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B1:
a, \(4x^2+y\left(y-4x\right)-9\)
\(=4x^2+y^2-4xy-9\)
\(=\left(x-y\right)^2-3^2\)
\(=\left(x-y+3\right)\left(x-y-3\right)\)
1.
b) \(a^2-b^2+a-b\)
\(=\left(a^2-b^2\right)+\left(a-b\right)\)
\(=\left(a-b\right)\left(a+b+1\right)\)
1)
a) \(x^3-5x^2+x-5=0\Rightarrow x^2.\left(x-5\right)+\left(x-5\right)\)
\(\Rightarrow\left(x^2+1\right).\left(x-5\right)=0\Rightarrow\orbr{\begin{cases}x^2+1=0\\x-5=0\end{cases}\Rightarrow}\orbr{\begin{cases}x^2=-1\left(sai\right)\\x=5\end{cases}}\)\(KL:x=5\)
b) \(x^4-2x^3+10x^2-20x=0\Rightarrow x^3.\left(x-2\right)+10x\left(x-2\right)\)
\(\Rightarrow\left(x-2\right).\left(x^3+10x\right)\Rightarrow\orbr{\begin{cases}x-2=0\\x^3+10x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x\left(x^2+10\right)=0\Rightarrow x=0\end{cases}}\)
Vì nếu x2 + 10 = 0 => x2 = -10 ( sai )
Vậy...
\(4.\)
\(a.A=5-8x-x^2\)
\(=-\left(16+8x+x^2\right)+21\)
\(=-\left(4+x\right)^2+21\le21\)
\(A_{max}=21\)
Dấu '='xảy ra khi \(x=-4\)
\(b.B=5-x^2+2x-4y^2-4y\)
\(=-\left(1-2x+x^2\right)-\left(4+4y+4y^2\right)+10\)
\(=-\left(1-x\right)^2-\left(2+2y\right)^2+10\le10\)
\(B_{max}=10\)
Dấu '=' xảy ra khi \(x=1;y=-1\)
\(5.\)
\(a.\) Ta có:\(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow a-b=0\Leftrightarrow a=b\left(1\right)\)
hay\(b-c=0\Leftrightarrow b=c\left(2\right)\)
hay\(c-a=0\Leftrightarrow c=a\left(3\right)\)
Từ \(\left(1\right),\left(2\right)\)và\(\left(3\right)\)suy ra:\(a=b=c\left(đpcm\right)\)
\(b.a^2-2a+b^2+4b+4c^2-4c+6=0\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2=0\)
\(\Leftrightarrow a-1=0\Leftrightarrow a=1\)
hay\(b+2=0\Leftrightarrow b=-2\)
hay\(2c-2=0\Leftrightarrow c=1\)
V...
^^
bạn phải tách từng câu ra. chứ kiểu này k ai trả lời cho đâu
2)
a)x2-y2=(x+y).(x-y)=(87+13).(87-13)=100.74=7400
b)x3-3x2+3x-1=(x-1)3=(101-1)3=1003=1000000
c)x3+9x2+27x+27=(x+3)3=(97+3)3=1003=1000000
4)
a)x2-6x+10=x2-6x+9+1=(x-3)2+1>=1>0 voi moi x
b)4x-x2-5= -(x2-4x+5)= -(x2-4x+4+1)= -(x-2)2 - 1<0 voi moi x
1)
\(\left(a+b\right)^5-a^5-b^5\)
\(=\left(a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5\right)-a^5-b^5\)
\(=5a^4b+10a^3b^2+10a^2b^3+5ab^4\)
\(=5ab\left(a^3+2a^2b+2ab^2+b^3\right)\)
\(=5ab.\left[\left(a^3+a^2b+ab^2\right)+\left(b^3+a^2b+ab^2\right)\right]\)
\(=5ab.\left[a.\left(a^2+ab+b^2\right)+b.\left(a^2+ab+b^2\right)\right]\)
\(=5ab.\left(a+b\right)\left(a^2+ab+b^2\right)\)
2)
\(\left(a+b\right)^7-a^7-b^7\)
\(=\left(a^7+7a^6b+21a^5b^2+35a^4b^3+35a^3b^4+21a^2b^5+7ab^6+b^7\right)-a^7-b^7\)
\(=7a^6b+21a^5b^2+35a^4b^3+35a^3b^4+21a^2b^5+7ab^6\)
\(=7ab.\left(a^5+3a^4b+5a^3b^2+5a^2b^3+3ab^4+b^5\right)\)
\(\ne7ab\left(a^3+2a^2b+2ab^2+b^3\right)\)
\(=7ab.\left[\left(a^3+a^2b+ab^2\right)+\left(b^3+a^2b+ab^2\right)\right]\)
\(=7ab.\left[a.\left(a^2+ab+b^2\right)+b.\left(a^2+ab+b^2\right)\right]\)
\(=7ab.\left(a+b\right)\left(a^2+ab+b^2\right)\)
\(a^2+b^2=a^2+b^2+2ab-2ab=\left(a+b\right)^2-2ab\)
Thay a + b = 5, ab = 1 vào biểu thức trên, ta được:
\(a^2+b^2=5^2-2.1=25-2=23\)\
=.= hk tốt!!
\(a^2+b^2=\left(a+b\right)^2-2ab=5^2-2.1=25-2=23\)\(23\)