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ND
2 tháng 6 2018
Ta có:\(a+2b+3c=0\Rightarrow\left(a+2b+3c\right)^2=a^2+4b^2+9c^2+2\left(2ab+3ac+6bc\right)=0\)
\(\Rightarrow20+2\left(2ab+3ac+6bc\right)=0\)
\(\Rightarrow2\left(2ab+3ac+6bc\right)=-20\)
\(\Rightarrow2ab+3ac+6bc=-10\)
\(\Rightarrow\left(2ab+3ac+6bc\right)^2=100\)
\(\Rightarrow4a^2b^2+9a^2c^2+36b^2c^2+6a^2bc+18abc^2+12ab^2c=100\)
\(\Rightarrow4a^2b^2+9a^2c^2+36b^2c^2+6abc\left(a+3c+2b\right)=100\)
\(\Rightarrow4a^2b^2+9a^2c^2+36b^2c^2+6abc.0=100\)
\(\Rightarrow4a^2b^2+9a^2c^2+36b^2c^2=100\)
Ta có: \(a^2+4b^2+9c^2=20\)
\(\Rightarrow\left(a^2+4b^2+9c^2\right)^2=400\)
\(\Rightarrow a^4+16b^4+81c^4+8a^2b^2+18a^2c^2+72b^2c^2=400\)
\(\Rightarrow a^4+16b^4+81c^4+2\left(4a^2b^2+9a^2c^2+36b^2c^2\right)=400\)
\(\Rightarrow a^4+16b^4+81c^4+2.100=400\)
\(\Rightarrow a^4+16b^4+81c^4=200\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 10 2020
Để đơn giản, đặt \(\left(a;-2b;3c\right)=\left(x;y;z\right)\Rightarrow\left\{{}\begin{matrix}x+y+z=0\\x^2+y^2+z^2=18\end{matrix}\right.\)
Ta cần tính \(P=x^4+y^4+z^4\)
\(xy+yz+zx=\frac{\left(x+y+z\right)^2-\left(x^2+y^2+z^2\right)}{2}=-9\)
\(\Rightarrow2\left(x^2y^2+y^2z^2+z^2x^2\right)=\left(xy+yz+zx\right)^2-2xyz\left(x+y+z\right)=81\)
\(x^4+y^4+z^4=\frac{\left(x^2+y^2+z^2\right)^2-2\left(x^2y^2+y^2z^2+z^2x^2\right)}{2}=\frac{18^2-81}{2}=\frac{243}{2}\)
L
2
D
datcoder
CTVVIP
10 tháng 9 2023
\(\left(a-b\right)^2-c^2=\left(a-b+c\right)\left(a-b-c\right)\)
\(\left(a+b\right)^2-4=\left(a+b\right)^2-2^2=\left(a+b+2\right)\left(a+b-2\right)\\ \left(a-2b\right)^2-4b^2=\left(a-2b\right)^2-\left(2b\right)^2=\left(a-2b+2b\right)\left(a-2b-2b\right)=a\left(a-4b\right)\\ \left(a+3b\right)^2-9b^2=\left(a+3b\right)^2-\left(3b\right)^2=\left(a+3b+3b\right)\left(a+3b-3b\right)=a\left(a+6b\right)\\ \left(a-5b\right)^2-16b^2=\left(a-5b\right)^2-\left(4b\right)^2=\left(a-5b+4b\right)\left(a-5b-4b\right)=\left(a-b\right)\left(a-9b\right)\)
Tất cả đều dùng hằng đẳng thức: \(a^2-b^2=\left(a+b\right)\left(a-b\right)\)
10 tháng 9 2023
a: =(a-b-c)(a-b+c)
b: =(a+b)^2-2^2
=(a+b+2)(a+b-2)
c: =(a-2b)^2-(2b)^2
=(a-2b-2b)(a-2b+2b)
=a(a-4b)
d: =(a+3b)^2-(3b)^2
=(a+3b-3b)(a+3b+3b)
=a(a+6b)
e: =(a-5b)^2-(4b)^2
=(a-5b-4b)(a-5b+4b)
=(a-9b)(a-b)
31 tháng 8 2019
a) (x - 1)(x + 1)(x2 + 1)(x4 + 1)(x8 + 1)
= (x2 - 1)(x2 + 1)(x4 + 1)(x8 + 1)
= (x4 - 1)(x4 + 1)(x8 + 1)
= (x8 - 1)(x8 + 1)
= x16 - 1
b) (a2 - 2b)(a2 + 2b)(a4 + 4b2)(a8 + 16b4)
= (a4 - 4b2)(a4 + 4b2)(a8 + 16b4)
= (a8 - 16b4)(a8 + 16b4)
= a16 - 256b8
MA
1
24 tháng 1 2023
Sẵn tiện mk chỉ cho bn luôn dạng này nhé.
Phân tích:
Với \(\alpha,\beta,\gamma>0\) thỏa \(\alpha< 2,\beta< 3,\gamma< 4\) ta có:
\(A=2a+3b+4c+\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}\)
\(=\left[\left(2-\alpha\right)a+\dfrac{3}{a}\right]+\left[\left(3-\beta\right)b+\dfrac{9}{2b}\right]+\left[\left(4-\gamma\right)c+\dfrac{4}{c}\right]+\left(\alpha a+\beta b+\gamma c\right)\)
\(\ge2\sqrt{3.\left(2-\alpha\right)}+2\sqrt{\dfrac{9}{2}.\left(3-\beta\right)}+2\sqrt{4.\left(4-\gamma\right)}+\left(\alpha a+\beta b+\gamma c\right)\)
Chọn \(\alpha,\beta,\gamma\) (thỏa đk trên) sao cho:
\(\left\{{}\begin{matrix}\left(2-\alpha\right)a=\dfrac{3}{a}\\\left(3-\beta\right)b=\dfrac{9}{2b}\\\left(4-\gamma\right)c=\dfrac{4}{c}\\\alpha=\dfrac{\beta}{2}=\dfrac{\gamma}{3}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=\sqrt{\dfrac{3}{2-\alpha}}\\b=\sqrt{\dfrac{9}{2\left(3-\beta\right)}}\\c=\sqrt{\dfrac{4}{\left(4-\gamma\right)}}\\\alpha=\dfrac{\beta}{2}=\dfrac{\gamma}{3}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=\sqrt{\dfrac{3}{2-\alpha}}\\b=\sqrt{\dfrac{9}{6-4\alpha}}\\c=\sqrt{\dfrac{4}{4-3\alpha}}\\\alpha=\dfrac{\beta}{2}=\dfrac{\gamma}{3}\end{matrix}\right.\)
Ta có: \(a+2b+3c\ge20\). Xác định điểm rơi: \(a+2b+3c=20\)
\(\Rightarrow\sqrt{\dfrac{3}{2-\alpha}}+2\sqrt{\dfrac{9}{6-4\alpha}}+3\sqrt{\dfrac{4}{4-3\alpha}}=20\)
Giải ra ta có \(\alpha=\dfrac{5}{4}\Rightarrow\beta=\dfrac{5}{2};\gamma=\dfrac{15}{4}\)
Lời giải:
Ta có: \(A=2a+3b+4c+\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}\)
\(=\left(\dfrac{3a}{4}+\dfrac{3}{a}\right)+\left(\dfrac{b}{2}+\dfrac{9}{2b}\right)+\left(\dfrac{c}{4}+\dfrac{4}{c}\right)+\left(\dfrac{5a}{4}+\dfrac{5b}{2}+\dfrac{15c}{4}\right)\)
\(\ge^{Cauchy}2\sqrt{\dfrac{3a}{4}.\dfrac{3}{a}}+2\sqrt{\dfrac{b}{2}.\dfrac{9}{2b}}+2\sqrt{\dfrac{c}{4}.\dfrac{4}{c}}+\dfrac{5}{4}\left(a+2b+3c\right)\)
\(=3+3+2+\dfrac{5}{4}\left(a+2b+3c\right)\)
\(\ge8+\dfrac{5}{4}.20=33\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\dfrac{3a}{4}=\dfrac{3}{a}\\\dfrac{b}{2}=\dfrac{9}{2b}\\\dfrac{c}{4}=\dfrac{4}{c}\\a+2b+3c=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=3\\c=4\end{matrix}\right.\)
Vậy \(MinA=33\), đạt được khi \(a=2;b=3;c=4\)