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a) ( x - 1 )2 - ( x - 1 )( x + 1 ) = 0
<=> x2 - 2x + 1 - ( x2 - 1 ) = 0
<=> x2 - 2x + 1 - x2 + 1 = 0
<=> 2 - 2x = 0
<=> 2x = 2
<=> x = 1
b) ( 2x - 1 )2 - ( 2x + 1 )2 = 0
<=> [ ( 2x - 1 ) - ( 2x + 1 ) ][ ( 2x - 1 ) + ( 2x + 1 ) ] = 0
<=> ( 2x - 1 - 2x - 1 )( 2x - 1 + 2x + 1 ) = 0
<=> -2.4x = 0
<=> -8x = 0
<=> x = 0
c) 25( x + 3 )2 + ( 1 - 5x )( 1 + 5x ) = 8
<=> 52( x + 3 )2 + 12 - 25x2 = 8
<=> [ 5( x + 3 ) ]2 + 1 - 25x2 = 8
<=> ( 5x + 15 )2 + 1 - 25x2 = 8
<=> 25x2 + 150x + 225 + 1 - 25x2 = 8
<=> 150x + 226 = 8
<=> 150x = -218
<=> x = -218/150 = -109/75
d) 9( x + 1 )2 - ( 3x - 2 )( 3x + 2 ) = 10
<=> 32( x + 1 )2 - ( 9x2 - 4 ) = 10
<=> [ 3( x + 1 ) ]2 - 9x2 + 4 = 10
<=> ( 3x + 3 )2 - 9x2 + 4 = 10
<=> 9x2 + 18x + 9 - 9x2 + 4 = 10
<=> 18x + 13 = 10
<=> 18x = -3
<=> x = -3/18 = -1/6
a) (x - 1)2 - (x - 1)(x + 1) = 0
=> (x - 1)2 - (x2 - 12) = 0
=> x2 - 2.x.1 + 12 - x2 + 1 = 0
=> x2 - 2x + 1 - x2 + 1 = 0
=> -2x + 1 + 1 = 0
=> -2x + 2 = 0
=> -2x = -2 => x = 1
b) (2x - 1)2 - (2x + 1)2 = 0
=> (2x - 1 - 2x + 1)(2x - 1 + 2x + 1) = 0
=> 0 = 0(đúng)
c) 25(x + 3)2 + (1 - 5x)(1 + 5x) = 8
=> 25(x2 + 2.x.3 + 32) + (12 - (5x)2) = 8
=> 25x2 + 150x + 225 + 1 - 25x2 = 8
=> 150x +225 + 1 = 8
=> 150x = -218
=> x = -109/75
d) 9(x + 1)2 - (3x - 2)(3x + 2) = 10
=> 9(x2 + 2x + 1) - [(3x)2 - 22 ] = 10
=> 9x2 + 18x + 9 - (9x2 - 4) = 10
=> 9x2 + 18x + 9 - 9x2 + 4 = 10
=> 18x + 9 + 4 = 10
=> 18x = -3
=> x = -1/6
Ta có: \(2x^2+2y^2-x-y-2xy+\frac{1}{2}=0\)
\(\Leftrightarrow\left(x^2+y^2-2xy\right)+\left(x^2-x+\frac{1}{4}\right)+\left(y^2-y+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}^2\right)=0\)
Nhận xét \(\left(x-y\right)^2\ge0;\left(x-\frac{1}{2}\right)^2\ge0;\left(y-\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow\hept{\begin{cases}\left(x-y\right)^2=0\\\left(x-\frac{1}{2}\right)^2=0\\\left(y-\frac{1}{2}\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-y=0\\x-\frac{1}{2}=0\\y-\frac{1}{2}=0\end{cases}\Leftrightarrow}x=y=\frac{1}{2}}\)
Bài 1
\(x^5+x^4+1=x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)
\(=\left(x^5+x^4+x^3\right)+\left(-x^3-x^2-x\right)+\left(x^2+x+1\right)\)
\(=x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^3-x+1\right)\left(x^2+x+1\right)\)
Bài 2
Ta có: \(\left(ax+b\right)\left(x^2+cx+1\right)=ax^3+bx^2+acx^2+bcx+ax+b\)
\(=ax^3+\left(b+ac\right)x^2+\left(bc+a\right)x+b=x^3-3x-2\)
\(\Rightarrow a=1\)
\(\Rightarrow b+ac=0\)
\(\Rightarrow bc+a=-3\)
\(\Rightarrow b=-2\)
Thay giá trị của \(a=1;b=-2\)vào \(b+ac=0\)ta được
\(\Leftrightarrow-2+c=0\Rightarrow c=2\)
Vậy \(a=1;b=-2;c=2\)
Bài 3
Ta có \(\left(x^4-3x^3+2x^2-5x\right)\div\left(x^2-3x+1\right)=x^2+1\left(dư-2x+1\right)\)
\(\Rightarrow b=2x-1\)
Bài 4 (cũng làm tương tự như bài 3 nhé )
Bài 5(bài nãy dễ nên bạn tự làm đi nhé)
Bài 6
\(\left(a+b\right)^2=2\left(a^2+b^2\right)\)
\(\Leftrightarrow a^2+2ab+b^2=2a^2+2b^2\)
\(\Leftrightarrow2a^2+2b^2-a^2-2ab-b^2=0\)
\(\Leftrightarrow a^2-2ab+b^2=0\)
\(\Leftrightarrow\left(a-b\right)^2=0\)\(\Rightarrow a-b=0\Rightarrow a=b\)
Bài 7
\(a^2+b^2+c^2=ab+ac+bc\)
\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2ac+2bc\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\Leftrightarrow a^2+a^2+b^2+b^2+c^2+c^2-2ab-2ac-2bc=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(\Rightarrow a-b=0\Rightarrow a=b\)
\(\Rightarrow b-c=0\Rightarrow b=c\)
\(\Rightarrow a-c=0\Rightarrow a=c\)
Vậy \(a=b=c\)
g) \(\left(2x-1\right)^2-\left(2x+4\right)^2=0\)
\(\Leftrightarrow\left(2x-1+2x+4\right)\left(2x-1-2x-4\right)=0\)
\(\Leftrightarrow-5\left(4x+3\right)=0\)
\(\Leftrightarrow4x+3=0\)
\(\Leftrightarrow4x=-3\)
\(\Leftrightarrow x=\frac{-3}{4}\)
Vậy tập nghiệm của pt là \(S=\left\{\frac{-3}{4}\right\}\)
h) \(\left(2x-3\right)\left(3x+1\right)-x\left(6x+10\right)=30\)
\(\Leftrightarrow3x\left(2x-3\right)+\left(2x-3\right)-6x^2-10x=30\)
\(\Leftrightarrow6x^2-9x+2x-3-6x^2-10x=30\)
\(\Leftrightarrow-9x+2x-3-10x=30\)
\(\Leftrightarrow-17x-3=30\)
\(\Leftrightarrow-17x=33\)
\(\Leftrightarrow x=\frac{-33}{17}\)
Vậy tập nghiệm của pt là \(S=\left\{\frac{-33}{17}\right\}\)
\(\left(2x^2-3x+1\right)\left(x^2-5\right)-\left(x^2-x\right)\left(2x^2-x-10\right)=5\)
\(\)=>\(2x^4-3x^3+x^2-10x^2+15x-5-2x^4+x^3+10x^2\)\(+2x^3-x^2\)-10x=5
=>(\(\left(2x^4-2x^4\right)+\left(-3x^3+x^3+2x^3\right)\)\(+\left(x^2-10x^2+10x^2-x^2\right)\)+(15x-10x)=5+5
=> 5x=10
=> x=2