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Câu 9.
a) Ta có: \(\left(a-1\right)^2\ge0\)(điều hiển nhiên)
\(\Leftrightarrow a^2-2a+1\ge0\)
\(\Leftrightarrow a^2+2a+1\ge4a\)
\(\Leftrightarrow\left(a+1\right)^2\ge4a\left(đpcm\right)\)
b) Áp dụng BĐT Cauchy cho 2 số không âm:
\(a+1\ge2\sqrt{a}\)
\(b+1\ge2\sqrt{b}\)
\(c+1\ge2\sqrt{c}\)
\(\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge8\sqrt{abc}=8\)(Vì abc = 1)
Câu 10.
a) Ta có: \(-\left(a-b\right)^2\le0\)(điều hiển nhiên)
\(\Leftrightarrow-a^2+2ab-b^2\le0\)
\(\Leftrightarrow a^2+2ab+b^2\le2a^2+2b^2\)
\(\Leftrightarrow\left(a+b\right)^2\le2\left(a^2+b^2\right)\)
b) \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
Có: \(2ab\le a^2+b^2;2bc\le b^2+c^2;2ac\le a^2+c^2\)(BĐT Cauchy)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac\le3\left(a^2+b^2+c^2\right)\)
Vậy \(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)
\(4.\)
\(a.A=5-8x-x^2\)
\(=-\left(16+8x+x^2\right)+21\)
\(=-\left(4+x\right)^2+21\le21\)
\(A_{max}=21\)
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...
^^
\(A=2\left(a^3+b^3\right)-3\left(a^2+b^2\right)\)
\(\Leftrightarrow A=2\left(a+b\right)\left(a^2-ab+b^2\right)-3\left(a^2+b^2\right)\)
\(\Leftrightarrow A=2\left(a^2-ab+b^2\right)-3\left(a^2+b^2\right)\)
\(\Leftrightarrow A=2a^2-2ab+2b^2-3a^2-3b^2\)
\(\Leftrightarrow A=-a^2-2ab-b^2\)
\(\Leftrightarrow A=-\left(a+b\right)^2\)
#)Giải :
Ta có : \(2\left(a^3+b^3\right)-3\left(a^2+b^2\right)\)
\(=2\left(\left(a+b\right)^3-3ab\left(a+b\right)\right)-3\left(\left(a+b\right)^2-2ab\right)\)
Thay a + b = 1 vao biểu thức, ta được :
\(2\left(1^3-3ab.1\right)-3\left(1^2-2ab\right)\)
\(=2\left(1-3ab\right)-3\left(1-2ab\right)\)
\(=2-6ab-3+6ab\)
\(=-1\)
Vậy \(A=-1\) khi a + b = 1
6) c) x3 - x2 + x = 1
<=> x3 - x2 + x - 1 = 0
<=> (x3 - x2) + (x - 1) = 0
<=> x2 (x - 1) + (x - 1) = 0
<=> (x - 1) (x2 + 1) = 0
=> x - 1 = 0 hoặc x2 + 1 = 0
* x - 1 = 0 => x = 1
* x2 + 1 = 0 => x2 = -1 => x = -1
Vậy x = 1 hoặc x = -1
Bài 5:
a) Đặt \(A=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=3^{32}-1\)
\(\Rightarrow A=\frac{3^{32}-1}{8}\)
b) (7x+6)2 + (5-6x)2 - (10-12x)(7x+6)
=(7x+6)2 + (5-6x)2 - 2(5-6x)(7x+6)
\(=\left(7x+6-5+6x\right)^2\)
\(=\left(13x+1\right)^2\)
a) \(A=100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=100+99+98+97+...+2+1\)
\(=\frac{\left(1+100\right).100}{2}=5050\)
b) \(B=3\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(=\left(4-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(=\left[\left(2^2-1\right)\left(2^2+1\right)\right]\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right).....\left(2^{64}+1\right)+1\)
Cứ tương tự như thế ......
\(B=2^{128}-1+1=2^{128}\)
c) \(C=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2+2ab-2bc-2ac-2\left(a^2+2ab+b^2\right)\)
\(=2a^2+2b^2+2c^2+4ab-2a^2-4ab-2b^2\)
\(=2c^2\)
Vậy C = 2c2
TA có \(\left(a+b+c\right)^2=0\Rightarrow ab+bc+ca=-\frac{1}{2}\Rightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\frac{1}{4}\)
=> \(a^2b^2+b^2c^2+c^2a^2=\frac{1}{4}\)
Mà \(\left(a^2+b^2+c^2\right)^2=1\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=1\)
=> \(a^4+b^4+c^4=\frac{1}{2}\)
^_^
Ta có: a+b+c=0 <=> (a+b+c)2=0 <=> a2+b2+c2+ 2( ab+ac+bc)=0 <=> 2(ab+ac+bc)= -1 ( vì a2+b2+c2=1) <=> ab+ac+bc= -1/2
=> (ab+ac+bc)2= 1/4 <=> a2b2+a2c2+b2c2+2abc(a+b+c)= 1/4 <=> 2(a2b2+a2c2+b2c2)= 1/2 ( vì a+b+c=0) (*)
Lại có: a2+b2+c2=1 <=> (a2+b2+c2)2=1 <=> a4+b4+c4+2(a2b2+a2c2+b2c2)=1 <=> a4+b4+c4= 1/2 ( vì (*))
Vậy,...