\(P=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(=\left(x^2+5x\right)^2-36\ge-36\)

Dấu " = " khi \(\left(x^2+5x\right)^2=0\Rightarrow x\left(x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Vậy MIN P = -36 khi x = 0 hoặc x = -5

Ta có: \(\left(x-y\right)\left(x-y\right)=\dfrac{3}{10}+\dfrac{3}{50}\)

\(\Rightarrow\left(x-y\right)^2=\dfrac{9}{25}\)

\(\Rightarrow x-y=\pm\dfrac{3}{5}\)

+) \(x-y=\dfrac{3}{5}\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{5}x=\dfrac{3}{10}\\\dfrac{3}{5}y=\dfrac{-3}{50}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-1}{10}\end{matrix}\right.\)

+) \(x-y=\dfrac{-3}{5}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{2}\\y=\dfrac{1}{10}\end{matrix}\right.\)

Vậy cặp số \(\left(x;y\right)\) là \(\left(\dfrac{1}{2};\dfrac{-1}{10}\right);\left(\dfrac{-1}{2};\dfrac{1}{10}\right)\)

Bài 1:

\(\left|x+1\right|\left(x^2-5\right)\left(x^2-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\left|x+1\right|=0\\x^2-5=0\\x^2-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=\sqrt{5}\\x=\pm2\end{matrix}\right.\)

Do \(x\in Z\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\\x=-2\end{matrix}\right.\)

Vậy...

Bài 3:

\(x^2-2xy+2y^2=0\)

\(\Rightarrow x^2-2xy+y^2+y^2=0\)

\(\Rightarrow\left(x-y\right)^2+y^2=0\)

Mà \(\left(x-y\right)^2+y^2\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\y^2=0\end{matrix}\right.\Rightarrow x=y=0\)

Vậy...

Bài 5,6 áp dụng t/c dãy tỉ số bằng nhau là ra

a, \(x^3-0,25x=0\)

\(\Leftrightarrow x\left(x^2-0,25\right)=0\)

\(\Leftrightarrow x\left(x-0,5\right)\left(x+0,5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=0,5\\x=-0,5\end{matrix}\right.\)

Vậy...

b, \(4x^2-9=0\)

\(\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)

Vậy...

c, \(x^2-10x=-25\)

\(\Leftrightarrow x^2-10x+25=0\)

\(\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x=5\)

Vậy...

a, Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{b+c+a}=1\)

\(\Rightarrow a=b=c\)

b, Ta có: \(a^2=bc\Rightarrow\dfrac{a}{c}=\dfrac{b}{a}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{a}{c}=\dfrac{b}{a}=\dfrac{a+b}{c+a}=\dfrac{a-b}{c-a}\)

\(\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)

\(\Rightarrowđpcm\)

a, \(\left(x^2+1\right)\left(x-2\right)=x^2+1\)

\(\Leftrightarrow x^3-2x^2+x-2-x^2-1=0\)

\(\Leftrightarrow x^3-3x^2+x-3=0\)

\(\Leftrightarrow x^2\left(x-3\right)+\left(x-3\right)=0\)

\(\Leftrightarrow\left(x^2+1\right)\left(x-3\right)=0\)

Mà \(x^2+1>0\Rightarrow x-3=0\Leftrightarrow x=3\)

Vậy x = 3

b, \(\left(x^2+1\right)\left(x-2\right)=-\left(x^2+1\right)\)

\(\Leftrightarrow x^3-2x^2+x-2+x^2+1=0\)

\(\Leftrightarrow x^3-x^2+x-1=0\)

\(\Leftrightarrow x^2\left(x-1\right)+\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^2+1\right)\left(x-1\right)=0\)

\(\Rightarrow x-1=0\left(x^2+1>0\right)\)

\(\Leftrightarrow x=1\)

Vậy x = 1

\(A=x^3-9x^2+27x-27\)

\(=\left(x-3\right)^3\)

Thay x = 5

\(\Rightarrow A=8\)

Vậy A = 8 khi x = 5

Bài 3:
Áp dụng bất đẳng thức \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}\) có:
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}+\dfrac{a+b+c}{a+b}-3\)

\(=\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)-3\)

\(\ge\left(a+b+c\right)\left(\dfrac{9}{2\left(a+b+c\right)}\right)-3\)

\(=\dfrac{9}{2}-3=1,5\)

Dấu " = " khi a = b = c

Bài 5:

Áp dụng bất đẳng thức AM - GM có:
\(a^2+b^2+c^2+d^2\ge2ab+2cd\ge4\sqrt{abcd}\)

Dấu " = " khi a = b = c = d = 1