X = 169

\(3\sqrt{x}+1=40\)

\(ĐKXĐ:x\ge0\)

\(pt\Leftrightarrow3\sqrt{x}=39\)

\(\Leftrightarrow\sqrt{x}=13\)

\(\Leftrightarrow x=169\)

2 ca thể áp chặt vào nhau tạo Cầu nối tế bào chất ,qua đây diễn ra sự trao đổi Nhân.

Nhớ cho mình 

Ta có:(3x-7)²⁰⁰⁹=(3x-7)²⁰⁰⁷

\(\Rightarrow\)(3x-7)²⁰⁰⁹-(3x-7)²⁰⁰⁷=0

​\(\Rightarrow\)​(3x-7)²⁰⁰⁷ [(3x-7)²-1]=0​​

\(\Rightarrow\)\(\orbr{\begin{cases}\left(3x-7\right)^{2007}=0\\\left(3x-7\right)^2-1=0\end{cases}}\)

\(\Rightarrow\)​\(\orbr{\begin{cases}3x-7=0\\\left(3x-7\right)^2=1\end{cases}}\)​

\(\Rightarrow\)\(\orbr{\begin{cases}3x=7\\3x-7=1\end{cases}}\)

​\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\3x=8\end{cases}}\)​

\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=\frac{8}{3}\end{cases}}\)

Vậy......

\(\left(3x-7\right)^{2009}=\left(3x-7\right)^{2007}\)

\(\Rightarrow\left(3x-7\right)^{2009}-\left(3x-7\right)^{2007}=0\)

\(\Rightarrow\left(3x-7\right)^{2007}\left[\left(3x-7\right)^2-1\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(3x-7\right)^{2007}=0\\\left(3x-7\right)^2-1=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}3x-7=0\\3x-7=\pm1\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x\in\left\{\frac{8}{3};2\right\}\end{cases}}\)

\(B=\left|x-33\right|+\left|x+34\right|+\left|x+35\right|\)

\(B=\left(\left|x-33\right|+\left|x+34\right|\right)+\left|x+35\right|\)

\(B=\left(\left|33-x\right|+\left|x+34\right|\right)+\left|x+35\right|\ge\left|33-x+x+34\right|+\left|x+35\right|=67+\left|x+35\right|\ge67\)

Dấu '' = '' xảy ra 

\(\Leftrightarrow\hept{\begin{cases}\left(33-x\right)\left(x+34\right)\ge0\\x+35=0\end{cases}}\Leftrightarrow\hept{\begin{cases}-34\le x\le33\\x=-35\end{cases}}\)

Vậy ..............................

B nhỏ nhất suy ra |x-33|+|x+34|+|x+35| nhỏ nhất

suy ra |x-33|+|x+34|+|x+35|=0

suy ra x-33+x+34+x+35=0

suy ra3x+36=0 suy ra 3x=-36

suy ra x=-12 ..................................................................................................................................................................................................

Ta có:

\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)

\(\Rightarrow\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{25}\)

\(\Rightarrow\frac{x^2}{4}=\frac{3y^2}{27}=\frac{z^2}{25}\)

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

\(\frac{x^2}{4}=\frac{3y^2}{27}=\frac{z^2}{25}=\frac{x^2+3y^2-z^2}{4+27-25}=\frac{30}{6}=5\)

\(\Rightarrow\)x²=20

         y²=45

         z²=125

Áp dụng .......................................

ta được: x/2=y/3=z/5=(x²+3y²-z²)/(2²⁺3*3^2-5²)=30/6=5

Vậy: x=10 

    y=15

    z=25

\(\frac{-9}{x+1}=\frac{\left(x+1\right)^2}{3}\)

\(\left(x+1\right)\cdot\left(x+1\right)^2=-9\cdot3\)

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

\(\left(x+1\right)=-3\)

\(x=-4\)

cảm ơn