\(\sqrt{1932932}+\sqrt[12]{14246}=1390,299248+2,218917107=1392,518165\)

Ta có: \(2x=3y\)=> \(\frac{x}{3}=\frac{y}{2}\)=> \(\frac{x}{6}=\frac{y}{4}\)(1)

           \(x=2z\)=> \(\frac{x}{2}=z\)=> \(\frac{x}{6}=\frac{z}{3}\)(2)

Từ(1) và (2) suy ra x/6 = y/4 = z/3

Áp dụng t/c của dãy tỉ số bằng nhau

ta có : \(\frac{x}{6}=\frac{y}{4}=\frac{z}{3}=\frac{x+y-z}{6+4-3}=\frac{21}{7}=3\)

=> \(\hept{\begin{cases}\frac{x}{6}=3\\\frac{y}{4}=3\\\frac{z}{3}=3\end{cases}}\)=> \(\hept{\begin{cases}x=3.6=18\\y=3.4=12\\z=3.3=9\end{cases}}\)

Vậy ...

Từ \(2x=3y\)\(\Rightarrow\frac{x}{3}=\frac{y}{2}\)\(\Rightarrow\frac{x}{3}.\frac{1}{2}=\frac{y}{2}.\frac{1}{2}\)\(\Rightarrow\frac{x}{6}=\frac{y}{4}\)(1)

Từ \(x=2z\)\(\Rightarrow\frac{x}{2}=\frac{z}{1}\)\(\Rightarrow\frac{x}{2}.\frac{1}{3}=\frac{z}{1}.\frac{1}{3}\)\(\Rightarrow\frac{x}{6}=\frac{z}{3}\)(2)

Từ (1) và (2) \(\Rightarrow\frac{x}{6}=\frac{y}{4}=\frac{z}{3}\)

\(\Rightarrow\frac{x}{6}=\frac{y}{4}=\frac{z}{3}=\frac{x+y-z}{6+4-3}=\frac{21}{7}=3\)

\(\Rightarrow x=18\); \(y=12\); \(z=9\)

Vậy \(x=18\), \(y=12\),\(z=9\)

ai giúp với plsss

Ta có: \(50>49\Rightarrow\sqrt{50}>\sqrt{49}\)

\(\Rightarrow\sqrt{50}>7\)

\(\Rightarrow-\sqrt{50}< -7\)

\(\Rightarrow1-\sqrt{50}< 1-7\)

\(\Rightarrow1-\sqrt{50}< -6\)

\(\Rightarrow\left(1-\sqrt{50}\right)^2< \left(-6\right)^2\)

\(\Rightarrow\left(1-\sqrt{50}\right)^2< 36\)

\(\Rightarrow\sqrt{\left(1-\sqrt{50}^2\right)}< \sqrt{36}\)

\(\Rightarrow\sqrt{\left(1-\sqrt{50}^2\right)}< 6\)

Ta có: \(\sqrt{50}>\sqrt{49}\)

\(\Rightarrow1-\sqrt{50}< 1-\sqrt{49}=1-7=-6< 6\)

\(\Rightarrow1-\sqrt{50}< 6\)

9x - 7i > 3 . \((3x-7u)\)

=> 9x - 71 > 9x - 21u

=> -7i > -21u

=> 7i < 21u

=> i < 3u

9x - 7i > 3 (3x - 7u)

9x-7i>9x-21u

-7i>-21u

i<3u

Ta có x+y+z = 0

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

  \(\left(+\right)x+z=-y\)

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

C= (x+y)(y+z)(x+z)=(-z)(-x)(-y)=-1(xyz)=-1. 2= -2

a) Ta có : M = 3 + 3²  + 3³ + ... + 3¹⁰⁰

=> M = (3 + 3²) + (3³ + 3⁴) + ... + (3⁹⁹ + 3¹⁰⁰)

=> M = 12 + 3²(3 + 3²) + ... + 3⁹⁸(3 + 3²)

=> M = 12 + 3².12 + ... + 3⁹⁸.12

=> M = 12(1 + 3² + ... + 3⁹⁸) \(⋮\)12

Do 12 = 3 . 4 \(⋮\)4 => M \(⋮\)4

b) Ta có: 2m + 3 = 3

=> 2m = 3 - 3

=> 2m = 0

=> m = 0 : 2

=> m = 0