=x^2\left(x-1\right)-2x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-2x+1\right)\)
\(=\left(x-1\right)^3\)
c) Ta có: \(1-9x+27x^2-27x^3\)
\(=1-3x-6x+18x^2+9x^2-27x^3\)
\(=\left(1-3x\right)-6x\left(1-3x\right)+9x^2\left(1-3x\right)\)
\(=\left(1-3x\right)\left(1-6x+9x^2\right)\)
\(=\left(1-3x\right)^3\)
d) Ta có: \(x^3+\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}\)
\(=x^3+3\cdot x^2\cdot\frac{1}{2}+3\cdot x\cdot\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3\)
\(=\left(x+\frac{1}{2}\right)^3\)
e) Ta có: \(27x^3-54x^2y+36xy^2-8y^3\)
\(=\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot2y+3\cdot3x\cdot\left(2y\right)^2-\left(2y\right)^3\)
\(=\left(3x-2y\right)^3\)
a)\(x^2-2xy+y^2+1=\left(x+y\right)^2+1\ge1>0\)
b)\(x-x^2-1=-\left(x^2-x+\frac{1}{4}\right)^2-\frac{3}{4}\le-\frac{3}{4}< 0\)
c)\(9x^2+12x+10=\left(9x^2+12x+4\right)+6=\left(3x+2\right)^2+6\ge6>0\)
d)\(3x^2-x+1=2x^2+\left(x^2-x+\frac{1}{4}\right)+\frac{3}{4}=2x^2+\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0`\)
\(Sửa:A=x^4-6x^3+13x^2-12x+2021\\ A=\left(x^4-6x^3+9x^2\right)+4\left(x^2-3x\right)+4+2017\\ A=\left(x^2-3x\right)^2+4\left(x^2-3x\right)+4+2017\\ A=\left(x^2-3x+2\right)^2+2017\ge2017\\ A_{min}=2017\Leftrightarrow x^2-3x+2=0\Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)