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Bài 1:
Ta có:
\(P=\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)+1\)
\(P=\left[\left(a+1\right)\left(a+4\right)\right]\cdot\left[\left(a+2\right)\left(a+3\right)\right]+1\)
\(P=\left(a^2+5a+4\right)\left(a^2+5a+6\right)+1\)
Đặt \(x=a^2+5a+5\) , khi đó:
\(P=\left(a-1\right)\left(a+1\right)+1\)
\(P=a^2-1+1\)
\(P=a^2=\left(x^2-5x+5\right)^2\)
Mà \(a\inℤ\Rightarrow x^2-5x+5\inℤ\)
=> P là số chính phương
\(\left(xy+yz+zx\right)^2+\left(x^2-yz\right)^2+\left(y^2-zx\right)^2+\left(z^2-xy\right)^2=x^2y^2+y^2z^2+z^2x^2+2xyz\left(x+y+z\right)+x^4-2x^2yz+y^2z^2+y^4-2y^2zx+z^2x^2+z^4-2z^2xy+x^2y^2=x^4+y^4+z^4+2\left(x^2y^2+y^2z^2+z^2x^2\right)=\left(x^2+y^2+z^2\right)^2=100^2=10000\)
\(1.\)
\(a,\left(a+b\right)^2=a^2+2ab+b^2\)
\(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+2ab+b^2\)
\(\Rightarrow\left(a+b\right)^2=\left(a-b\right)^2+4ab\left(đpcm\right)\)
a) \(x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)(luôn dương)
b) \(x^2-x+\frac{1}{2}=x^2-x+\frac{1}{4}+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2+\frac{1}{4}>0\)(luôn dương)
Answer:
Câu 1:
\(\left(5x-x-\frac{1}{2}\right)2x\)
\(=\left(4x-\frac{1}{2}\right)2x\)
\(=4x.2x-\frac{1}{2}.2x\)
\(=8x^2-x\)
\(\left(x^3+4x^2+3x+12\right)\left(x+4\right)\)
\(=x\left(x^3+4x^2+3x+12\right)+4\left(x^3+4x^2+3x+12\right)\)
\(=x^4+4x^3+3x^2+12x+4x^3+16x^2+12x+48\)
\(=x^4+\left(4x^3+4x^3\right)+\left(3x^2+16x^2\right)+\left(12x+12x\right)+48\)
\(=x^4+8x^3+19x^2+24x+48\)
Ta thay \(x=99\) vào phân thức \(\frac{x^2+1}{x-1}\): \(\frac{\left(99\right)^2+1}{99-1}=\frac{9802}{98}=\frac{4901}{49}\)
Ta thay \(x=4\) vào phân thức \(\frac{x^2-x}{2\left(x-1\right)}\) : \(\frac{4^2-4}{2.\left(4-1\right)}=\frac{12}{6}=2\)
\(\left(x+y\right)^2-\left(x-y\right)^2\)
\(= (x²+2xy+y²)-(x²-2xy+y²)\)
\(= x²+2xy+y²-x²+2xy-y²\)
\(= 4xy\)
\(4x^2+4x+1=\left(2x+1\right)^2=\left(2.2+1\right)^2=25\)
Câu 2:
\(x^2+x=0\)
\(\Rightarrow x\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
\(x^2.\left(x-1\right)+4-4x=0\)
\(\Rightarrow x^2.\left(x-1\right)+4\left(1-x\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x^2-4\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\)
Trường hợp 1: \(x-1=0\Rightarrow x=1\)
Trường hợp 2: \(x-2=0\Rightarrow x=2\)
Trường hợp 3: \(x+2=0\Rightarrow x=-2\)
Câu 3: Bạn xem lại đề bài nhé.
\(A=100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(A=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)
\(A=1.199+1.195+...+3.1\)
\(A=3+7+...+195+199\)
Tổng A có: \(\frac{199-3}{4}+1=50\)( số hạng)
\(\Rightarrow A=\frac{\left(199+3\right).50}{2}=5050\)
Mấy ý kia chốc về lm nốt
\(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(B=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(B=\left(2^8-1\right)...\left(2^{64}+1\right)+1\)
\(B=2^{64}-1+1\)
\(B=2^{64}\)
=\(=\frac{2^2-1}{2^2}\cdot\frac{3^2-1}{3^2}\cdot...\cdot\frac{2009^2-1}{2009^2}=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot...\cdot\frac{2009\cdot2010}{2009\cdot2009}\)
Áp dụng HĐT ( a^2 - b^2)
\(=\frac{1.2.3.4.3.5.4.6....2009.2010}{2.2.3.3.4.4....2009.2009}=\frac{1.2010}{2.2009}=\frac{1005}{2009}\)
\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\cdot\cdot\cdot\frac{2009^2-1}{2009^2}=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3\cdot5}{4.4}\cdot\cdot\cdot\cdot\frac{2008.2010}{2009}=\frac{1}{2}\cdot\frac{2010}{2009}=\frac{1005}{2009}\)
2, \(\frac{x^2}{2}+\frac{y^2}{3}+\frac{z^2}{4}=\frac{x^2+y^2+z^2}{5}\)
<=>\(\left(\frac{x^2}{2}-\frac{x^2}{5}\right)+\left(\frac{y^2}{3}-\frac{y^2}{5}\right)+\left(\frac{z^2}{4}-\frac{z^2}{5}\right)=0\)
<=>\(\frac{3}{10}x^2+\frac{2}{15}y^2+\frac{1}{20}z^2=0\)
<=>x=y=z=0
4,
a, \(\frac{1}{x\left(x^2+1\right)}=\frac{a}{x}+\frac{bx+c}{x^2+1}\)
=>\(\frac{1}{x\left(x^2+1\right)}=\frac{ax^2+a+bx^2+cx}{x\left(x^2+1\right)}=\frac{\left(a+b\right)x^2+cx+a}{x\left(x^2+1\right)}\)
Đồng nhất 2 phân thức ta được:
\(\hept{\begin{cases}a+b=0\\c=0\\a=1\end{cases}\Leftrightarrow\hept{\begin{cases}b=-1\\c=0\\a=1\end{cases}}}\)
b,a=1/4,b=-1/4
c, a=-1,b=1,c=1