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b)(2x - 1)^2 - (2x + 5) (2x - 5 ) = 18
4x 2 -4x+1-4x 2+25=18
26-4x=18
4x=8
x=2
a,27x-18=2x-3x^2
<=> 3x^2-2x+27-18x=0
<=> 3x^2-20x+27=0
\(\Delta\)= 20^2-4-12.27
tính \(\Delta\)rồi tìm x1 ,x2
BÀI 1:
a) \(x^4+2x^2y+y^2=\left(x^2+y\right)^2\)
b) \(\left(2a+b\right)^2-\left(2b+a\right)^2=\left(2a+b+2b+a\right)\left(2a+b-2b-a\right)\)
\(=\left(3a+3b\right)\left(a-b\right)=3\left(a+b\right)\left(a-b\right)\)
c) \(\left(a^3-b^3\right)+\left(a-b\right)^2=\left(a-b\right)\left(a^2+ab+b^2\right)+\left(a-b\right)^2\)
\(=\left(a-b\right)\left[a^2+ab+b^2+\left(a-b\right)\right]=\left(a-b\right)\left(a^2+ab+b^2+a-b\right)\)
d) \(\left(x^2+1\right)^2-4x^2=\left(x^2+1-2x\right)\left(x^2+1+2x\right)=\left(x-1\right)^2\left(x+1\right)^2\)
e) \(\left(y^3+8\right)+\left(y^2-4\right)=\left(y+2\right)\left(y^2-y+2\right)\)
f) \(1-\left(x^2-2xy+y^2\right)=1-\left(x-y\right)^2=\left(1-x+y\right)\left(1+x-y\right)\)
g) \(x^4-1=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\)
h) ktra lại đề
m) \(\left(x-a\right)^4-\left(x+a\right)^4=-8ax\left(a^2+x^2\right)\)
a) \(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{ab^2-ac^2-b^3+bc^2}\)
\(=\frac{a^2b-a^2c+b^2c-b^2a+c^2\left(a-b\right)}{ab^2-b^3-ac^2+bc^2}\)
\(=\frac{\left(a^2b-b^2a\right)+\left(b^2c-a^2c\right)+c^2\left(a-b\right)}{b^2\left(a-b\right)-c^2\left(a-b\right)}\)
\(=\frac{ab\left(a-b\right)+c\left(b^2-a^2\right)+c^2\left(a-b\right)}{\left(b^2-c^2\right)\left(a-b\right)}\)
\(=\frac{ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)}{\left(b-c\right)\left(b+c\right)\left(a-b\right)}\)
\(=\frac{ab-c\left(a+b\right)+c^2}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{ab-ac+c^2-bc}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{a\left(b-c\right)-c\left(b-c\right)}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{\left(b-c\right)\left(a-c\right)}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{a-b}{b+c}\)
\(VT=\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)\)
\(=x\left(x^3+x^2y+xy^2+y^3\right)-y\left(x^3+x^2y+xy^2+y^3\right)\)
\(=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4\)
\(=\left(x^4-y^4\right)+\left(x^3y-x^3y\right)+\left(x^2y^2-x^2y^2\right)+\left(xy^3-xy^3\right)\)
\(=x^4-y^4=VP\)
\(VT=\left(a+b\right)^2-\left(a-b\right)^2=4ab\)
\(=\left(a^2+2ab+b^2\right)-\left(a^2-2ab+b^2\right)\)
\(=a^2+2ab+b^2-a^2+2ab-b^2\)
\(=\left(a^2-a^2\right)-\left(b^2+b^2\right)+\left(2ab+2ab\right)\)
\(=4ab=VP\)
Câu a :
\(VT=\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)\)
Nhân 2 vế lại ta được \(x^4-y^4=VP\)
\(\Rightarrowđpcm\)
Câu b :
\(VT=\left(a+b\right)^2-\left(a-b\right)^2=\left(a+b-a+b\right)\left(a+b+a-b\right)=2b.2a=4ab=VP\)
\(\Rightarrowđpcm\)
ta có: \(\left(a-b+c\right)^2-\left(a+b+c\right)^2\)
VT \(=\left(a-b+c\right)\left(a-b+c\right)-\left(a+b+c\right)\left(a+b+c\right)\)
\(=a^2-ab+ac-ab+b^2-bc+ac-bc+c^2-a^2-ab-ac-ab-b^2-bc-ac-c-c^2\)
= \(-4ab-4bc=VT\left(đpcm\right)\)
a ) \(\left(a-b+c\right)^2-\left(a+b+c\right)^2\)
\(=\left(a-b+c-a-b-c\right)\left(a-b+c+a+b+c\right)\)
\(=-2b\left(2a+2c\right)\)
\(=-4ab-4bc\left(đpcm\right)\)
b ) \(6,3-5x+x^2\)
\(=x^2-5x+\dfrac{63}{10}\)
\(=x^2-5x+\dfrac{25}{4}+\dfrac{1}{20}\)
\(=\left(x-\dfrac{5}{2}\right)^2+\dfrac{1}{20}\ge\dfrac{1}{20}>0\forall x\left(đpcm\right)\)
:D