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\(\text{1) }\dfrac{x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^2-1}\\ =\dfrac{\left(x^7+x^6\right)+\left(x^5+x^4\right)+\left(x^3+x^2\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(x^6+x^4+x^2+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6+x^4+x^2+1}{x-1}\)
\(\text{3) }\dfrac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\\ =\dfrac{\left(x^2-2xy+y^2\right)+\left(2xz-2yz\right)+z^2}{\left(x^2-2xy+y^2\right)-z^2}\\ =\dfrac{\left(x-y\right)^2+2\left(x-y\right)z+z^2}{\left(x-y\right)^2-z^2}\\ =\dfrac{\left(x-y+z\right)^2}{\left(x-y+z\right)\left(x-y-z\right)}\\ =\dfrac{x-y+z}{x-y-z}\)
a: \(A=4\cdot15^2-70^2=-4000\)
b: \(B=x^2+2x\left(y+1\right)+\left(y+1\right)^2\)
\(=\left(x+y+1\right)^2\)
\(=100^2=10000\)
c: \(C=b^2-3b+a^2+3a-2ab\)
\(=\left(a-b\right)^2+3\left(a-b\right)\)
\(=\left(a-b\right)\left(a-b+3\right)\)
\(=\left(-5\right)\cdot\left(-5+3\right)=\left(-5\right)\cdot\left(-2\right)=10\)
d: \(D=\left(x-y\right)^3+3xy\left(x-y\right)+3xy\)
\(=\left(-1\right)^3-3xy+3xy\)
=-1
B3;a,ĐKXĐ:\(x\ne\pm4\)
A=\(\left(\dfrac{4}{x-4}-\dfrac{4}{x+4}\right)\dfrac{x^2+8x+16}{32}=\left(\dfrac{4x+16}{x^2-16}-\dfrac{4x-16}{x^2-16}\right)\dfrac{x^2+2.4x+4^2}{32}=\left(\dfrac{4x+16-4x+16}{x^2-16}\right)\dfrac{\left(x+4\right)^2}{32}=\left(\dfrac{32}{x^2-16}\right)\dfrac{\left(x+4\right)^2}{32}=\dfrac{32\left(x+4\right)^2}{32.\left(x-4\right)\left(x+4\right)}=\dfrac{x+4}{x-4}\\ \\ \\ \\ \\ \\ b,Tacó\dfrac{x+4}{x-4}=\dfrac{1}{3}\Leftrightarrow3x+12=x-4\Leftrightarrow x=-8\left(TM\right)c,TAcó\dfrac{x+4}{x-4}=3\Leftrightarrow x+4=3x-12\Leftrightarrow x=8\left(TM\right)\)
1) \(\left(x-3\right)\left(x-5\right)+44\)
\(=x^2-3x-5x+15+44\)
\(=x^2-8x+59\)
\(=x^2-2.x.4+4^2+43\)
\(=\left(x-4\right)^2+43\ge43>0\)
\(\rightarrowĐPCM.\)
2) \(x^2+y^2-8x+4y+31\)
\(=\left(x^2-8x\right)+\left(y^2+4y\right)+31\)
\(=\left(x^2-2.x.4+4^2\right)-16+\left(y^2+2.y.2+2^2\right)-4+31\)
\(=\left(x-4\right)^2+\left(y+2\right)^2+11\ge11>0\)
\(\rightarrowĐPCM.\)
3)\(16x^2+6x+25\)
\(=16\left(x^2+\dfrac{3}{8}x+\dfrac{25}{16}\right)\)
\(=16\left(x^2+2.x.\dfrac{3}{16}+\dfrac{9}{256}-\dfrac{9}{256}+\dfrac{25}{16}\right)\)
\(=16\left[\left(x+\dfrac{3}{16}\right)^2+\dfrac{391}{256}\right]\)
\(=16\left(x+\dfrac{3}{16}\right)^2+\dfrac{391}{16}>0\)
-> ĐPCM.
4) Tương tự câu 3)
5) \(x^2+\dfrac{2}{3}x+\dfrac{1}{2}\)
\(=x^2+2.x.\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{9}+\dfrac{1}{2}\)
\(=\left(x+\dfrac{1}{3}\right)^2+\dfrac{7}{18}>0\)
-> ĐPCM.
6) Tương tự câu 5)
7) 8) 9) Tương tự câu 3).
\(P=27y^3+9y^2+y+\dfrac{1}{27}=\left(3y+3\right)^3\)
Với \(y=\dfrac{2}{3}\) ta có:
\(P=\left(3.\dfrac{2}{3}+3\right)^3=5^3=125\)
\(Q=x^2+4y^2-2x+10+4xy-4y\)
\(=\left(x^2-2x+4xy\right)+4y^2-4y+10\)
\(=\left[x^2-2x\left(1-2y\right)+\left(1-2y\right)^2\right]+4y^2-4y+10-\left(1-2y\right)^2\)\(=\left(x+2y-1\right)^2+4y^2-4y+10-1+4y-4y^2\)\(=\left(x+2y-1\right)^2+9\)
Với \(x+2y=5\) , ta có:
\(Q=\left(5-1\right)^2+9=25\)
Bài 1:
a: \(A=\dfrac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}=\dfrac{x^3\left(x+1\right)+\left(x+1\right)}{x^4-x^3+x^2+x^2-x+1}\)
\(=\dfrac{\left(x+1\right)\left(x^3+1\right)}{\left(x^2-x+1\right)\left(x^2+1\right)}=\dfrac{\left(x+1\right)^2}{x^2+1}\)
Để A=0 thì x+1=0
hay x=-1
b: \(B=\dfrac{x^4-5x^2+4}{x^4-10x^2+9}=\dfrac{\left(x^2-1\right)\left(x^2-4\right)}{\left(x^2-1\right)\left(x^2-9\right)}=\dfrac{x^2-4}{x^2-9}\)
Để B=0 thi (x-2)(x+2)=0
=>x=2 hoặc x=-2
Bài 1 . Chia :( x3 + 5x2 - 4x - 20) cho ( x2 + 3x - 10) ta được x+ 2
Chia :( x3 + 5x2 - 4x - 20) cho ( x2 + 7x + 10) ta được x - 2
Do đó , ta có :
\(\dfrac{1}{x^2+3x-10}=\dfrac{x+2}{\left(x^2+3x-10\right)\left(x+2\right)}=\dfrac{x+2}{x^3+5x^2-4x-20}\)
Và : \(\dfrac{x}{x^2+7x+10}=\dfrac{x\left(x-2\right)}{\left(x^2+7x+10\right)\left(x-2\right)}=\dfrac{x^2-2x}{x^3+5x^2-4x-20}\)
Bài 2 . a) Ta có :
\(\dfrac{x-1}{x^3+1}\)( giữ nguyên)
\(\dfrac{2x}{x^2-x+1}=\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{2x^2+2x}{x^3+1}\)
\(\dfrac{2}{x+1}=\dfrac{2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{2x^2-2x+2}{x^3+1}\)
b) Ta có MTC = x2( y - z)2
Ta có :
\(\dfrac{x+y}{x\left(y-z\right)^2}=\dfrac{x^2+xy}{x^2\left(y-z\right)^2}\)
\(\dfrac{y}{x^2\left(y-z\right)^2}\)( giữ nguyên )
\(\dfrac{z}{x^2}=\dfrac{z\left(y-z\right)^2}{x^2\left(y-z\right)^2}\)
Lời giải:
a) \(\frac{45x(3-x)}{15(x-3)^3}=\frac{-45x(x-3)}{15(x-3)^3}=\frac{-3x}{(x-3)^2}\)
b) \(\frac{36(x-2)^3}{32-16x}=\frac{36(x-2)^3}{-16(x-2)}=\frac{-9}{4}(x-2)^2\)
c) \(\frac{x^2-xy}{5y^2-5xy}=\frac{x(x-y)}{-5y(x-y)}=\frac{x}{-5y}\)
d) \(\frac{y^2-x^2}{x^3-3x^2y+3xy^2-y^3}=\frac{-(x^2-y^2)}{(x-y)^3}=\frac{-(x-y)(x+y)}{(x-y)^3}=\frac{-(x+y)}{(x-y)^2}\)
Lời giải:
a.
$27A=x^3-9x^2+162x-27=(x-3)^3+135x$
$=(303-3)^3+135.303=27040905$
$A=1001515$
b.
$B=2[(x+y)^3-3xy(x+y)]-3[(x+y)^2-2xy]$
$=2(1-3xy)-3(1-2xy)=2-6xy-3+6xy=-1$
c.
$C=x^3+y^3+3xy(x+y)=(x+y)^3=1^3=1$