ai giải thích cho em nguyên lí bù trừ với ak
\(\left|A\text{h}ợpB\text{h}ợpC\right|=\left|A\right|+\left|B\right|+\left|C\right|-\left|AgiaoB\right|-\left|BgiaoC\right|-\left|CgiaoA\right|+\left|AgiaoBgiaoC\right|\)
Mạnh mẽ hơn Nesbitt?
Với a, b, c là các số thực sao cho: \(a+b+c>0,\text{ }ab+bc+ca>0,\text{ }\left(a+b\right)\left(b+c\right)\left(c+a\right)>0\) thì:
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-\frac{3}{2}\ge\left(\Sigma ab\right)\left(\Sigma\frac{1}{\left(a+b\right)^2}\right)-\frac{9}{4}\)
Chứng minh: \(4\left(a+b+c\right)\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\cdot\left(\text{VT}-\text{VP}\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(c+a\right)\left[\Sigma\left(ab+bc-2ca\right)^2+\left(ab+bc+ca\right)\Sigma\left(a-b\right)^2\right]\)
\(+\left(a+b+c\right)\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2\ge0\)
Bất đẳng thức trên đúng với mọi số thực a, b, c. Ai có thể chứng minh?
a, \(\text{[}\left(x-y\right)^3+3\left(x-y\right)\text{]}:\dfrac{1}{3}\left(x-y\right)\)
b, \(\left(8x^3-27y^3\right):\left(2x-3y\right)\)
c, \(\text{[}5\left(x+2y\right)^6-6\left(x+2y\right)^5\text{]}:2\left(x+2y\right)^4\)
a: \(=\left(x-y\right)^3:\dfrac{1}{3}\left(x-y\right)+3\left(x-y\right):\dfrac{1}{3}\left(x-y\right)\)
=3(x-y)^2+9
b: \(=\dfrac{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}{2x-3y}=4x^2+6xy+9y^2\)
c: \(=\dfrac{5\left(x+2y\right)^6}{2\left(x+2y\right)^4}-\dfrac{6\left(x+2y\right)^5}{2\left(x+2y\right)^4}=\dfrac{5}{2}\left(x+2y\right)^2-3\left(x+2y\right)\)
Bài 31 : Tính :
a) \(\left(2^{-1}+3^{-1}\right):\left(2^{-1}-3^{-1}\right)+\left(2^{-1}.2^0\right):2^3\)
b) \(\left(-\frac{1}{3}\right)^{-1}-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2\)
c) \(\text{[}\left(0,1\right)^2\text{]}^0+\text{[}\left(\frac{1}{7}\right)^1\text{]}^2.\frac{1}{49}.\text{[}\left(2^3\right)^3:2^5\text{]}\)
Mong các cao nhân giúp ak , đang cần gấp
a) \(\frac{81}{16}\)
b) \(\frac{-31}{8}\)
c) \(\frac{2417}{2401}\)
Bài 31:
a) \(\left(2^{-1}+3^{-1}\right):\left(2^{-1}-3^{-1}\right)+\left(2^{-1}.2^0\right):2^3\)
\(=\left(\frac{1}{2}+\frac{1}{3}\right):\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{2}.1\right):8\)
\(=\frac{5}{6}:\frac{1}{6}+\frac{1}{2}:8\)
\(=5+\frac{1}{16}\)
\(=\frac{81}{16}.\)
c) \(\left[\left(0,1\right)^2\right]^0+\left[\left(\frac{1}{7}\right)^1\right]^2.\frac{1}{49}.\left[\left(2^3\right)^3:2^5\right]\)
\(=1+\frac{1}{49}.\frac{1}{49}.16\)
\(=1+\frac{1}{2401}.16\)
\(=1+\frac{16}{2401}\)
\(=\frac{2417}{2401}.\)
Chúc bạn học tốt!
Cho a, b, c là ba số dương thỏa mãn: \(\dfrac{\text{2b+c-a}}{a}=\dfrac{\text{2c-b+a}}{b}=\dfrac{\text{ 2a+b-c}}{c}\)
Tính giá trị biểu thức: P = \(\dfrac{\left(3a-2b\right)\left(3b-2c\right)\left(3a-2c\right)}{\left(3a-c\right)\left(3b-a\right)\left(3c-b\right)} \)
Vì \(a,b,c>0\Rightarrow a+b+c\ne0\)
Áp dụng tc dtsbn:
\(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\\ \Rightarrow\left\{{}\begin{matrix}2b+c-a=2a\\2c-b+a=2b\\2a+b-c=2c\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3a-2b=c\\3b-2c=a\\3c-2a=b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3a-c=2b\\3b-a=2c\\3c-b=2a\end{matrix}\right.\\ \Rightarrow P=\dfrac{abc}{2a\cdot2b\cdot2c}=\dfrac{1}{8}\)
Gọi a, b, c là 3 số đôi một nguyên tố cùng nhau.Chứng minh rằng:
a)\(\text{ƯCLN}\left(a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right),\left(ab\right)\left(bc\right)\left(ca\right)\right)=1\)
b)\(\text{ƯCLN}\left(ab+bc+ca,abc\right)=1\)
Giải các phương trình sau:
a \(\left(x+2\right)\left(x+\text{4}\right)\left(x+6\right)\left(x+8\right)+16=0\)
b \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)
c \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4=0\)
d \(\left(x^2-3x+2\right)\left(x^2+15x+56\right)+8=0\)
b: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24=0\)
\(\Leftrightarrow x^2+7x+6=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)
Cho a,b,c là các số thực bất kì. CMR:
\(\left|a\right|+\left|b\right|+\left|c\right|+\left|a+b+c\right|\ge\left|a+b\right|+\left|b+c\right|+\left|c+a\right|\)
Một bạn nào đó (không biết rõ tên) đã giải như sau:
Ta có: \(\left|a\right|+\left|b\right|+\left|c\right|+\left|a+b+c\right|\ge\left|a+b\right|+\left|a+c\right|+\left|b\right|+\left|c\right|\)
\(\ge\left|a+b\right|+\left|b+c\right|+\left|c+a\right|\)
Em hãy cho nhận xét về lời giải trên! Nếu sai hãy sửa lại cho đúng!
\(P=\sqrt{3a^2+2ab+3b^2}+...+\sqrt{3c^2+2ac+3a^2}\)
ta có:
\(\sqrt{3a^2+2ab+3b^2}\ge\sqrt{2}\left(a+b\right)\Leftrightarrow3a^2+2ab+3b^2\ge2a^2+4ab+2b^2\Leftrightarrow\left(a-b\right)^2\ge0\left(đ\right)\)
\(\text{tương tự suy ra:}P\ge2\sqrt{2}\left(a+b+c\right)\)
Tìm x, biết:
a) \(x\left(x-1\right)-x^2+2\text{x}=5\)
b) \(8\left(x-2\right)-2\left(3\text{x}-4\right)=2\)
c) \(\left(3\text{x}+2\right)\left(x-1\right)-3\left(x+1\right)\left(x-2\right)=4\)
d) \(\left(3\text{x}-5\right)\left(7-5\text{x}\right)-\left(5\text{x}+2\right)\left(2-3\text{x}\right)=4\)