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a)
$A=(1^2-2^2)+(3^2-4^2)+....+(2003^2-2004^2)+2005^2$
$=(1-2)(1+2)+(3-4)(3+4)+....+(2003-2004)(2003+2004)+2005^2$
$=-(1+2)-(3+4)-...-(2003+2004)+2005^2$
$=-(1+2+3+...+2004)+2005^2=-\frac{2004.2005}{2}+2005^2$
$=2005^2-1002.2005=2005(2005-1002)=2011015$
b)
$B=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^8-1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^{16}-1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^{32}-1)(2^{32}+1)-2^{64}$
$=2^{64}-1-2^{64}=-1$
c) Do $x=16$ nên $x-16=0$
$R(x)=x^4-17x^3+17x^2-17x+20$
$=(x^4-16x^3)-(x^3-16x^2)+x^2-16x-x+20$
$=x^3(x-16)-x^2(x-16)+x(x-16)-x+20$
$=x^3.0-x^2.0+x.0-x+20=-x+20=-16+20=4$
d) Do $x=12$ nên $x-12=0$. Khi đó:
$S(x)=(x^{10}-12x^9)-(x^9-12x^8)+(x^8-12x^7)-....+(x^2-12x)-x+10$
$=x^9(x-12)-x^8(x-12)+x^7(x-12)-....+x(x-12)-x+10$
$=(x-12)(x^9-x^8+x^7-....+x)-x+10$
$=0-x+10=-x+10=-12+10=-2$
Bài 1:
\(a, \dfrac{1}{2}x(2-x)=x-\dfrac{1}{2}x^2\)
\(b, \dfrac{x-5}{5-x}\)\(=-\dfrac{x-5}{x-5}\)\(=-1\)
Bài 2:
\(a, x+y-x^2+y^2=(x+y)-(x^2-y^2)=(x+y)-(x-y)(x+y)\)
\(=(x+y)(1-x+y)\)
\(b, x(x-3)+3x-1=0 \)
\(⇔x^2-3x+3x-1=0 \)
\(⇔x^2-1=0 \)
\(⇔(x-1)(x+1)=0 \)
\(⇔\left[\begin{array}{} x-1=0\\ x+1=0 \end{array}\right.\)
\(⇔\left[\begin{array}{} x=1\\ x=-1 \end{array}\right.\)
Bài 3:
\(a,A=\dfrac{x(x+2)-x(x-2)+8}{x^2-4}:\dfrac{4}{x-2}\)
\(A=\dfrac{4x+8}{(x-2)(x+2)}.\dfrac{x-2}{4}\)
\(A=\dfrac{4(x+2)}{(x-2)(x+2)}.\dfrac{x-2}{4}\)
\(A=1\)
\(b, B=(1-\dfrac{a+b}{a-b})(1-\dfrac{2b}{a+b})\)
\(B=\dfrac{-2b}{a-b}.\dfrac{a-b}{a+b}\)
\(B=\dfrac{-2b}{a+b}\)
Bài 4:
\(C=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)\)
\(C=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)\)
\(C=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)\)
\(C=(2^8-1)(2^8+1)(2^{16}+1)(2^{32}+1)\)
\(C=(2^{16}-1)(2^{16}+1)(2^{32}+1)\)
\(C=(2^{32}-1)(2^{32}+1)=2^{64}-1\)
b) Đặt \(x-7=a\) ta có:
\(\left(a+1\right)^4+\left(a-1\right)^4=16\)
\(\Leftrightarrow\)\(a^4+4a^3+6a^2+4a+1+a^4-4a^3+6a^2-4a+1=16\)
\(\Leftrightarrow\)\(2a^4+12a^2+2-16=0\)
\(\Leftrightarrow\)\(2\left(a^4+6a^2-7\right)=0\)
\(\Leftrightarrow\)\(a^4+6a^2-7=0\)
\(\Leftrightarrow\)\(\left(a-1\right)\left(a+1\right)\left(a^2+7\right)=0\)
Vì \(a^2+7>0\) nên \(\orbr{\begin{cases}a-1=0\\a+1=0\end{cases}}\)
Thay trở lại ta có: \(\orbr{\begin{cases}x-8=0\\x-6=0\end{cases}}\) \(\Leftrightarrow\)\(\orbr{\begin{cases}x=8\\x=6\end{cases}}\)
Vậy...
câu a tự quy đồng cùng mẫu rồi làm thôi :"))
b) \(\left[x.\left(x-1\right)\right].\left[\left(x-2\right).\left(x+1\right)\right]=24\)
\(\Leftrightarrow\left(x^2-x\right).\left(x^2-x-2\right)=24\)
Đặt \(x^2-x=k\), ta có:
\(k.\left(k-2\right)=24\)
\(\Leftrightarrow k^2-2k+1=25\)
\(\Leftrightarrow\left(k-1\right)^2=5^2\Leftrightarrow\orbr{\begin{cases}k-1=5\\k-1=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}k=6\\k=-4\end{cases}}}\)
\(k=6\Rightarrow x^2-x=6\Rightarrow x^2-x-6=0\)
\(\Rightarrow x^2-3x+2x-6=0\Rightarrow x.\left(x-3\right)+2.\left(x-3\right)=0\)
\(\Rightarrow\left(x+2\right).\left(x-3\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
\(k=-4\Rightarrow x^2-x+4=0\Rightarrow x^2-x+\frac{1}{4}+\frac{15}{4}=0\Rightarrow\left(x-\frac{1}{2}\right)^2=-\frac{15}{4}\left(\text{loại}\right)\)
c)\(x^4+2x^3+5x^2+4x-12=0\)
\(\Leftrightarrow x^4+2x^3+2x^2+4x+3x^2-12=0\)
\(\Leftrightarrow x^3.\left(x+2\right)+2x.\left(x+2\right)+3.\left(x^2-2^2\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(x^3+5x-6\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(x^3-x^2+x^2-x+6x-6\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left[x^2.\left(x-1\right)+x.\left(x-1\right)+6.\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right).\left(x-1\right).\left(x^2+x+6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}\text{vì }x^2+x+6>0\left(\text{tự c/m}\right)}\)
p/s: bn tự kết luận nha :))
a, \(\Leftrightarrow\left(x+1+x-2\right)\left[\left(x+1\right)^2-\left(x+1\right)\left(x-2\right)+\left(x-2\right)^2\right]-\left(2x-1\right)^3=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2+2x+1-x^2+x+2+x^2-4x+4\right)-\left(2x-1\right)^3=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2-x+7-\left(2x-1\right)^2\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2-x+7-4x^2+4x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(-3x^2+3x+6\right)=0\)
\(\Leftrightarrow-3\left(2x-1\right)\left(x^2-x-2\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x+1\right)\left(x-2\right)=0\)
=>x=1/2 hoặc x=-1 hoặc x=2
Vậy pt có tập nghiệm là S={1/2;-1;2}
b, \(x^4=24x+32\Leftrightarrow x^4-24x-32=0\)
\(\Leftrightarrow x^4-2x^3-4x^2+2x^3-4x^2-8x+8x^2-16x-32=0\)
\(\Leftrightarrow x^2\left(x^2-2x-4\right)+2x\left(x^2-2x-4\right)+8\left(x^2-2x-4\right)=0\)
\(\Leftrightarrow\left(x^2-2x-4\right)\left(x^2+2x+8\right)=0\)
\(\Leftrightarrow x^2-2x-4=0\) (vì x^2+2x+8 > 0)
\(\Leftrightarrow\left(x-1\right)^2-5=0\Leftrightarrow\left(x-1\right)^2=5\Leftrightarrow x-1=\pm\sqrt{5}\Leftrightarrow x=1\pm\sqrt{5}\)
Vậy...
c, \(\left(x-6\right)^4+\left(x-8\right)^4=16\)
Đặt x-6=t => x-8=t-2
Ta có: \(t^4+\left(t-2\right)^4=16\Leftrightarrow t^4+t^4-8t^3+24t^2-32t+16=16\)
\(\Leftrightarrow2t^4-8t^3+24t^2-32t=0\Leftrightarrow t^4-4t^3+12t^2-16t=0\)
\(\Leftrightarrow t^4-2t^3-2t^3+4t^2+8t^2-16t=0\)
\(\Leftrightarrow t^3\left(t-2\right)-2t^2\left(t-2\right)+8t\left(t-2\right)=0\)
\(\Leftrightarrow\left(t-2\right)\left(t^3-2t^2+8t\right)=0\Leftrightarrow\left(t-2\right)t\left(t^2-2t+8\right)=0\)
Mà t^2-2t+8=(t-1)^2+7 > 0
\(\Rightarrow\orbr{\begin{cases}t-2=0\\t=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x-6-2=0\\x-6=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=8\\x=6\end{cases}}}\)
Vậy...
a)(x+1)(x+2)(x+3)(x+4)+1
=(x+1)(x+4)(x+2)(x+3)+1
=(x2+5x+4)(x2+5x+6)+1
Đặt a=(x2+5x+4) thì (x2+5x+4)(x2+5x+6)+1
= a.(a+2)+1
=a2+2a+1
=(a+1)2
Thay: =(x2+5x+4+1)2
=(x2+5x+5)2
b)(x+2)(x+4)(x+6)(x+8)+16
=(x+2)(x+8)(x+4)(x+6)+16
=(x2+10x+16)(x2+10x+24)+16
Đặt a=(x2+10x+16) thì (x2+10x+16)(x+5x+24)+1
= a.(a+8)+16
=a2+8x+16
=(a+4)2
Thay: =(x2+10x+16+4)2
=(x2+5x+20)2
a)(x+1)(x+2)(x+3)(x+4)+1
=[(x+1)(x+4][(x+2)(x+3)]+1
=(x2+5x+4)(x2+5x+6)+1
Đặt a=(x2+5x+4)
Ta có: (x2+5x+4)(x2+5x+6)+1
= a.(a+2)+1
=a2+2a+1
=(a+1)2
=(x2+5x+4+1)2
=(x2+5x+5)2
b)(x+2)(x+4)(x+6)(x+8)+16
=(x+2)(x+8)(x+4)(x+6)+16
=(x2+10x+16)(x2+10x+24)+16
Đặt a=(x2+10x+16)
Ta có:(x2+10x+16)(x+5x+24)+1
= a.(a+8)+16
=a2+8x+16
=(a+4)2
=(x2+10x+16+4)2
=(x2+5x+20)2
Mk yêu bé Shin-Conan lém