Cho a + b = 1. Tính A = 2\(\left(a^3+b^3\right)\) - 3\(\left(a^2+b^2\right)\).
Những câu hỏi liên quan
cho
\(A=\dfrac{1}{2}+\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+\left(\dfrac{3}{2}\right)^4+...+\left(\dfrac{3}{2}\right)^{2021}\)
\(B=\left(\dfrac{3}{2}\right)^{2013}:2\)
tính B-A
Ta có \(A=\dfrac{1}{2}+\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+\left(\dfrac{3}{2}\right)^4+...+\left(\dfrac{3}{2}\right)^{2021}\left(1\right)\)
\(\Rightarrow\dfrac{3}{2}A=\dfrac{3}{4}+\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+\left(\dfrac{3}{2}\right)^4+...+\left(\dfrac{3}{2}\right)^{2013}\left(2\right)\)
Lấy (2) - (1) ta được:
\(\dfrac{3}{2}A-A=\left(\dfrac{3}{2}\right)^{2013}+\dfrac{3}{4}-\dfrac{1}{2}-\dfrac{3}{2}\)
\(\dfrac{1}{2}A=\left(\dfrac{3}{2}\right)^{2013}+\dfrac{1}{4}\Rightarrow A=\dfrac{3^{2013}}{2^{2012}}+\dfrac{1}{2}\)
Vậy \(B-A=\dfrac{3^{2013}}{2^{2014}}-\dfrac{3^{2013}}{2^{2012}}+\dfrac{5}{2}\)
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Rút gọn :
a,Aleft(3+1right)left(3^2+1right)left(3^4+1right)...left(3^{64}+1right)
b,B-1^2+2^2-3^2+4^2-...-99^2+100^2
c,C-1^2+2^2-3^2+4^2-...+left(-1right)^ncdot n^2
d,D3cdotleft(2^2+1right)left(2^4+1right)...left(2^{64}+1right)+1
e,Eleft(a+b+cright)^2+left(a+b-cright)^2-2left(a+bright)^2
g,Gleft(a+b+c+dright)^2+left(a+b-c-dright)^2+left(a+c-b-dright)^2+left(a+d-b-cright)^2
h,Hleft(a+b+cright)^3-left(b+c-aright)^3-left(a+c-bright)^3+left(a+b-cright)^3
i,Ileft(a+bright)^3+left(b+cright)^3+left(c...
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Rút gọn :
\(a,A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ b,B=-1^2+2^2-3^2+4^2-...-99^2+100^2\\ c,C=-1^2+2^2-3^2+4^2-...+\left(-1\right)^n\cdot n^2\\ d,D=3\cdot\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\\ e,E=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\\ g,G=\left(a+b+c+d\right)^2+\left(a+b-c-d\right)^2+\left(a+c-b-d\right)^2+\left(a+d-b-c\right)^2\\ h,H=\left(a+b+c\right)^3-\left(b+c-a\right)^3-\left(a+c-b\right)^3+\left(a+b-c\right)^3\\ i,I=\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3-3\left(a+b\right)\left(c+b\right)\left(c+a\right)\)
Mọi người ơi, giúp mk vs, đc câu nào hay câu ấy ! Help me!!!!!!!!!!!!!!!!!!
a/ \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(\Rightarrow2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}\)
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e) ta dể dàng thấy được : \(a^2+b^2=\left(a+b\right)^2-2ab\)
\(\Rightarrow E=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)
\(=\left(2a+2b\right)^2-2\left(a+b+c\right)\left(a+b-c\right)-2\left(a+b\right)^2\)
\(=4\left(a+b\right)^2-2\left(\left(a+b\right)^2-c^2\right)-2\left(a+b\right)^2\)
\(=4\left(a+b\right)^2-2\left(a+b\right)^2+2c^2-2\left(a+b\right)^2=2c^2\)
g) củng sử dụng cái trên ta có : \(G=\left(a+b+c+d\right)^2+\left(a+b-c-d\right)^2+\left(a+c-b-d\right)^2+\left(a+d-b-c\right)^2\)
\(=\left(2a+2b\right)^2-2\left(a+b+c+d\right)\left(a+b-c-d\right)+\left(2a-2b\right)^2-2\left(a+c-b-d\right)\left(a+d-b-c\right)\)
\(=4\left(a+b\right)^2+4\left(a-b\right)^2-2\left(\left(a+b\right)^2-\left(c+d\right)^2\right)-2\left(\left(a-b\right)^2-\left(c-d\right)^2\right)\)
\(=4\left(\left(a+b\right)^2+\left(a-b\right)^2\right)-2\left(\left(a+b\right)^2+\left(a-b\right)^2\right)+2\left(\left(c+d\right)^2+\left(c-d\right)^2\right)\)
\(=2\left(\left(a+b\right)^2+\left(a-b\right)^2\right)+2\left(\left(c+d\right)^2+\left(c-d\right)^2\right)\)\(=2\left(\left(2a\right)^2-2\left(a+b\right)\left(a-b\right)\right)+2\left(\left(2c\right)^2-2\left(c+d\right)\left(c-d\right)\right)\)
\(=2\left(4a^2-2\left(a^2-b^2\right)\right)+2\left(4c^2-2\left(c^2-d^2\right)\right)\)
\(=2\left(2a^2+2b^2\right)+2\left(2c^2+2d^2\right)=4\left(a^2+b^2+c^2+d^2\right)\)
bn đăng nhiều quá nên mk làm câu nào hay câu đó nha
mà nè mấy câu a;b;c;d hình như trên mạng có bn lên đó tìm nha .
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Cho a+b+c\(a^3+b^3+c^3=3abc\) áp dụng tính B=\(\frac{\left(a^2-b^2\right)^3+\left(b^2-c^2\right)^3+\left(c^2-a^2\right)^3}{\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3}\)
1. Cho a2 - b2 - c2 =3abc
Tính H = \(\left(1-\frac{a}{b}\right)\left(1-\frac{b}{c}\right)\left(1-\frac{c}{a}\right)\)
2. Cho a - b + c = - 4
Tính B = \(\frac{a^3-b^3+c^3+3abc}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)
Cho ab=1 và a+b≠0. Tính
\(P=\frac{1}{\left(a+b\right)^3}\left(\frac{1}{a^3}+\frac{1}{b^3}\right)+\frac{3}{\left(a+b\right)^4}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+\frac{6}{\left(a+b\right)^5}\left(\frac{1}{a}+\frac{1}{b}\right)\)
Cho a,b,c khác 1 và a+b+c=3. Tính\(A=\frac{\left(a-1\right)^2}{\left(b-1\right)\left(c-1\right)}+\frac{\left(b-1\right)^2}{\left(c-1\right)\left(a-1\right)}+\frac{\left(c-1\right)^2}{\left(a-1\right)\left(b-1\right)}\)
cho \(a+b+c=0\) tính \(\dfrac{a^3+b^3+c^3}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
cho a,b không âm thỏa mãn \(\left(a-b\right)^2=a+b+2\)
CMR: \(\left(1+\dfrac{a^3}{\left(b+1\right)^3}\right)\left(1+\dfrac{b^3}{\left(a+1\right)^3}\right)\le9\)
\(GT\Leftrightarrow a^2+b^2-2ab=a+b+2\)
\(\Leftrightarrow a^2+a+b^2+b=2\left(ab+a+b+1\right)\)
\(\Leftrightarrow a\left(a+1\right)+b\left(b+1\right)=2\left(a+1\right)\left(b+1\right)\)
\(\Leftrightarrow\dfrac{a}{b+1}+\dfrac{b}{a+1}=2\)
Đặt \(\left(\dfrac{a}{b+1};\dfrac{b}{a+1}\right)=\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}x;y\ge0\\x+y=2\end{matrix}\right.\)
\(\Rightarrow0\le xy\le1\)
\(P=\left(1+x^3\right)\left(1+y^3\right)=1+x^3+y^3+x^3y^3\)
\(P=1+\left(x+y\right)^3-3xy\left(x+y\right)+\left(xy\right)^3\)
\(P=\left(xy\right)^3-6xy+9=xy\left[\left(xy\right)^2-6\right]+9\le9\)
Dấu "=" xảy ra khi \(xy=0\Leftrightarrow\left(a;b\right)=\left(0;2\right);\left(2;0\right)\)
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8 tháng 1 2021 lúc 16:23
Cho a-b+c=-4. Tính B = \(\dfrac{a^3-b^3+c^3+3abc}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)
\(B=\dfrac{a^3+c^3+3ac\left(a+c\right)-b^3-3ac\left(a+c\right)+3abc}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)
\(=\dfrac{\left(a+c\right)^3-b^3-3ac\left(a+c-b\right)}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)
\(=\dfrac{\left(a+c-b\right)\left[\left(a+c\right)^2+b\left(a+c\right)+b^2\right]-3ac\left(a+c-b\right)}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)
\(=\dfrac{\left(a+c-b\right)\left(a^2+b^2+c^2+ab+bc-ac\right)}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)
\(=\dfrac{-2\left(2a^2+2b^2+2c^2+2ab+2bc-2ca\right)}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)
\(=\dfrac{-2\left[\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2\right]}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}=-2\)
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