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Phương Ann Nhã Doanh đề bài khó wá Mashiro Shiina Đinh Đức Hùng
Nguyễn Huy Tú Lightning Farron Akai Haruma
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Ta có: \(x+y+z=by+cz+ax+cz+ax+by=2\left(ax+by+cz\right)\)Thay \(z=ax+by\)
\(\Rightarrow x+y+z=2\left(z+cz\right)=2z\left(1+c\right)\)
\(\Rightarrow\dfrac{1}{1+c}=\dfrac{2z}{x+y+z}\)
Tương tự:\(\left\{{}\begin{matrix}\dfrac{1}{1+a}=\dfrac{2x}{x+y+z}\\\dfrac{1}{1+b}=\dfrac{2y}{x+y+z}\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)Vậy A=2
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Đặt x/a=y/b=z/c=k
=>x=ak; y=bk; z=ck
\(\dfrac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\dfrac{a^2k^2+b^2k^2+c^2k^2}{a^4k^2+b^4k^2+c^4k^2}=\dfrac{1}{a^2+b^2+c^2}\)
Cho x = by + cz ; y = ax + cz; z = ax + by.
\(CMR:A=\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}=2\)
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Ta có
\(x-y=\left(by+cz\right)-\left(ax+cz\right)=by-ax\)
\(\Leftrightarrow x\cdot\left(a+1\right)=y\cdot\left(b+1\right)\)
\(y-z=\left(ax+cz\right)-\left(ax+by\right)=cz-by\)
\(\Leftrightarrow z\cdot\left(c+1\right)=y\cdot\left(b+1\right)\)
\(x-z=\left(by+cz\right)-\left(ax+by\right)=cz-ax\)
\(\Leftrightarrow x\cdot\left(a+1\right)=z\cdot\left(c+1\right)\)
\(\Rightarrow x\cdot\left(a+1\right)=z\cdot\left(c+1\right)=y\left(b+1\right)\)
Đặt \(x\cdot\left(a+1\right)=z\cdot\left(c+1\right)=y\left(b+1\right)=k\)
\(\Rightarrow\left\{{}\begin{matrix}a+1=\dfrac{k}{x}\\b+1=\dfrac{k}{y}\\c+1=\dfrac{k}{z}\end{matrix}\right.\)
Thay vào A, ta có :
\(A=\dfrac{1}{\dfrac{k}{x}}+\dfrac{1}{\dfrac{k}{y}}+\dfrac{1}{\dfrac{k}{z}}\)
\(=\dfrac{x}{k}+\dfrac{y}{k}+\dfrac{z}{k}\)
=\(\dfrac{x+y+z}{k}\)
Vì z = ax + by; x = cz + by; y = ax + cz nen :
\(k=z\cdot\left(c+1\right)=cz+z=cz+ax+by\)
\(\Rightarrow A=\dfrac{2\cdot\left(ax+by+czz\right)}{ax+by+cz}=2\)
⇒ĐPCM
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Vì ax + by =2c
ax + cz =2b
by + cz = 2a
=>Ta có ax + by + cz =a+b+c
=> ax + 2a=a+b+c
và 2c + cz =a+b+c
và 2b+ by =a+b+c
=> \(x=\dfrac{b+c-a}{a}\); \(y=\dfrac{a+c-b}{b}\);\(z=\dfrac{b+a-c}{c}\)
=> \(x+2=\dfrac{b+c+a}{a}\); \(y+2=\dfrac{a+c+b}{b}\);\(z+2=\dfrac{b+a+c}{c}\)
=>\(M=\dfrac{1}{x+2}+\dfrac{1}{y+2}+\dfrac{1}{z+2}=\dfrac{a+b+c}{a+b+c}=1\)
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x/3=c/a=>ax=3c
y/d=3/b=>by=3d
ax+by=3c+3d=3(c+d)=3.27=34
\(\sqrt{ax+by}=\sqrt{3^4}=3^2=9\)
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Từ đề bài ta có:
\(x+y+z=2\left(ax+by+cz\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x+y+z=2\left(ax+x\right)\\x+y+z=2\left(by+y\right)\\x+y+z=2\left(cz+z\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+z=2x\left(1+a\right)\\x+y+z=2y\left(1+b\right)\\x+y+z=2z\left(1+c\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{1+a}=\dfrac{2x}{x+y+z}\\\dfrac{1}{1+b}=\dfrac{2y}{x+y+z}\\\dfrac{1}{1+c}=\dfrac{2z}{x+y+z}\end{matrix}\right.\)
\(\Rightarrow Q=\dfrac{2x}{x+y+z}+\dfrac{2y}{x+y+z}+\dfrac{2z}{x+y+z}\)
\(=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)
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a) Ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\)
=> ad = bc
Ta có : (a + 2c)(b + d)
= a(b + d) + 2c(b + d)
= ab + ad + 2cb + 2cd (1)
Ta có : (a + c)(b + 2d)
= a(b + 2d) + c(b + 2b)
= ab + a2d + cb + c2b
= ab + c2d + ad + c2b (Vì ad = cd) (2)
Từ (1),(2) => (a + 2c)(b + d) = (a + c)(b + 2d) (ĐPCM)
Sửa đề bài : P = \(\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}\)
Ta có : \(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}\)
=> \(\dfrac{y+z+t}{x}=\dfrac{z+t+x}{y}=\dfrac{t+x+y}{z}=\dfrac{x+y+z}{t}\)
=> \(\dfrac{y+z+t}{x}+1=\dfrac{z+t+x}{y}+1=\dfrac{t+x+y}{z}+1=\dfrac{x+y+z}{t}+1\)=> \(\dfrac{y+z+t+x}{x}=\dfrac{z+t+x+y}{y}=\dfrac{t+x+y+z}{z}=\dfrac{x+y+z+t}{t}\)TH1: x + y + z + t # 0
=> x = y = z = t
Ta có : P = \(\dfrac{x+y}{z+t}=\dfrac{y+z}{t+x}=\dfrac{z+t}{x+y}=\dfrac{t+x}{y+z}\)
P = \(\dfrac{x+x}{x+x}+\dfrac{x+x}{x+x}+\dfrac{x+x}{x+x}+\dfrac{x+x}{x+x}\)
P = 1 + 1 + 1 + 1 = 4
TH2 : x + y + z + t = 0
=> x + y = -(z + t)
y + z = -(t + x)
z + t = -(x + y)
t + x = -(y + z)
Ta có : P = \(\dfrac{x+y}{z+t}=\dfrac{y+z}{t+x}=\dfrac{z+t}{x+y}=\dfrac{t+x}{y+z}\)
P = \(\dfrac{-\left(z+t\right)}{z+t}=\dfrac{-\left(t+x\right)}{t+x}=\dfrac{-\left(x+y\right)}{x+y}=\dfrac{-\left(y+z\right)}{y+z}\)
P = (-1) + (-1) + (-1) + (-1)
P = -4
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