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Ta có: 5x2 + 5y2 + 8xy - 2x + 2y + 2 = 0
\(\Leftrightarrow\)(4x2 + 8xy + 4y2) + (x2 - 2x + 1) + (y2 + 2y + 1) = 0
\(\Leftrightarrow\)(2x + 2y)2 + (x - 1)2 + (y + 1)2 = 0
\(\Leftrightarrow\)\(\hept{\begin{cases}2x+2y=0\\x-1=0\\y+1=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x+y=0\\x=1\\y=-1\end{cases}}\)
Thay x = 1; y = -1; x + y = 0 vào M ta được:
M = 0 + (1 + 2)2008 + ( - 1 + 1)2009
= 0 + 32008 + 0 = 32008
\(a,\)\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)\(\Leftrightarrow14+2\left(ab+bc+ac\right)=0\)\(\Rightarrow\left(ab+bc+ac\right)^2=49\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=49\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2=49\)
Ta có: \(a^2+b^2+c^2=14\Rightarrow\left(a^2+b^2+c^2\right)=196\)\(\Leftrightarrow a^{^{ }4}+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=196\)\(\Leftrightarrow\)\(a^4+b^4+c^4=98\)
Ta có:
\(\frac{x}{x^2+x+1}=-\frac{1}{4}\Rightarrow x^2+x+1=-4x\)
\(\Rightarrow x^2+5x+1=0\Rightarrow x^2=5x+1\)
Với x2=5x+1 ta được:
\(P=\frac{2x\left(5x+1\right)^2+10\left(5x+1\right)^2+2x\left(5x+1\right)-7\left(5x+1\right)-35x+2009}{2029+60x+11\left(5x+1\right)-5x\left(5x+1\right)-\left(5x+1\right)^2}\)
\(P=\frac{2x\left(25x^2+10x+1\right)+10\left(25x^2+10x+1\right)+10x^2+2x-35x-7-35x+2009}{2029+60x+55x+11-25x^2-5x-\left(25x^2+10x+1\right)}\)
\(P=\frac{50x^3+20x^2+2x+250x^2+100x+10+10x^2+2x-35x-7-35x+2009}{2029+60x+55x+11-25x^2-5x-25x^2-10x-1}\)
\(P=\frac{50x^3+280x^2+34x+2012}{2039+100x-50x^2}\)
\(P=\frac{50x\left(5x+1\right)+280\left(5x+1\right)+34x+2012}{2039+100x-50\left(5x+1\right)}\)
\(P=\frac{250x^2+50x+1400x+280+34x+2012}{2039+100x-250x-50}\)
\(P=\frac{250\left(5x+1\right)+50x+1400x+280+34x+2012}{1989-150x}\)
\(P=\frac{1250x+250+50x+1400x+280+34x+2012}{1989-150x}\)
\(.\)M= bn ghi lại đề nha ^.^
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\left[\left(a^2+2ab+b^2\right)-2ab\right]+6a^2b^2\left(a+b\right)\)
\(=1^3-3ab.1+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2.1\)
\(=1-3ab+3ab\left(1-2ab\right)+6a^2b^2\)
\(M=1-3ab+3ab-6a^2b^2+6a^2b^2\)\(=1\)
k cho mình nha bn thanks nhìu <3 <3 (^3^)
2. \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)(1)
Đặt \(x^2+5x+4=t\)
(1) = \(t.\left(t+2\right)-24\)
\(=t^2+2t+1-25\)
\(=\left(t+1\right)^2-25\)
\(=\left(t+1-5\right)\left(t+1+5\right)\)
\(=\left(t-4\right)\left(t+6\right)\)(2)
Thay \(t=x^2+5x+4\)vào (2) ta có:
(2) = \(\left(x^2+5x+4-4\right)\left(x^2+5x+4+6\right)\)
\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)\(=x\left(x+5\right)\left(x^2+5x+10\right)\)
k mình nha bn <3 thanks
1/Ta có: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)=81\)
\(\Rightarrow M=ab+bc+ca=\frac{\left(81-141\right)}{2}\)
Bài 1:
\(Q=x^4+2x^2+2\left(x^2+1\right)\left(x^2+6x-1\right)+\left(x^2+6x-1\right)^2\)
\(Q=\left[\left(x^2+6x-1\right)^2+2\left(x^2+6x-1\right)\left(x^2+1\right)+\left(x^4+2x^2+1\right)\right]-1\)
\(Q=\left[\left(x^2+6x-1\right)^2+2\left(x^2-6x+1\right)\left(x^2+1\right)+\left(x^2+1\right)^2\right]-1\)
\(Q=\left(x^2+6x-1+x^2+1\right)^2-1\)
\(Q=\left(2x^2+6x\right)^2-1\)
\(Q=99^2-1\)
\(Q=9800\)
Bài 2:
Đặt \(A=\left(2+1\right)\left(2^2+1\right)...\left(x^{64}+1\right)+1\)
\(\left(2-1\right)\cdot A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)...\left(2^{64}+1\right)+1\)
\(1\cdot A=\left(2^2-1\right)\left(2^2+1\right)...\left(2^{64}+1\right)+1\)
\(A=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(A=\left(2^{64}-1\right)\left(2^{64}+1\right)+1\)
\(A=2^{128}-1^2+1\)
\(A=2^{128}\left(đpcm\right)\)
Bài 3:
Để C là số nguyên thì x2 - 3 ⋮ x - 2
<=> x (x - 2) + 2x - 3 ⋮ x - 2
mà x (x - 2) ⋮ x - 2
=> 2x - 3 ⋮ x - 2
<=> 2 (x - 2) + 3 ⋮ x - 2
mà 2 (x - 2) ⋮ x - 2
=> 3 ⋮ x - 2
=> x - 2 thuộc Ư(3) = { 1; 3; -1; -3 }
Ta có bảng :
| x-2 | 1 | 3 | -1 | -3 |
| x | 3 | 5 | 1 | -1 |
Vậy x thuộc { -1; 1; 3; 5 }
A = 12 - 22 + 32 - 42 + . . . - 20082 + 20092
= (12 - 22) + (32 - 42) + ... + (20072 - 20082) + 20092
= -3 + (-7) + ... + (-4015) + 20092
Bí rồi bạn ơi, cái đề mình thấy sai sai hay mình sai ta?
\(A=1^2-2^2+3^2-4^2+...+2009^2\)
\(=1+\left(3^2-2^2\right)+\left(5^2-4^2\right)+...+\left(2009^2-2008^2\right)\)
\(=1+\left(3-2\right)\left(2+3\right)+\left(5-4\right)\left(4+5\right)+...+\left(2009-2008\right)\left(2008+2009\right)\)
\(=1+\left(2+3\right)+\left(4+5\right)+...+\left(2008+2009\right)\)
\(=1+2+3+4+...+2009\)
\( =\frac{\left(2009+1\right)2009}{2}=2019045\)