Ta có:

\(1+x^2=xy+yz+zx+x^2=\left(x+y\right)\left(x+z\right)\)

\(1+y^2=xy+yz+xz+y^2=\left(y+z\right)\left(x+y\right)\)

\(1+z^2=xy+yz+xz+z^2=\left(x+z\right)\left(y+z\right)\)

Thay vào A được:

\(P=x\sqrt{\frac{\left(y+z\right)\left(x+y\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}+y\sqrt{\frac{\left(x+z\right)\left(y+z\right)\left(x+y\right)\left(x+z\right)}{\left(y+z\right)\left(x+y\right)}}\)\(+z\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(x+y\right)}{\left(x+z\right)\left(y+z\right)}}\)

\(=x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)

\(=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\)

\(=xy+xz+xy+yz+xz+zy\)

\(=2\left(xy+yz+xz\right)\)

\(=2\)(do xy+yz+xz=1)

=>Đpcm

Dạng toán này rất nhiều bạn hỏi rồi: thay \(xy+yz+zx=1\) vào các căn thức rồi phân tích đa thức thành nhân tử.

a) \(4\left(xy+yz+zx\right)=x^2+y^2+z^2+2\left(xy+yz+zx\right)=\left(x+y+z\right)^2\) là bình phương 1 số hữu tỉ => 4(xy+yz+zx) cũng là bp số hữu tỉ mà 4=2² => xy+yz+zx là bp 1 số hữu tỉ 

b) \(x^2+y^2+z^2=2\left(xy+yz+zx\right)\)\(\Leftrightarrow\)\(\left(x+y\right)^2+z^2=4xy+2yz+2zx\)

\(\Leftrightarrow\)\(\left(x+y\right)^2-2z\left(x+y\right)+z^2=4xy\)\(\Leftrightarrow\)\(\left(x+y-z\right)^2=4xy\)

Do (x+y-z)² là bình phương 1 số hữu tỉ => 4xy là bp số hữu tỉ => xy là bp số hữu tỉ 

a) Ta có : \(1+x^2=xy+yz+zx+x^2=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(z+x\right)\)

b) \(\Sigma\left(x\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\right)=\Sigma\left(x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right).\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\right)\)

\(=\Sigma\left(x\left(y+z\right)\right)=xy+xz+xy+yz+zx+zy=2\left(xy+yz+zx\right)=2\)

a: \(\left|a-2b+3\right|^{2023}>=0\forall a,b\)

\(\left(b-1\right)^{2024}>=0\forall b\)

Do đó: \(\left|a-2b+3\right|^{2023}+\left(b-1\right)^{2024}>=0\forall a,b\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}a-2b+3=0\\b-1=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}b=1\\a=2b-3=2\cdot1-3=-1\end{matrix}\right.\)

Thay a=-1 và b=1 vào P, ta được:

\(P=\left(-1\right)^{2023}\cdot1^{2024}+2024=2024-1=2023\)

Câu hỏi của Nguyễn Phong - Toán lớp 8 - Học toán với OnlineMath

1111111111111111111

\(VT=\Sigma\frac{xy+yz+zx}{xy}=3+\Sigma\frac{z\left(x+y\right)}{xy}\)

Đến đây để ý \(\frac{1}{2}\left[\frac{z\left(x+y\right)}{xy}+\frac{y\left(z+x\right)}{zx}\right]\ge\sqrt{\frac{\left(z+x\right)\left(x+y\right)}{x^2}}\left(\text{AM - GM}\right)\)

Là xong.

ếu

bài lớp mấy mà khó dữ

Ta có : \(xy+yz+zx=1\)

\(\Rightarrow\hept{\begin{cases}1+x^2=xy+yz+zx+x^2=\left(x+y\right)\left(x+z\right)\\1+y^2=xy+yz+zx+y^2=\left(y+x\right)\left(y+z\right)\\1+z^2=xy+yz+zx+z^2=\left(z+x\right)\left(z+y\right)\end{cases}}\)

Do đó :

\(\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}=\sqrt{\frac{\left(y+x\right)\left(y+z\right)\left(z+x\right)\left(z+y\right)}{\left(x+y\right)\left(x+z\right)}}=\sqrt{\left(y+z\right)^2}\)\(=y+z\)

\(x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}=x\left(y+z\right)\)

Hoàn toàn tương tự :

\(y\sqrt{\frac{\left(1+z^2\right)\left(1+x^2\right)}{1+y^2}}=y\left(z+x\right)\)

\(z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}=z\left(x+y\right)\)

Do đó :

\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)\)

\(=2\left(xy+yz+zx\right)=2\)

Từ xy+yz+zx=1 (giả thiết) 

\(\Rightarrow\hept{\begin{cases}1+x^2=xy+yz+zx+x^2\\1+y^2=xy+yz+zx+y^2\\1+z^2=xy+yz+zx+z^2\end{cases}}\Leftrightarrow\hept{\begin{cases}1+x^2=y\left(x+z\right)+x\left(x+z\right)\\1+y^2=z\left(x+y\right)+y\left(x+y\right)\\1+z^2=y\left(x+z\right)+z\left(x+z\right)\end{cases}}.\)

\(\Leftrightarrow\hept{\begin{cases}1+x^2=\left(x+z\right)\left(x+y\right)\\1+y^2=\left(x+y\right)\left(y+z\right)\\1+z^2=\left(x+z\right)\left(y+z\right)\end{cases},}\)Khi đó thế vào biểu thức A đã cho ta có:

\(A=x\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(y+z\right)}{\left(x+z\right)\left(x+y\right)}}+y\sqrt{\frac{\left(x+z\right)\left(y+z\right)\left(x+z\right)\left(x+y\right)}{\left(x+y\right)\left(y+z\right)}}\)

    \(+z\sqrt{\frac{\left(x+z\right)\left(x+y\right)\left(x+y\right)\left(y+z\right)}{\left(x+z\right)\left(y+z\right)}}\)

\(=x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)

\(=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\)(vì x,y,z là các số dương)

\(=2\left(xy+yz+xz\right)=2.1=2\)

Lời giải:

$xy+yz+xz=1$
$\Rightarrow x^2+1=x^2+xy+yz+xz=(x+y)(x+z)$

Tương tự: $y^2+1=(y+z)(y+x); z^2+1=(z+x)(z+y)$

Khi đó:

\(\sum \sqrt{\frac{(x^2+1)(y^2+1)}{z^2+1}}=\sum \sqrt{\frac{(x+y)(x+z)(y+x)(y+z)}{(z+x)(z+y)}}=\sum \sqrt{(x+y)^2}\)

$=\sum (x+y)=2(x+y+z)$