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a) Ta có:
\(\frac{2a+b}{a+b}+\frac{2b+c}{b+c}+\frac{2c+d}{c+d}+\frac{2d+a}{d+a}=6\)
\(\Leftrightarrow\left[\left(\frac{2a+b}{a+b}-1\right)+\left(\frac{2b+c}{b+c}-1\right)-1\right]+\left[\left(\frac{2c+d}{c+d}-1\right)+\left(\frac{2d+a}{d+a}-1\right)-1\right]=0\)
\(\Leftrightarrow\left(\frac{a}{a+b}+\frac{b}{b+c}-1\right)+\left(\frac{c}{c+d}+\frac{d}{d+a}-1\right)=0\)
\(\Leftrightarrow\left(\frac{a.\left(b+c\right)}{\left(a+b\right).\left(b+c\right)}+\frac{b.\left(a+b\right)}{\left(a+b\right).\left(b+c\right)}-\frac{\left(a+b\right).\left(b+c\right)}{\left(a+b\right).\left(b+c\right)}\right)+\left(\frac{c.\left(d+a\right)}{\left(c+d\right).\left(d+a\right)}+\frac{d.\left(c+d\right)}{\left(c+d\right).\left(d+a\right)}-\frac{\left(c+d\right).\left(d+a\right)}{\left(c+d\right).\left(d+a\right)}\right)=0\)
\(\Leftrightarrow\left(\frac{ab+ac}{\left(a+b\right).\left(b+c\right)}+\frac{ab+b^2}{\left(a+b\right).\left(b+c\right)}-\frac{ab+ac+b^2+bc}{\left(a+b\right).\left(b+c\right)}\right)+\left(\frac{cd+ac}{\left(c+d\right).\left(d+a\right)}+\frac{cd+d^2}{\left(c+d\right).\left(d+a\right)}-\frac{cd+ac+d^2+ad}{\left(c+d\right).\left(d+a\right)}\right)=0\)
\(\Leftrightarrow\left(\frac{ab+ac+ab+b^2-ab-ac-b^2-bc}{\left(a+b\right).\left(b+c\right)}\right)+\left(\frac{cd+ac+cd+d^2-cd-ac-d^2-ad}{\left(c+d\right).\left(d+a\right)}\right)=0\)
\(\Leftrightarrow\frac{ab-bc}{\left(a+b\right).\left(b+c\right)}+\frac{cd-ad}{\left(c+d\right).\left(d+a\right)}=0\)
\(\Leftrightarrow\frac{ab-bc}{\left(a+b\right).\left(b+c\right)}=-\frac{cd-ad}{\left(c+d\right).\left(d+a\right)}\)
\(\Leftrightarrow\frac{ab-bc}{\left(a+b\right).\left(b+c\right)}=\frac{ad-cd}{\left(c+d\right).\left(d+a\right)}\)
\(\Leftrightarrow\frac{b.\left(a-c\right)}{\left(a+b\right).\left(b+c\right)}=\frac{d.\left(a-c\right)}{\left(c+d\right).\left(d+a\right)}\)
\(\Leftrightarrow\frac{b}{\left(a+b\right).\left(b+c\right)}=\frac{d}{\left(c+d\right).\left(d+a\right)}\) (vì \(a;b;c;d\) là số nguyên dương).
\(\Leftrightarrow b\left(c+d\right).\left(d+a\right)=d\left(a+b\right).\left(b+c\right)\)
\(\Leftrightarrow\left(bc+bd\right).\left(d+a\right)=\left(ad+bd\right).\left(b+c\right)\)
\(\Leftrightarrow bcd+abc+bd^2+abd=abd+acd+b^2d+bcd\)
\(\Leftrightarrow bd^2+abc=b^2d+acd\)
\(\Leftrightarrow bd^2-b^2d=acd-abc\)
\(\Leftrightarrow bd.\left(d-b\right)=ac.\left(d-b\right)\)
\(\Leftrightarrow bd.\left(d-b\right)-ac.\left(d-b\right)=0\)
\(\Leftrightarrow\left(d-b\right).\left(bd-ac\right)=0\)
Vì \(a;b;c;d\) là số nguyên dương.
\(\Rightarrow d-b>0\)
\(\Rightarrow d-b\ne0.\)
\(\Leftrightarrow bd-ac=0\)
\(\Leftrightarrow bd=ac.\)
Lại có:
\(A=abcd\)
\(\Rightarrow A=ac.bd\)
\(\Rightarrow A=ac.ac\)
\(\Rightarrow A=\left(ac\right)^2.\)
\(\Rightarrow A=abcd\) là số chính phương (đpcm).
Chúc bạn học tốt!
Đặt \(\left(\frac{a-b}{c},\frac{b-c}{a},\frac{c-a}{b}\right)\rightarrow\left(x,y,z\right)\)
Khi đó:\(\left(\frac{c}{a-b},\frac{a}{b-c},\frac{b}{c-a}\right)\rightarrow\left(\frac{1}{x},\frac{1}{y},\frac{1}{z}\right)\)
Ta có:
\(P\cdot Q=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3+\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)
Mặt khác:\(\frac{y+z}{x}=\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\cdot\frac{c}{a-b}=\frac{b^2-bc+ac-a^2}{ab}\cdot\frac{c}{a-b}\)
\(=\frac{c\left(a-b\right)\left(c-a-b\right)}{ab\left(a-b\right)}=\frac{c\left(c-a-b\right)}{ab}=\frac{2c^2}{ab}\left(1\right)\)
Tương tự:\(\frac{x+z}{y}=\frac{2a^2}{bc}\left(2\right)\)
\(=\frac{x+y}{z}=\frac{2b^2}{ac}\left(3\right)\)
Từ ( 1 );( 2 );( 3 ) ta có:
\(P\cdot Q=3+\frac{2c^2}{ab}+\frac{2a^2}{bc}+\frac{2b^2}{ac}=3+\frac{2}{abc}\left(a^3+b^3+c^3\right)\)
Ta có:\(a+b+c=0\)
\(\Rightarrow\left(a+b\right)^3=-c^3\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
Khi đó:\(P\cdot Q=3+\frac{2}{abc}\cdot3abc=9\)
Bài 3: y hệt bài mình đã từng đăng Câu hỏi của Thắng Nguyễn - Toán lớp 9 - Học toán với OnlineMath- trước mình có ghi lời giải mà lâu ko xem giờ quên r` :)
1) Đặt n+1 = k^2
2n + 1 = m^2
Vì 2n + 1 là số lẻ => m^2 là số lẻ => m lẻ
Đặt m = 2t+1
=> 2n+1 = m^2 = (2t+1)^2
=> 2n+1 = 41^2 + 4t + 1
=> n = 2t(t+1)
=> n là số chẵn
=> n+1 là số lẻ
=> k lẻ
+) Vì k^2 = n+1
=> n = (k-1)(k+1)
Vì k -1 và k+1 là 2 số chẵn liên tiếp
=> (k+1)(k-1) chia hết cho *
=> n chia hết cho 8
+) k^2 + m^2 = 3a + 2
=> k^2 và m^2 chia 3 dư 1
=> m^2 - k^2 chia hết cho 3
m^2 - k^2 = a
=> a chia hết cho 3
Mà 3 và 8 là 2 số nguyên tố cùng nhau
=> a chia hết cho 24
Câu 1, Quy đồng mẫu của 2 về lấy MTC là (x-y)(y-z)(z-x).
Câu 2, Chỉ có thể xảy ra khi a+b+c=x+y+z=x/a+y/b+z/c=0
2) Có: \(a+b+c=0\)
\(\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ac\right)\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ac\right)^2\)
\(\Leftrightarrow VT=4\left[\left(ab\right)^2+\left(bc\right)^2+\left(ac\right)^2-2abc\left(a+b+c\right)\right]\)
\(\Leftrightarrow VT=4\left(ab\right)^2+4\left(ac\right)^2+4\left(bc\right)^2\)
Có: \(a+b+c=0\Rightarrow a+b=-c\Leftrightarrow\left(a+b\right)^2=c^2\Leftrightarrow2ab=c^2-a^2-b^2\)
Tương tự:...
\(VT=\text{Σ}_{cyc}\left(c^2-a^2-b^2\right)^2=2\left(a^4+b^4+c^4\right)=VP\)
2) 1/x - 1/y - 1/z = 1
=> (1/x - 1/y - 1/z)^2 = 1
<=> 1/x^2 + 1/y^2 + 1/z^2 - 2/xy - 2/xz + 2/yz = 1
<=> 1/x^2 + 1/y^2 + 1/z^2 - 2.(1/xy + 1/xz - 1/yz) = 1
<=> 1/x^2 + 1/y^2 + 1/z^2 - 2.(z+y-x/xyz) = 1
<=> 1/x^2 + 1/y^2 + 1/z^2 - 2.0 = 1
<=> 1/x^2 + 1/y^2 + 1/z^2 = 1 (đpcm)
3/ Ta có:
\(x+y+z=0\)
\(\Rightarrow x^2=\left(y+z\right)^2;y^2=\left(z+x\right)^2;z^2=\left(x+y\right)^2\)
\(a+b+c=0\)
\(\Rightarrow a+b=-c;b+c=-a;c+a=-b\)
\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)
\(\Leftrightarrow ayz+bxz+cxy=0\)
Ta có:
\(ax^2+by^2+cz^2=a\left(y+z\right)^2+b\left(z+x\right)^2+c\left(x+y\right)^2\)
\(=x^2\left(b+c\right)+y^2\left(c+a\right)+z^2\left(a+b\right)+2\left(ayz+bzx+cxy\right)\)
\(=-ax^2-by^2-cz^2\)
\(\Leftrightarrow2\left(ax^2+by^2+cz^2\right)=0\)
\(\Leftrightarrow ax^2+by^2+cz^2=0\)
1/ Đặt \(a-b=x,b-c=y,c-z=z\)
\(\Rightarrow x+y+z=0\)
Ta có:
\(\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
\(=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2\left(x+y+z\right)}{xyz}\)
\(=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\)
1) Sửa lại:Cho x,y,z dương nhé!
\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=x\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+y\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+z\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(=1+\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+1+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}+1=\left(1+1+1\right)+\left(\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}\right)\)
\(=3+\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)+\left(\frac{z}{x}+\frac{x}{z}\right)\)
Vì x,y,z là các số dương ,ta áp dụng bất đẳng thức Cô-Si:
\(\frac{x}{y}+\frac{y}{x}\ge2\sqrt{\frac{x}{y}.\frac{y}{x}}=2\)
\(\frac{y}{z}+\frac{z}{y}\ge2\sqrt{\frac{y}{z}.\frac{z}{y}}=2\)
\(\frac{z}{x}+\frac{x}{z}\ge2\sqrt{\frac{z}{x}.\frac{x}{z}}=2\)
Do đó \(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge3+2+2+2=9\)
Dấu "=" xảy ra <=> \(x=y=z\)
câu 2) mk chịu
câu 2 đề sai . sửa số 3 thành số 2 . neu sua thanh co 2 thi co the ap dung bdt cosi hoac trebusep