x nhân (7,3+2,7) = 10

x nhân 10  = 10

x = 10 : 10

x = 1

k mk nha

\(x^2+4x+y^2-2xy+x^2+4=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2+4x+4\right)=0\)

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

vì \(\left(x-y\right)^2\ge0;\left(x+2\right)^2\ge0\)nên

\(\Rightarrow\orbr{\begin{cases}x+2=0\\x-y=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\y=-2\end{cases}\Rightarrow}x=y=-2}\)

dễ mà x = 2

     và  x = -10/3

3^x ???

=>3^x*8=24

=>3^x=3

=>x=1

x=25 y=9 k cho mình nha bạn 

5/x = 1/5 = y/45

x=25;y=9

Ta có:

\(\frac{5}{x}=\frac{1}{5}\)\(\Leftrightarrow x=5\times5:1=25\)

\(\frac{y}{45}=\frac{1}{5}\)\(\Leftrightarrow y=45\times1:5=9\)

Mình đang cần gấp mong các bạn giúp mình nhé

Lời giải:
Ta có:

$(2x-y)+(2y-z)+(2z-x)=1+2+3$
$2x-y+2y-z+2z-x=6$

$x+y+z=6$

\(PTK_{hc}=1,9375\times PTK_S=1,9375\times32=62\text{đ}vC\)

\(2\times NTK_X+1\times NTK_O=62\text{đ}vC\)

\(2\times NTK_X+16=62\)

\(2\times NTK_X=62-16\)

\(2\times NTK_X=46\)

\(NTK_X=\frac{46}{2}\)

\(NTK_X=23\text{đ}vC\)

=> Na

PTK_{hợp chất}=1,9375 * 32 =   62  (đvC)

PTKoxi = 16 (đvC) => PTK_2x = 62 - 16 = 46 (đvC)

=> PTK_x = 46 : 2 = 23 (đvC) => NTK = 23 (đvC)

=> X là nguyên tố NATRI

a) Ta có: \(x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)

\(< =>2x^2+2y^2\ge x^2+2xy+y^2\)

\(< =>x^2+y^2\ge2xy\)

\(< =>x^2-2xy+y^2\ge0\)

\(< =>\left(x-y\right)^2\ge0\) (luôn đúng)

Dấu "=" xảy ra <=> x=y

=>(đpcm).

a. \(x^2+y^2-\frac{\left(x+y\right)^2}{2}\ge0\)

\(\Leftrightarrow2x^2+2y^2-\left(x+y\right)^2\ge0\)

\(\Leftrightarrow2x^2+2y^2-x^2-2xy-y^2\ge0\)

\(\Leftrightarrow x^2-2xy+y^2=\left(x+y\right)^2\ge0\) (Luôn đúng)

Hay \(x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\left(Dfcm\right)\)

b. \(ab-\frac{\left(a+b\right)^2}{4}\le0\)

\(\Leftrightarrow4ab-a^2-2ab-b^2\le0\)

\(\Leftrightarrow-\left(a^2-2ab+b^2\right)=-\left(a-b\right)^2\le0\) (Luôn đúng)

Hay \(ab\le\frac{\left(a+b\right)^2}{4}\)

a)\(< =>\frac{2x^2+2y^2-x^2-2xy-y^2}{2}\ge0\)<=>\(\frac{x^2-2xy+y^2}{2}\ge0< =>\frac{\left(x-y\right)^2}{2}\ge0\left(lđ\right)\)

b)<=>\(\frac{4ab-a^2-2ab-b^2}{4}\le0< =>\frac{-\left(a^2-2ab+b^2\right)}{4}\le0\)<=>\(\frac{-\left(a-b\right)^2}{4}\le0\left(lđ\right)\)