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a) Ta có: \(\left(x^2+8x+7\right)\left(x+3\right)\left(x+5\right)+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
\(=\left(x^2+8x\right)^2+22\left(x^2+8x\right)+105+15\)
\(=\left(x^2+8x\right)^2+22\left(x^2+8x\right)+120\)
\(=\left(x^2+8x\right)^2+12\left(x^2+8x\right)+10\left(x^2+8x\right)+120\)
\(=\left(x^2+8x\right)\left(x^2+8x+12\right)+10\left(x^2+8x+12\right)\)
\(=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)
\(=\left(x+2\right)\left(x+6\right)\left(x^2+8x+10\right)\)
b) Ta có: \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)
\(=\left(12x^2+11x\right)^2+\left(12x^2+11x\right)-2-4\)
\(=\left(12x^2+11x\right)^2+\left(12x^2+11x\right)-6\)
\(=\left(12x^2+11x\right)^2+3\left(12x^2+11x\right)-2\left(12x^2+11x\right)-6\)
\(=\left(12x^2+11x\right)\left(12x^2+11x+3\right)-2\left(12x^2+11x+3\right)\)
\(=\left(12x^2+11x+3\right)\left(12x^2+11x-2\right)\)
c) Ta có: \(\left(x^2+2x\right)^2+9x^2+18x+20\)
\(=\left(x^2+2x\right)^2+9\left(x^2+2x\right)+20\)
\(=\left(x^2+2x\right)^2+5\left(x^2+2x\right)+4\left(x^2+2x\right)+20\)
\(=\left(x^2+2x\right)\left(x^2+2x+5\right)+4\left(x^2+2x+5\right)\)
\(=\left(x^2+2x+5\right)\left(x^2+2x+4\right)\)
\(5x.\left(x-10\right)-2x+2x+20\)
\(=5x^2-50x+20\)
\(=5\left(x^2-10x+5^2-21\right)\)
\(=5\left[\left(x-5\right)^2-\left(\sqrt{21}\right)^2\right]\)
\(=5\left(x-5-\sqrt{21}\right)\left(x-5+\sqrt{21}\right)\)
\(a\left(a-b\right)^2\left(a+b\right)-\left(b-a\right)^2\left(a^2-5ab+b^2\right)\)
\(=a\left(a-b\right)^2\left(a+b\right)-\left(a-b\right)^2\left(a^2-5ab+b^2\right)\)
\(=\left(a-b\right)^2\left[a.\left(a+b\right)-a^2+5ab-b^2\right]\)
\(=\left(a-b\right)^2\left[a^2+ab-a^2+5ab-b^2\right]\)
\(=\left(a-b\right)^2\left(6ab-b^2\right)\)
Sửa đề: \(\left(a-b\right)^2-\left(b-a\right)\left(a-3b\right)\)
\(=\left(a-b\right)^2+\left(a-b\right)\left(a-3b\right)\)
\(=\left(a-b\right)\left(a-b+a-3b\right)\)
\(=\left(a-b\right)\left(2a-4b\right)\)
\(=2.\left(a-b\right)\left(a-2b\right)\)
Tham khảo nhé~
1. a) $(5-2x)^2-16=0$
$=>(5-2x)^2-4^2=0$
$=>(5-2x-4)(5-2x+4)=0$
$=>(1-2x)(9-2x)=0$
\(=>\left[{}\begin{matrix}1-2x=0=>x=0,5\\9-2x=0=>x=4,5\end{matrix}\right.\)
b) $x^2-4x=29$
$=>x^2-4x-29=0$
$=>(x^2-4x+4)-33=0$
$=>(x-2)^2-(\sqrt{33})^2=0$
$=>(x-2-\sqrt{33})(x-2+\sqrt{33})=0$
\(=>\left[{}\begin{matrix}x-2-\sqrt{33}=0=>x=\sqrt{33}+2\\x-2+\sqrt{33}=0=>x=2-\sqrt{33}\end{matrix}\right.\)
Bài 1:
a) \(\left(5-2x\right)^2-16=0\) (1)
\(\Leftrightarrow\left(5-2x\right)^2=16\)
\(\Leftrightarrow5-2x=\pm4\)
\(\Leftrightarrow\left[{}\begin{matrix}5-2x=4\\5-2x=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{1}{2};\dfrac{9}{2}\right\}\)
b) \(x^2-4x=29\) (2)
\(\Leftrightarrow x^2-4x-29=0\)
\(\Leftrightarrow x=\dfrac{4\pm2\sqrt{33}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4+2\sqrt{33}}{2}\\x=\dfrac{4-2\sqrt{33}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2+\sqrt{33}\\x=2-\sqrt{33}\end{matrix}\right.\)
Vậy tập nghiệm phương trình (2) là \(S=\left\{2-\sqrt{33};2+\sqrt{33}\right\}\)
c) \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\) (3)
\(\Leftrightarrow x^3-9x^2+27x-27-\left(x^3-27\right)+9\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow x^3-9x^2+27x-27-\left(x^3-27\right)+9x^2+18x+9=15\)
\(\Leftrightarrow x^3+27x-27-x^3+27+18x+9=15\)
\(\Leftrightarrow45x+9=15\)
\(\Leftrightarrow45x=15-9\)
\(\Leftrightarrow45x=6\)
\(\Leftrightarrow x=\dfrac{2}{15}\)
Vậy tập nghiệm phương trình (3) là \(S=\left\{\dfrac{2}{15}\right\}\)
d) \(2\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(2x-3\right)+x\left(x^2+8\right)=\left(x+1\right)\left(x^2-x+1\right)\)(4)
\(\Leftrightarrow2\left(x^2-25\right)-\left(2x^2-3x+4x-6\right)+x^3-8x=x^3+1\)
\(\Leftrightarrow2x^2-50-\left(2x^2+x-6\right)+x^3-8x=x^3+1\)
\(\Leftrightarrow2x^2-50-2x^2-x+6-8x=1\)
\(\Leftrightarrow-44-9x=1\)
\(\Leftrightarrow-9x=1+45\)
\(\Leftrightarrow-9x=45\)
\(\Leftrightarrow x=-5\)
Vậy tập nghiệm phương trình (4) là \(S=\left\{-5\right\}\)
Ta có:
P(x)=x2+(x+2)2+(x+3)2+...+(x+98)2−[(x+1)2+(x+3)2+...+(x+99)2]
=[x2-(x+1)2]+[(x+2)2-(x+3)2]+[(x+3)2-(x+4)2]+...+[(x+98)2-(x+99)2]
=(x-x-1)(x+x+1)+(x+2-x-3)(x+2+x+3)+...+(x+98-x-99)(x+98+x+99)
=-(2x+1)-(2x+5)-....-(2x+197)
=(-2x-2x-...-2x)+(-1-5-...-197)
Vì đa thức trên có \(\dfrac{197-1}{4}+1=50\text{ số hạng => -2x có 50 hạng tử}\)
Nên ta có:
=(-2x*50)+\(\left(\dfrac{\left(-197-1\right)\cdot50}{2}\right)\)
=-100x-4950
Mà P(x)=ax+b =>{a=-100; b=-4950}
Vậy a-b= -100-(-4950)= 4850 (Hihi! Mình tự làm nên ko biết đúng hay ko?)
a) Ta có : \(a^2+1=a^2+ab+bc+ac=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)
Tương tự : \(b^2+1=\left(b+a\right)\left(b+c\right)\) ; \(c^2+1=\left(c+a\right)\left(c+b\right)\)
Suy ra \(\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\)
Vậy \(A=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}=1\)
b) Ta có ; \(a^2+2bc-1=a^2+2bc-\left(ab+bc+ac\right)=a^2-ab+bc-ac=a\left(a-b\right)-c\left(a-b\right)\)
\(=\left(a-b\right)\left(a-c\right)\)
Tương tự : \(b^2+2ac-1=\left(a-b\right)\left(c-b\right)\) ; \(c^2+2ab-1=\left(a-c\right)\left(b-c\right)\)
Suy ra \(\left(a^2+2bc-1\right)\left(b^2+2ac-1\right)\left(c^2+2ab-1\right)=\left(a-b\right)^2.\left(c-a\right)^2.\left[-\left(b-c\right)^2\right]\)
Vậy : \(B=\frac{-\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)}=-1\)
ta có 2(a+b)=240
=> a+b=120=> a=120-b
ta có (a+15)(b+20)=6000
=> (120-b+15)(b+20)=6000
=> (135-b)(b+20)=6000
=> 135b+2700-b^2-20b=6000
=> 115b+2700-b^2=6000
=> 115b-3300-b^2=0
=> -(b^2-115b+3300)=0
=> -(b^2-60b-55b+3300)=0
=> -[b(b-60)-55(b-60)]=0
=> -[(b-60)(b-55)]=0
=> -(b-60)(b-55)=0
=> -(b-60)=0 hoac b-55=0
=> -b+60=0 , b=55
=> -b=-60=> b=60, b=55
thê b vào a ta được a=120-b=120-60=60
a=120-b=120-55=65
ta có 2(a+b)=240=> a=120-b
đề bài cho (a+15)(b+20)=6000
=> (120-b+15)(b+20)=6000
=> (135-b)(b+20)=6000
=> 135b+2700-b^2-20b=6000
=> -115b-3300-b^2=0
=> -(b^2+115+3300)=0
=> -(b^2+55b+60b+3300)=0
=> -[(b+55)+60(b+55)]=0
=> -(b+55)(b+60)=0
=> -(b+55)=0 hoac b+60=0
=> b=-55. , b=-60
thê b vào a ta có a=120-b=120+55=175 va a=120+60=180