\(A=a^4\left(b-c\right)+b^4\left(c-a\right)+c^4\left(a-b\right)\)với a;b;c phân biệt
CMR: \(A\ne0\) với mọi a;b;c
Lm đc cho kẹo
Những câu hỏi liên quan
Cho a, b, c là ba sô thực dương thỏa mãn điều kiện \(a+b+c+\sqrt{abc}=4\). Tính gt biểu thức:
\(A=\sqrt{a\left(4-b\right)\left(4-c\right)}+\sqrt{b\left(4-a\right)\left(4-c\right)}+\sqrt{c\left(4-a\right)\left(4-b\right)}-\sqrt{abc}\)
Giúp với!!
\(a+b+c+\sqrt{abc}=4\Rightarrow4a+4b+4c+4\sqrt{abc}=16\Rightarrow16-4b-4c=4a+4\sqrt{abc}\)
\(\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a\left(16-4b-4c+bc\right)}=\sqrt{a\left(4a+4\sqrt{abc}+bc\right)}\)
\(=\sqrt{4a^2+4a\sqrt{abc}+abc}=\sqrt{\left(2a+\sqrt{abc}\right)^2}=2a+\sqrt{abc}\)
Tương tự : \(\sqrt{b\left(4-a\right)\left(4-c\right)}=2b+\sqrt{abc}\); \(\sqrt{c\left(4-a\right)\left(4-b\right)}=2c+\sqrt{abc}\)
\(\Rightarrow A=2a+2b+2c+3\sqrt{abc}-\sqrt{abc}=2\left(a+b+c+\sqrt{abc}\right)=8\)
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cho a,b,c là 3 số thực dương thỏa mãn điều kiện a+b+c+\(\sqrt{abc}\)=4.
tính giá trị của biểu thức: A=\(\sqrt{a\left(4-b\right)\left(4-c\right)}+\sqrt{b\left(4-c\right)\left(4-a\right)}+\sqrt{c\left(4-a\right)\left(4-b\right)}-\sqrt{abc}\)
Ta có: \(a+b+c+\sqrt{abc}=4\)
\(\Rightarrow4a+4b+4c+4\sqrt{abc}=16\)
\(\Rightarrow4a+4\sqrt{abc}=16-4b-4c\)
\(\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a\left(16-4b-4c+bc\right)}=\sqrt{a\left(4a+4\sqrt{abc}+bc\right)}\)
\(=\sqrt{4a^2+4a\sqrt{abc}+abc}=\sqrt{\left(2a+\sqrt{abc}\right)^2}=\left|2a+\sqrt{abc}\right|=2a+\sqrt{abc}\)
Tương tự:
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{b\left(4-a\right)\left(4-c\right)}=2b+\sqrt{abc}\\\sqrt{c\left(4-a\right)\left(4-b\right)}=2c+\sqrt{abc}\end{matrix}\right.\)
\(\Rightarrow A=\sqrt{a\left(4-b\right)\left(4-c\right)}+\sqrt{b\left(4-c\right)\left(4-a\right)}+\sqrt{c\left(4-a\right)\left(4-b\right)}-\sqrt{abc}=2a+2b+2c+3\sqrt{abc}-\sqrt{abc}=2\left(a+b+c+\sqrt{abc}\right)=8\)
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Ta có \(\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a\left(a+c+\sqrt{abc}\right)\left(4-c\right)}\)
\(=\sqrt{\left(a^2+ac+a\sqrt{abc}\right)\left(4-c\right)}\\ =\sqrt{4a^2+ac\left(4-\sqrt{abc}-a-c\right)+4a\sqrt{abc}}\\ =\sqrt{4a^2+4a\sqrt{abc}+abc}=\sqrt{\left(2a+\sqrt{abc}\right)^2}\\ =2a+\sqrt{abc}\left(a,b,c>0\right)\)
Cmtt \(\sqrt{b\left(4-c\right)\left(4-a\right)}=2b+\sqrt{abc};\sqrt{c\left(4-b\right)\left(4-a\right)}=2c+\sqrt{abc}\)
\(\Rightarrow A=2\left(a+b+c\right)+3\sqrt{abc}-\sqrt{abc}=2\left(a+b+c\right)+2\sqrt{abc}\\ A=2\left(a+b+c+\sqrt{abc}\right)=2\cdot4=8\)
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Cmr nếu a+b+c=0 thì:
a) \(10\left(a^7+b^7+c^7\right)=7\left(a^2+b^2+c^2\right)\left(a^5+b^5+c^5\right)\)
b) \(a^5\left(b^2+c^2\right)+b^5\left(c^2+a^2\right)+c^5\left(a^2+b^2\right)=\dfrac{1}{2}\left(a^3+b^3+c^3\right)\left(a^4+b^4+c^4\right)\)
Cho a,b,c>0. CMR: \(\frac{a^4}{\left(a+b\right)\left(a^2+b^2\right)}+\frac{b^4}{\left(b+c\right)\left(b^2+c^2\right)}+\frac{c^4}{\left(c+a\right)\left(c^2+a^2\right)}\ge\frac{a+b+c}{4}\)
Phân tích đa thức sau thành nhân tử:
a) a^2left(b-cright)+b^2left(c-aright)+c^2left(a-bright)
b) a^3left(b-cright)+b^3left(c-aright)+c^3left(a-bright)
c) a^4left(b-cright)+b^4left(c-aright)+c^4left(a-bright)
d) left(a+bright)left(a^2-b^2right)+left(b+cright)left(b^2-c^2right)+left(c+aright)left(c^2-a^2right)
e) a.left(b+cright)^2left(b-cright)+bleft(c+aright)^2left(c-aright)+c^2left(a+bright)^2.left(a-bright)
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Phân tích đa thức sau thành nhân tử:
a) \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
b) \(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)
c) \(a^4\left(b-c\right)+b^4\left(c-a\right)+c^4\left(a-b\right)\)
d) \(\left(a+b\right)\left(a^2-b^2\right)+\left(b+c\right)\left(b^2-c^2\right)+\left(c+a\right)\left(c^2-a^2\right)\)
e) \(a.\left(b+c\right)^2\left(b-c\right)+b\left(c+a\right)^2\left(c-a\right)+c^2\left(a+b\right)^2.\left(a-b\right)\)
a) \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2b-a^2c+c^2a-c^2b+b^2\left(c-a\right)\)
\(=\left(a^2b-c^2b\right)-\left(a^2c-c^2a\right)-b^2\left(a-c\right)\)
\(=b\left(a^2-c^2\right)-ac\left(a-c\right)-b^2\left(a-c\right)\)
\(=b\left(a-c\right)\left(a+c\right)-ac\left(a-c\right)-b^2\left(a-c\right)\)
\(=\left(a-c\right)\left[b\left(a+c\right)-ac-b^2\right]\)
\(=\left(a-c\right)\left(ab+bc-ac-b^2\right)\)
\(=\left(a-c\right)\left[\left(ab-b^2\right)+\left(bc-ac\right)\right]\)
\(=\left(a-c\right)\left[b\left(a-b\right)+c\left(b-a\right)\right]\)
\(=\left(a-c\right)\left[b\left(a-b\right)-c\left(a-b\right)\right]\)
\(=\left(a-c\right)\left(a-b\right)\left(b-c\right)\)
b) \(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)
\(=a^3b-a^3c+c^3a-c^3b+b^3\left(c-a\right)\)
\(=\left(a^3b-c^3b\right)-\left(a^3c-c^3a\right)-b^3\left(a-c\right)\)
\(=b\left(a^3-c^3\right)-ac\left(a^2-c^2\right)-b^3\left(a-c\right)\)
\(=b\left(a-c\right)\left(a^2+ac+c^2\right)-ac\left(a-c\right)\left(a+c\right)-b^3\left(a-c\right)\)
\(=\left(a-c\right)\left[b\left(a^2+ac+c^2\right)-ac\left(a+c\right)-b^3\right]\)
\(=\left(a-c\right)\left(ba^2+abc+bc^2-a^2c-ac^2-b^3\right)\)
\(=\left(a-c\right)\left[\left(ba^2-a^2c\right)+\left(abc-ac^2\right)+\left(bc^2-b^3\right)\right]\)
\(=\left(a-c\right)\left[a^2\left(b-c\right)+ac\left(b-c\right)+b\left(c^2-b^2\right)\right]\)
\(=\left(a-c\right)\left[a^2\left(b-c\right)+ac\left(b-c\right)-b\left(b^2-c^2\right)\right]\)
\(=\left(a-c\right)\left[a^2\left(b-c\right)+ac\left(b-c\right)-b\left(b-c\right)\left(b+c\right)\right]\)
\(=\left(a-c\right)\left(b-c\right)\left[a^2+ac-b\left(b+c\right)\right]\)
\(=\left(a-c\right)\left(b-c\right)\left(a^2+ac-b^2-bc\right)\)
\(=\left(a-c\right)\left(b-c\right)\left[\left(a-b\right)\left(a+b\right)+c\left(a-b\right)\right]\)
\(=\left(a-c\right)\left(b-c\right)\left(a-b\right)\left(a+b+c\right)\)
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Rút gọn phân thức sau:
\(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{a^4\left(b^2-c^2\right)+b^4\left(c^2-a^2\right)+c^4\left(a^2-b^2\right)}\)
Giúp mk với
\(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{a^4\left(b^2-c^2\right)+b^4\left(c^2-a^2\right)+c^4\left(a^2-b^2\right)}=\frac{a^2\left(b-c\right)+b^2c-b^2a+c^2a-c^2b}{a^4\left(b^2-c^2\right)+b^4c^2-b^4a^2+c^4a^2-c^4b^2}\)
\(=\frac{a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b^2-c^2\right)}{a^4\left(b^2-c^2\right)+b^2c^2\left(b^2-c^2\right)-a^2\left(b^4-c^4\right)}=\frac{\left(b-c\right)\left[a^2+bc-a\left(b+c\right)\right]}{\left(b^2-c^2\right)\left[a^4+b^2c^2-a^2\left(b^2+c^2\right)\right]}\)
\(=\frac{\left(b-c\right)\left(a^2-ab+bc-ac\right)}{\left(b^2-c^2\right)\left(a^4+b^2c^2-a^2b^2-a^2c^2\right)}=\frac{a\left(a-b\right)-c\left(a-b\right)}{\left(b+c\right)\left[a^2\left(a^2-b^2\right)-c^2\left(a^2-b^2\right)\right]}\)
\(=\frac{\left(a-b\right)\left(a-c\right)}{\left(b+c\right)\left(a^2-b^2\right)\left(a^2-c^2\right)}=\frac{1}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Tìm giá trị lớn nhất của :a) A left(sqrt{a}+sqrt{b}right)^2 với a,b 0 và a + b le1b) B left(sqrt{a}+sqrt{b}right)^4+left(sqrt{a}+sqrt{c}right)^4+left(sqrt{a}+sqrt{d}right)^4+left(sqrt{b}+sqrt{c}right)^4+left(sqrt{b}+sqrt{d}right)^4+left(sqrt{c}+sqrt{d}right)^4với a,b,c,d 0 và a + b + c + d le1
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Tìm giá trị lớn nhất của :
a) A = \(\left(\sqrt{a}+\sqrt{b}\right)^2\) với a,b > 0 và a + b \(\le\)1
b) B = \(\left(\sqrt{a}+\sqrt{b}\right)^4+\left(\sqrt{a}+\sqrt{c}\right)^4+\left(\sqrt{a}+\sqrt{d}\right)^4+\left(\sqrt{b}+\sqrt{c}\right)^4+\left(\sqrt{b}+\sqrt{d}\right)^4+\left(\sqrt{c}+\sqrt{d}\right)^4\)
với a,b,c,d > 0 và a + b + c + d \(\le\)1
a) \(A=\left(\sqrt{a}+\sqrt{b}\right)^2\le\left(\sqrt{a}+\sqrt{b}\right)^2+\left(\sqrt{a}-\sqrt{b}\right)^2=2a+2b\le2\)
Vậy GTLN của A là 2 \(\Leftrightarrow\hept{\begin{cases}\sqrt{a}=\sqrt{b}\\a+b=1\end{cases}\Leftrightarrow a=b=\frac{1}{2}}\)
b) Ta có : \(\left(\sqrt{a}+\sqrt{b}\right)^4\le\left(\sqrt{a}+\sqrt{b}\right)^4+\left(\sqrt{a}-\sqrt{b}\right)^4=2\left(a^2+b^2+6ab\right)\)
Tương tự : \(\left(\sqrt{a}+\sqrt{c}\right)^4\le2\left(a^2+c^2+6ac\right)\)
\(\left(\sqrt{a}+\sqrt{d}\right)^4\le2\left(a^2+d^2+6ad\right)\)
\(\left(\sqrt{b}+\sqrt{c}\right)^4\le2\left(b^2+c^2+6bc\right)\)
\(\left(\sqrt{b}+\sqrt{d}\right)^4\le2\left(b^2+d^2+6bd\right)\)
\(\left(\sqrt{c}+\sqrt{d}\right)^4\le2\left(c^2+d^2+6cd\right)\)
Cộng các vế lại, ta được :
\(B\le6\left(a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bd+2cd+2bc\right)=6\left(a+b+c+d\right)^2\)
\(\Rightarrow B\le6\)
Vậy GTLN của B là 6 \(\Leftrightarrow\hept{\begin{cases}\sqrt{a}=\sqrt{b}=\sqrt{c}=\sqrt{d}\\a+b+c+d=1\end{cases}}\Leftrightarrow a=b=c=d=\frac{1}{4}\)
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Cho a,b,c >0 và abc=1. Tìm min:
\(P=\dfrac{a^4+b^4}{\left(a^2+b^2\right)\left(a+b\right)}+\dfrac{b^4+c^4}{\left(b^2+c^2\right)\left(b+c\right)}+\dfrac{a^4+c^4}{\left(a^2+c^2\right)\left(a+c\right)}\)
Cho a,b,c dương và abc=1
CMR: \(\frac{a^4}{2\left(b+c\right)^2}+\frac{b^4}{2\left(a+c\right)^2}+\frac{c^4}{2\left(a+b\right)^2}+\frac{1}{c^2\left(a+c\right)\left(a+b\right)}+\frac{1}{b^2\left(a+b\right)\left(b+c\right)}+\frac{1}{a^2\left(a+c\right)\left(a+b\right)}\ge\frac{1}{8}\)
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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