cho tam giác ABC vuông tại A.Đường cao AH cắt đường phân giác BD tại I. chứng minh rằng
a,IA.BH=IH.BA
b,IA=ID
c,\(\frac{HI}{IA}=\frac{AD}{DC}\)
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M=x^2-2/x^2+1
=x^2+1-3/x^2+1
=1- 3/x^2+1
M đạt gtnn khi 3/x^2+1 đạt gtln
=>x^2+1=1
=>x^2=0
=>x=0.
Khi đó M=1- 3/1+1 = 1-3+1 = -2+1 = -1
Vậy Mmin=-1 khi x=0
\(4.\)
\(a.A=5-8x-x^2\)
\(=-\left(16+8x+x^2\right)+21\)
\(=-\left(4+x\right)^2+21\le21\)
\(A_{max}=21\)
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) Ta có: \(A=4x^2+4x+11\)
\(\Rightarrow A=4x^2+2x+2x+11\)
\(\Rightarrow A=2x.\left(2x+1\right)+\left(2x+1\right)+10\)
\(\Rightarrow A=\left(2x+1\right).\left(2x+1\right)+10\)
\(\Rightarrow A=\left(2x+1\right)^2+10\)
Ta lại có: \(\left(2x+1\right)^2\ge0\forall x\inℝ\)
\(\Rightarrow A\ge10\)
Dấu "=" xảy ra \(\Leftrightarrow\left(2x+1\right)^2=0\)
\(\Rightarrow2x+1=0\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=\frac{-1}{2}\)
Vậy \(A_{min}=10\Leftrightarrow x=\frac{-1}{2}\)
a, a^3 + b^3=(a + b)^3 - 3a2b - 3ab2=(a + b)^3 - 3ab(a + b)
b, a^3 + b^3 + c^3 - 3abc= (a + b)^3 + c3 - 3ab(a + b)-3abc
=(a + b + c)\([\)(a + b)2- (a + b)c +c2\(]\)- 3ab(a + b + c)
=(a + b + c)(a2 + 2ab + b2 - ac - bc + c2 - 3ab)
=(a + b + c)(a2 + b2 + c2 - ab - bc- ca)
\(a.A=100^2-99^2+98^2-97^2+...+2^2-1\)
\(=100+99+98+97+...+2+1\)
\(=\frac{\left(100+1\right).100}{2}=5050\)(công thức tính dãy số hạng)
\(b.B=3\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^{64}-1\right)\left(2^{64}+1\right)+1\)
\(=2^{4096}-1+1\)
\(=2^{4096}\)
\(c.\)Đặt\(a+b=d\)
Thay vào \(C\)ta được:
\(C=\left(d+c\right)^2+\left(d-c\right)^2-2d^2\)
\(=d^2+2dc+c^2+d^2-2dc+c^2-2d^2\)
\(=2c^2\)
Ta có: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)
\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ca\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2ab+2bc+2ca\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(\frac{3}{2}\right)^2\Leftrightarrow a^2+b^2+c^2\ge\frac{3}{4}\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}a=b=c\\a+b+c=\frac{3}{2}\end{cases}\Leftrightarrow}a=b=c=\frac{1}{2}\)
\(\Leftrightarrow\)(x2+1)2+x(x2+1)+2x(x2+1)+2x2=0
\(\Leftrightarrow\)(x2+1)(x2+1+x)+2x(x2+1+x)=0
\(\Leftrightarrow\)(x2+1+x)(x2+1+2x)=0
\(\Leftrightarrow\)(x2+x+1)(x+1)2=0
Vì x2+x+1=(x2+x+\(\frac{1}4\))+\(\frac{3}4\)=(x+\(\frac{1}2\))2+\(\frac{3}{4}\)\(\ge\)\(\frac{3}{4} \)>0 nên x+1=0, x=1.
\(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)
\(x^4+2x^2+1+3x^3+3x+2x^2=0\)
\(x^4+3x^3+4x^2+3x+1=0\)
\(x^4+2x^3+x^3+2x^2+2x^2+x+2x+1=0\)
\(\left(x+1\right)\left(x^3+2x^2+1+2x\right)=0\)
\(\left(x+1\right)\left(x+1\right)\left(x^2+x+1\right)=0\)
\(\left(x+1\right)^2\left(x^2+x+1\right)=0\)
\(\orbr{\begin{cases}x+1=0\Rightarrow x=\left(-1\right)\\x^2+x+1=0\end{cases}}\)
x^2 +x +1 =0 (vô lí)
\(\Rightarrow x=\left(-1\right)\)