\(\dfrac{x}{a}=\dfrac{m-\dfrac{x}{2}}{m}\) 

\(\Rightarrow xm=a\left(m-\dfrac{x}{2}\right)\)

\(\Rightarrow xm=am-\dfrac{ax}{2}\)

\(\Rightarrow2xm=2am-ax\)

\(\Rightarrow2xm+ax=2am\)

\(\Rightarrow x\left(2m+a\right)=2am\)

\(\Rightarrow x=\dfrac{2am}{a+2m}\)

vâng em cảm ơn ạ

(-25).68+34.(250)

=25.68+34.250

=25.68+34.25.10

=25.(68+34).10

=25.100.10=2500

\(\sqrt{x-2016}+\sqrt{y-2017}+\sqrt{z-2018}+3024=\frac{1}{2}\left(x+y+z\right)\)

\(\Leftrightarrow2\left(\sqrt{x-2016}+\sqrt{y-2017}+\sqrt{z-2018}+3024\right)=x+y+z\)

\(\Leftrightarrow2\sqrt{x-2016}+2\sqrt{y-2017}+2\sqrt{z-2018}+6048=x+y+z\)

\(\Leftrightarrow x-2\sqrt{x-2016}+y-2\sqrt{y-2017}+z-2\sqrt{z-2018}+6048=0\)

\(\Leftrightarrow x-2016-2\sqrt{x-2016}+1+y-2017+2\sqrt{y-2017}+1+z-2018-2\sqrt{z-2018}+1=0\)

\(\Leftrightarrow\left(\sqrt{x-2016}-1\right)^2+\left(\sqrt{y-2017}-1\right)^2+\left(\sqrt{z-2018}-1\right)^2=0\)

\(ĐK:x\ge2016;y\ge2017;z\ge2018\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2016}-1=0\\\sqrt{y-2017}-1=0\\\sqrt{z-2018}-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}\sqrt{x-2016}=1\\\sqrt{y-2017}=1\\\sqrt{z-2018}=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2017\\y=2018\\z=2019\end{cases}}}\)

nhân đôi 2 vế rồi chuyển vế trái sang vế phải, ta có:

\(\left(\sqrt{x-2016}-1\right)^2\) + \(\left(\sqrt{y-2017}-1\right)^2\)

+ \(\left(\sqrt{z-2018}-1\right)^2\)

= 0

a) \(\left(2x+1\right)^2-4\left(x+2\right)^2=12\)

\(\Leftrightarrow4x^2+4x+1-4\left(x^2+4x+4\right)=12\)

\(\Leftrightarrow4x^2+4x+1-4x^2-16x-16-12=0\)

\(\Leftrightarrow-12x-27=0\)

\(\Leftrightarrow x=\frac{-9}{4}\)

b) xem lại đề

c) \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x-3\right)\left(3-x\right)=1\)

\(\Leftrightarrow x^3-27-x\left(x-3\right)^2=1\)

\(\Leftrightarrow x^3-27-x\left(x^2-6x+9\right)-1=0\)

\(\Leftrightarrow x^3-28-x^3+6x^2-9x=0\)

\(\Leftrightarrow6x^2-9x-28=0\)

\(\Leftrightarrow6\left(x^2-\frac{3}{2}x-\frac{14}{3}\right)=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\frac{3}{4}+\frac{9}{16}-\frac{251}{48}=0\)

\(\Leftrightarrow\left(x-\frac{3}{4}\right)^2=\frac{251}{48}=\left(\pm\sqrt{\frac{251}{48}}\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{3}{4}=\sqrt{\frac{251}{48}}=\frac{\sqrt{753}}{12}\\x-\frac{3}{4}=-\sqrt{\frac{251}{48}}=\frac{-\sqrt{753}}{12}\end{matrix}\right.\)

\(\Leftrightarrow x=\frac{\pm\sqrt{753}}{12}+\frac{3}{4}=\frac{9\pm\sqrt{753}}{12}\)

d) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-19\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+19=0\)

\(\Leftrightarrow12x+15=0\)

\(\Leftrightarrow x=\frac{-5}{4}\)

Theo giả thiết:

\(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=3ab+3bc+3ca\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)

\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Dễ thấy \(VT\ge0\forall a;b;c\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\)\(\Leftrightarrow a=b=c\)(đpcm)

Vì nó bé hơn thì nó bé hơn

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cái câu cmr này á

được không hãy giúp mình đi nha hứa sẽ cho like