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Xét : \(\frac{1}{\text{k}\left(\text{k}+1\right)}=\frac{1}{\text{k}}-\frac{1}{\text{k}+1}\)
\(\Leftrightarrow VT=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+...+\left(\frac{1}{n}-\frac{1}{n+1}\right)=1-\frac{1}{n+1}< 1\)
\(\Rightarrow\) ĐPCM
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n.\left(n+1\right)}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\)
\(=1-\frac{1}{n+1}\)
mà \(n\inℕ^∗\Rightarrow\frac{1}{n+1}>0\)
\(\Rightarrow1-\frac{1}{n+1}< 1\)
( đpcm )
Đây là toán lp 8 sao?
?_?
\(a,=64x^3-48x^2+12x-1-\left(64x^3+12x-48x^2-9\right)\)
\(=\left(64x^3-64x^3\right)+\left(48x^2-48x^2\right)+\left(12x-12x\right)+\left(9-1\right)\)
\(=8\) => ko phụ thuộc vào biến x
\(b,=2\left(x+y\right)\left(x^2-xy+y^2\right)-3\left(x^2+y^2\right)\)
thay x+y=1 vào
\(=2\left(x^2-xy+y^2\right)-3\left(x^2+y^2\right)\)
\(=2x^2-2xy+2y^2-3x^2-3y^2\)
\(=-\left(x^2+2xy+y^2\right)=-\left(x+y\right)^2=-1\) =>ko phụ thuộc vào biến
\(c,=x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-1\right)\)
\(=6x^2+2-6x^2+6=8\)
\(d,\frac{\left(2x+5\right)^2+\left(5x-2\right)^2}{x^2+1}=\frac{4x^2+20x+25+25x^2-20x+4}{x^2+1}=\frac{29\left(x^2+1\right)}{x^2+1}=29\)
\(x\left(x-1\right)+y\left(y-1\right)+z\left(z-1\right)< =\frac{4}{3}\Rightarrow x^2-x+y^2-y+z^2-z< =\frac{4}{3}\)
\(\Rightarrow3x^2-3x+3y^2-3y+3z^2-3z< =4\Rightarrow3\left(x^2+y^2+z^2\right)-3\left(x+y+z\right)< =4\)
\(3\left(x^2+y^2+z^2\right)=\left(1+1+1\right)\left(x^2+y^2+z^2\right)>=\left(x+y+z\right)^2\)
\(\Rightarrow\left(x+y+z\right)^2-3\left(x+y+z\right)< =3\left(x^2+y^2+z^2\right)-3\left(x+y+z\right)< =4\)
\(\Rightarrow\left(x+y+z\right)^2-3\left(x+y+z\right)< =4\Rightarrow\left(x+y+z\right)\left(x+y+z-3\right)< =4\)
\(x+y+z>4\Rightarrow x+y+z-3>1\Rightarrow\left(x+y+z\right)\left(x+y+z-3\right)>4\cdot1=4\)(loại)
\(x+y+z=4\Rightarrow x+y+z-3=1\Rightarrow\left(x+y+z\right)\left(x+y+z-3\right)=4\cdot1=4\left(tm\right)\)
\(x+y+z< 4\Rightarrow x+y+z-3< 1\Rightarrow\left(x+y+z\right)\left(x+y+z-3\right)< 4\cdot1=4\left(tm\right)\)
\(\Rightarrow x+y+z< =4\)thì \(\left(x+y+z\right)\left(x+y+z-3\right)< =4\)
dấu = xảy ra khi \(x=y=z=\frac{4}{3}\)
vậy \(x\left(x-1\right)+y\left(y-1\right)+z\left(z-1\right)< =\frac{4}{3}\Rightarrow x+y+z< =4\)dấu = xảy ra khi \(x=y=z=\frac{4}{3}\)
\(x,y,z>0\Rightarrow\left(x+y\right)+z>=2\sqrt{\left(x+y\right)z}\Rightarrow1>=2\sqrt{\left(x+y\right)z}\Rightarrow1>=4\left(x+y\right)z\)(bđt cosi)
\(M=\frac{x+y}{xyz}=\frac{1\left(x+y\right)}{xyz}>=\frac{4\left(x+y\right)z\left(x+y\right)}{xyz}=\frac{4\left(x+y\right)^2z}{xyz}>=4\cdot\frac{\left(2\sqrt{xy}\right)^2z}{xyz}=\frac{4\cdot4xyz}{xyz}=4\cdot4=16\)
dấu = xảy ra khi \(\hept{\begin{cases}x=y=\frac{1}{4}\\z=\frac{1}{2}\end{cases}}\)
vậy min M là 16 khi \(x=y=\frac{1}{4}:z=\frac{1}{2}\)
c)x2-2xy+y2+3x-3y-10
=(x-y)2+3(x-y)-10
=(x-y)2+2(x-y).3/2+9/4-49/4
=(x-y+3/2)2-(7/2)2
=(x-y+3/2+7/2)(x-y+3/2-7/2)
=(x-y+5)(x-y-2)
a Đặt \(x^2\)=t[t\(\ge\)0}
6t^2-11t+3=6t^2-3t-9t+3=2t[3t-1] -3[3t-1]=[3t-1][2t-3]=[3x^2-1][2x^2-3]
b Đặt x^2+x=t[t\(\ge\)0]
t^2+3t+2=[t+1][t+2]
Đến đó Dương làm tương tự như câu a nhé
\(\text{ab = -18 và a - b = 9}\)
\(=>\text{ab = -3 . 6 = 3 . -6}\)
\(\text{Vậy a + b với a = -3 ; b = 6 thì : }\)
\(\text{a + b = -3 + 6 = 3}\)
\(\text{Vậy a + b với a = 3 ; b = -6 thì : }\)
\(\text{a + b = 3 + -6 = -3}\)
\(\left(a+b\right)^2=\left(a-b\right)^2+4ab=9^2+4.\left(-18\right)=9\)
\(\Rightarrow a+b=3\)
1/
\(B=\frac{\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-3^{16}}{4}\)
\(=\frac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-3^{16}}{4}\)
\(=\frac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)-3^{16}}{4}\)
\(=\frac{\left(3^8-1\right)\left(3^8+1\right)-3^{16}}{4}\)
\(=\frac{3^{16}-1-3^{16}}{4}=\frac{-1}{4}\)
2/
a, (x-5)2-(x+3)2=1
<=>(x-5+x+3)(x-5-x-3)=1
<=>-16.(x-1)=1
<=>x-1=-1/16
<=>x=15/16
b, (2x-1)2-(2x-3)2=4
<=>(2x-1+2x-3)(2x-1-2x+3)=4
<=>-8(x-1)=4
<=>x-1=-1/2
<=>x=1/2
