ÁP dụng t/c dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{1,2}=\dfrac{y}{1,8}\)=\(\dfrac{x+y}{1,2+1,8}\)=\(\dfrac{15}{3}\)=5

Vậy x=5.1,2=6

       y=5.1,8=9

\(\dfrac{x}{1,2}=\dfrac{y}{1,8}=\dfrac{x+y}{1,2+1,8}=\dfrac{15}{3}=5\\ \Rightarrow\left\{{}\begin{matrix}x=6\\y=9\end{matrix}\right.\)

Áp dụng t/c dtsbn ta có:

\(\dfrac{x}{1,2}=\dfrac{y}{1,8}=\dfrac{x+y}{1,2+1,8}=\dfrac{15}{3}=5\)

\(\dfrac{x}{1,2}=5\Rightarrow x=6\\ \dfrac{y}{1,8}=5\Rightarrow y=9\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x+y}{2+3}=-\dfrac{15}{5}=-3\)

=>x=-6; y=-9

`# \text {Ryo}`

`x/2 = y/3` và `x + y = -15`

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

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x+y}{2+3}=\dfrac{-15}{5}=-3\)

`=> x/2 = y/3 = -3`

\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot\left(-3\right)=-6\\y=3\cdot\left(-3\right)=-9\end{matrix}\right.\)

Vậy, `x = -6; y = -9.`

Lời giải:

Áp dụng BĐT AM-GM:

$x^2+y^2\geq 2\sqrt{x^2y^2}=2|xy|\geq 2xy$

$\Rightarrow 3(x^2+y^2)\geq 6xy$

$x^2+9\geq 2\sqrt{9x^2}=2|3x|\geq 6x$

$y^2+9\geq 2\sqrt{9y^2}=2|3y|\geq 6y$

Cộng theo vế các BĐT trên:

$4(x^2+y^2)+18\geq 6(xy+x+y)=90$

$\Rightarrow x^2+y^2=18$

Vậy $A_{\min}=18$ khi $(x,y)=(3,3)$

cái này x,y phải là số thực dương chứ nhỉ

\(xy+x+y=15< =>x\left(y+1\right)+\left(y+1\right)=16\)

\(< =>\left(x+1\right)\left(y+1\right)=16\)

đặt \(\left\{{}\begin{matrix}x+1=a\\y+1=b\end{matrix}\right.\)\(=>a.b=16\)

Ta có:

 \(a^2-2ab+b^2\ge0\)

=> \(a^2+b^2+2ab-4ab\ge0\)\(=>\left(a+b\right)^2\ge4ab\)\(< =>\left(x+y+2\right)^2\ge4.16=64\)

\(=>x+y+2\ge\sqrt{64}=>x+y\ge\sqrt{64}-2=6\)

\(=>\left(x+y\right)^2=6^2=36\)

lại có \(\left(x-y\right)^2\ge0=>\left(x+y\right)^2+\left(x-y\right)^2\ge36\)

\(< =>x^2+2xy+y^2+x^2-2xy+y^2\ge36\)

\(< =>2\left(x^2+y^2\right)\ge36=>x^2+y^2\ge18\)

dấu"=" xảy ra<=>x=y=3=>Min A=18

Chứng minh P>3 nha!!! 

Chứng minh P>3 nha!!! 

bài2 \(x\times\dfrac{15}{16}-x\times\dfrac{4}{16}=2\) 

     \(x\times\dfrac{11}{16}=2\) 

     \(x=2:\dfrac{11}{16}\) 

    \(x=\dfrac{32}{11}\)

Bài 1 : 

 \(\dfrac{x}{16}\times\left(2017-1\right)=2\)

          \(\dfrac{x}{16}\times2016=2\)

                      \(\dfrac{x}{16}=\dfrac{2}{2016}\)

                         \(x=\dfrac{2}{2016}\times16\)

                         \(x=\dfrac{1}{63}\)

1- (5\(\dfrac{4}{9}\) +x+7\(\dfrac{7}{18}\)) : 15\(\dfrac{3}{4}\) = 0 

1- (\(\dfrac{49}{9}+x+\dfrac{133}{18}\)) : \(\dfrac{63}{4}=0\) 

 (\(\dfrac{49}{9}+\dfrac{133}{18}\)+\(x\) ) : \(\dfrac{63}{4}\) = 1 - 0 

       (\(\dfrac{77}{6}\) + \(x\) ) : \(\dfrac{63}{4}\) = 1

         \(\dfrac{77}{6}+x\)         = 1 x \(\dfrac{63}{4}\) 

          \(\dfrac{77}{6}\) + \(x\)          = \(\dfrac{63}{4}\)

                  \(x\)            = \(\dfrac{63}{4}\) - \(\dfrac{77}{6}\)

                  \(x=\) \(\dfrac{35}{12}\)

a) Ta có: \(\dfrac{x-2}{15}+\dfrac{x-3}{14}+\dfrac{x-4}{13}+\dfrac{x-5}{12}=4\)

\(\Leftrightarrow\dfrac{x-2}{15}-1+\dfrac{x-3}{14}-1+\dfrac{x-4}{13}-1+\dfrac{x-5}{12}-1=0\)

\(\Leftrightarrow\dfrac{x-17}{15}+\dfrac{x-17}{14}+\dfrac{x-17}{13}+\dfrac{x-17}{12}=0\)

\(\Leftrightarrow\left(x-17\right)\left(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}\right)=0\)

mà \(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}>0\)

nên x-17=0

hay x=17

Vậy: x=17

b) Ta có: \(\dfrac{x+1}{19}+\dfrac{x+2}{18}+\dfrac{x+3}{17}+...+\dfrac{x+18}{2}+18=0\)

\(\Leftrightarrow\dfrac{x+1}{19}+1+\dfrac{x+2}{18}+1+\dfrac{x+3}{17}+1+...+\dfrac{x+18}{2}+1=0\)

\(\Leftrightarrow\dfrac{x+20}{19}+\dfrac{x+20}{18}+\dfrac{x+20}{17}+...+\dfrac{x+20}{2}=0\)

\(\Leftrightarrow\left(x+20\right)\left(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}\right)=0\)

mà \(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}>0\)

nên x+20=0

hay x=-20

Vậy: x=-20

mai nah

\(\frac{x+1}{18}+\frac{x+2}{17}=\frac{x+3}{16}+\frac{x+4}{15}\)

\(\Rightarrow\frac{x+1}{18}+1+\frac{x+2}{17}+1=\frac{x+3}{16}+1+\frac{x+4}{15}+1\)

\(\Rightarrow\frac{x+19}{18}+\frac{x+19}{17}-\frac{x+19}{16}-\frac{x+19}{15}=0\)

\(\Rightarrow\left(x+19\right).\left(\frac{1}{18}+\frac{1}{17}-\frac{1}{16}-\frac{1}{15}\right)=0\)

Vì \(\frac{1}{18}+\frac{1}{17}-\frac{1}{16}-\frac{1}{15}\ne0\)

Nên \(x+19=0\)

\(\Rightarrow x=-19\)

Vậy x = -19