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C=\(\left(x-1\right)x^2-4x\left(x-1\right)+4\left(x-1\right)=\left(x-1\right)\left(x^2-4x+4\right)=\left(x-1\right)\left(x-2x-2x+4\right)\)
C= \(\left(x-1\right)\left(x-2\right)\left(x-2\right)\)
bạn thay x vào rồi tính là được
B=\(x\left(2x-y\right)-z\left(y-2x\right)=x\left(2x-y\right)+z\left(2x-y\right)=\left(2x-y\right)\left(x+z\right)\)
bạn thay x,y,z tính là ok
Bài a mình k chắc lắm nhưng nghĩ là thay vào rồi tính
\(\dfrac{x\left(x-8\right)+3\left(x+6\right)}{\left(x+6\right)\left(x-8\right)}=\dfrac{-12x+33}{\left(x+6\right)\left(x-8\right)}\left(đk:x\ne-6;8\right)\)
\(x^2-8x+3x+18=-12x+33\)
\(x^2-5x+18+12x-33=0\)
\(x^2+7x+15=0\)
\(\text{∆}=7^2-4.15=-11< 0\)
⇒ pt vô nghiệm
đk : x khác -6 ; 8
\(x^2-8x+3x+18=-12x+33\Leftrightarrow x^2+7x-25=0\)
\(\Leftrightarrow x=\dfrac{-7\pm\sqrt{149}}{2}\)
a) \(x^7+x^5+1\)
\(=x^7-x+x^5-x^2+x^2+x+1\)
\(=x\left(x^6-1\right)+x^2\left(x^3-1\right)+\left(x^2+x+1\right)\)
\(=x\left(x^3+1\right)\left(x^3-1\right)+x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=x\left(x^3+1\right)\left(x-1\right)\left(x^2+x+1\right)+x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)]
\(=\left(x^2+x+1\right)\left[x\left(x^3+1\right)\left(x-1\right)+x^2\left(x-1\right)+1\right]\)
\(=\left(x^2+x+1\right)\left[x\left(x^4-x^3+x-1\right)+x^3-x^2+1\right]\)
\(=\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+x^3-x^2+1\right)\)
\(=\left(x^2+x+1\right)\left(x^5-x^4+x^3-x+1\right)\)
b) \(x^5-x^4-1\)
\(=x^5-x^4+x^3-x^3+x^2-x-x^2+x-1\)
\(=x^3\left(x^2-x+1\right)-x\left(x^2-x+1\right)-\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^3-x-1\right)\)
i) (x - 1)(5x + 3) = (3x - 8)(x - 1)
<=> 5x2 + 3x - 5x - 3 = 3x2 - 3x - 8x + 8
<=> 5x2 - 2x - 3 = 3x2 - 11x + 8
<=> 5x2 - 2x - 3 - 3x2 + 11x - 8 = 0
<=> 2x2 + 9x - 11 = 0
<=> 2x2 + 11x - 2x - 11 = 0
<=> x(2x + 11) - (2x + 11) = 0
<=> (x - 1)(2x + 11) = 0
<=> x - 1 = 0 hoặc 2x + 11 = 0
<=> x = 0 hoặc x = -11/2
m) 2x(x - 1) = x2 - 1
<=> 2x2 - 2x = x2 - 1
<=> 2x2 - 2x - x2 + 1 = 0
<=> x2 - 2x + 1 = 0
<=> (x - 1)2 = 0
<=> x - 1 = 0
<=> x = 1
n) (2 - 3x)(x + 11) = (3x - 2)(2 - 5x)
<=> 2x + 22 - 3x2 - 33x = 6x - 15x2 - 4 + 10x
<=> -31x + 22 - 3x2 = 16x - 15x2 - 4
<=> 31x - 22 + 3x2 + 16x - 15x2 - 4 = 0
<=> 47x - 18 - 12x2 = 0
<=> -12x2 + 47x - 26 = 0
<=> 12x2 - 47x + 26 = 0
<=> 12x2 - 8x - 39x + 26 = 0
<=> 4x(3x - 2) - 13(3x - 2) = 0
<=> (4x - 13)(3x - 2) = 0
<=> 4x - 13 = 0 hoặc 3x - 2 = 0
<=> x = 13/4 hoặc x = 2/3
i) (x - 1)(5x + 3) = (3x - 8)(x - 1)
<=> 5x2 + 3x - 5x - 3 = 3x2 - 3x - 8x + 8
<=> 5x2 - 2x - 3 = 3x2 - 11x + 8
<=> 5x2 - 2x - 3 - 3x2 + 11x - 8 = 0
<=> 2x2 + 9x - 11 = 0
<=> 2x2 + 11x - 2x - 11 = 0
<=> x(2x + 11) - (2x + 11) = 0
<=> (x - 1)(2x + 11) = 0
<=> x - 1 = 0 hoặc 2x + 11 = 0
<=> x = 0 hoặc x = -11/2
m) 2x(x - 1) = x2 - 1
<=> 2x2 - 2x = x2 - 1
<=> 2x2 - 2x - x2 + 1 = 0
<=> x2 - 2x + 1 = 0
<=> (x - 1)2 = 0
<=> x - 1 = 0
<=> x = 1
n) (2 - 3x)(x + 11) = (3x - 2)(2 - 5x)
<=> 2x + 22 - 3x2 - 33x = 6x - 15x2 - 4 + 10x
<=> -31x + 22 - 3x2 = 16x - 15x2 - 4
<=> 31x - 22 + 3x2 + 16x - 15x2 - 4 = 0
<=> 47x - 18 - 12x2 = 0
<=> -12x2 + 47x - 26 = 0
<=> 12x2 - 47x + 26 = 0
<=> 12x2 - 8x - 39x + 26 = 0
<=> 4x(3x - 2) - 13(3x - 2) = 0
<=> (4x - 13)(3x - 2) = 0
<=> 4x - 13 = 0 hoặc 3x - 2 = 0
<=> x = 13/4 hoặc x = 2/3
\(A=3\left(x-3\right)\left(x+7\right)+\left(x+4\right)^2+48\)
\(A=3\left(x^2-4x-21\right)+\left(x^2+8x+16\right)+48\)
\(A=\left(3x^2+x^2\right)-\left(12x-8x\right)-\left(21-16-48\right)\)
\(A=4x^2-4x+43\)
\(A=\left(4x^2-4x+1\right)+42\)
\(A=\left(2x+1\right)^2+42\)
Thay \(x=\frac{1}{2}\) vao A ta duoc:
\(A=\left(2\cdot\frac{1}{2}+1\right)^2+42=46\)
\(A=3\left(x-3\right)\left(x-7\right)+\left(x+4\right)^2+48\)
\(=3x^2-13x+63+x^2+8x+16+48\)
\(=4x^2-5x+127\)
\(4\cdot0,25-5\cdot0,5+127=1-1+127=127\)